TS Inter 1st Year

Telangana TSBIE TS Inter 1st Year English Study Material Reading Comprehension Passages from Short Stories Exercise Questions and Answers.

TS Inter 1st Year English Reading Comprehension Passages from Short Stories

Q.No. 6 (4 out of 6 Questions: 4 × 1 = 4)

Comprehension is a very important skill. It refers to the process of understanding the information presented in a written or oral text. In the Intermediate Public Examination (English) the students’ ability to understand a printed text is tested. This is, therefore, classified as ‘reading comprehension’. Understanding a spoken text is known as ‘listening comprehension’.

Comprehension needs a careful study of the given text. Apart from the explicit meaning there may be implicit meanings and inferences.

In the Public examination, questions of various types follow the given text. Understanding the questions properly is as or rather more, important as understanding the text. Reading the questions first, therefore, will be of some help in improving one’s comprehension skills.

Careful observation of the way the given language is used and regular practice will ensure mastering this all important skill-comprehension.

Exercises

Read the following passages and answer the questions given after them.

PLAYING THE GAME

1. Do his best! Of course he would. For Alan was playing in the school cricket match and was mightily proud of being chosen to play. He had practised bowling with his father for weeks now, and Daddy said he was shaping well. Daddy was nearly as excited as Alan over the match and he promised that if Alan’s side won he would buy him a bicycle.

Questions :
i) Do his best ! Of course he would. Who would be doing well ?
ii) What was he proud of ?
iii) How did he practise cricket ?
iv) What was his father’s promise ?
v) Write the antonym of win’.
vi) ‘Daddy was nearly as excited as Alan over the match ‘Here the adverb, ‘nearly’ means … Choose the answer.
a) almost
b) completely
c) quietly
d) happily
vii) What is the informal name of father mentioned in the passage ?
viii) Write the noun form of practise’.
Answers:
i) Alan
ii) He was proud of being chosen to play.
iii) He practised bowling with his father.
iv) Alan’s father promised him that if their side won the match, he would buy him a bicycle.
v) lose
vi) (a) almost
vii) daddy
viii) practice

2. “Where do you live, sir ?” called Alan at last in the old man’s ear.
“Up the road and some way round the corner, “he answered in his thin, weak voice. ” I should be so much obliged if you could see me home. You look a very kind little boy.” See him home ? And they were still a long way off! If only he could walk a little faster. Why, the teams would be already on the field, and the captain would be wondering why he did not come.

Questions :
i) What did Alan ask in the old man’s ear ?
ii) Where was the old man’s house ?
iii) What request did the old man make to Alan ?
iv) Why was Alan in a hurry ?
v) If only he could walk a little faster. Why did Alan want the old man to walk faster ?
vi) Why would the captain be wondering ?
vii) I should be so much obliged if you could see me home. Obliged in the sentence means ………. …………… Choose the answer.
a) forceful
b) thankful
c) polished
d) all the above
viii) Pick out the word from the passage which is the antonym of strong.
Answers:
i) “Where do you live, sir ?”
ii) up the road and some way round the corner.
iii) The old man requested Alan to see him home.
iv) He was in a hurry to reach the ground in time to play in the match.
v) because, he wanted to reach the ground in time
vi) The captain would be wondering why Alan did not come.
vii) (b) thankful
viii) weak

3. And in the classroom next morning the boys gave Alan three loud cheers, as only schoolboys can, for in some mysterious fashion they too had learned all about his kind act.

Questions :
i) From which story is this passage taken ?
ii) Who is the writer of the story from which this passage is taken ?
iii) Where and when did the schoolboys meet ?
iv) What did they do ?
v) Why did the boys give Alan three loud cheers ?
vi) What does the phrase three loud cheers mean ?
vii) How did the boys come to know about Alan’s kind act ?
viii) Pick out the word from the passage which means strange.
Answers:
i) “Playing the Game”
ii) Arthur Henry Mee
iii) in the classroom, the next morning.
iv) They gave Alan three loud cheers.
v) As they learned about Alan’s kind act in some mysterious way.
vi) Compliments conveyed through rhythmic claps in a joyous way
vii) in some mysterious fashion
viii) mysterious

4. “Bravo, Alan,” he said, patting his little son on the back. “But, Daddy,” began Alan. But his father interrupted him. “It’s all right, old man,” he said. “You see, I came up behind that policeman and he told me what had happened. So I knew you were playing the game although it wasn’t on the cricket pitch. So I went back into the High Street and bought the bicycle I promised you. It’s a beauty. And, Alan, we’re proud of you, your Mother and I.” [Revision Test-I]

Questions :
i) Who is the writer of the story from which this passage is taken ?
ii) Why did Alan’s father pat on his back ?
iii) How did Alan’s father come to know what had happened ?
iv) I came up behind that policeman. Did Alan’s father go to the spot where Alan helped the old man ?
v) Alans’ father says, ” ……………….. although it wasn’t on the cricket pitch.” Where did Alan play the game ?
vi) Why did Alan’s father buy the bicycle ?
vii) Why were they proud of Alan ?
viii) When would you use the expression, bravo ?
Answers:
i) Arthur Henry Mee
ii) to encourage and to appreciate what Alan had done
iii) through the policeman behind whom Alan’s father came
iv) Yes.
v) in the real world-in life-on the meadow
vi) to support and appreciate Alan’s service activities
vii) because Alan ‘played the game’ in its true sense
viii) When we want to appreciate someone’s achievement we use the word ‘bravo’.

THE FIVE BOONS OF LIFE

5. The man considered long, then chose Love; and did not mark the tears that rose in the fairy’s eyes.
After many, many years the man sat by a coffin, in an empty home. And he communed with himself, saying : “One by one they have gone away and left me; and now she lies here, the dearest and the last. Desolation after desolation has swept over me; for each hour of happiness the treacherous trader, Love, as sold me I have paid a thousand hours of grief. Out of my heart of hearts I curse him.”
Questions :
i) The man chose the gift Love at once, without thinking. Write true or false.
ii) With whom did the man commune ?
iii) The word rose used as a verb in the passage means came forth. As a noun, it means :
a) stood up
b) a flower
c) a fruit
iv) How is his sadness or loss expressed in the passage ?
v) Whom did the man call a treacherous trader ?
vi) ‘Out of my heart of hearts I curse him’… whom does the word him refer to ?
vii) Find the antonym of the word bless from the passage. ‘
viii) Write the word from the passage which means the box in which a dead body is buried or cremated.
Answers:
i) false
ii) He communed (talked) with himself,
iii) (b) a flower
iv) by saying to himself how love has left him in grief
v) Love
vi) Love
vii) curse
viii) coffin

6. “Choose yet again” It was he fairy’s voice.
“Two gifts remain. And do not despair. In the beginning there was but one that was precious and it is still here”.
“Wealth – which is power ! How blind 1 was !” said the man.
“Now, at last, life will be worth the living. I will spend, squander, dazzle. These mockers and despisers will crawl in the dirt before me, and I will feed my hungry heart with their envy.

Questions :
i) How many gifts had the man already chosen ?
ii) Did the man choose the precious gift before ? How do you know ?
iii) Who does the word I refer to in the sentence, “How blind I was !” ?
iv) With what would his life be worth the living ?
v) How would the man feed his hungry heart ?
vi) The man said that he was blind to the fact. What was the poet ?
vii) What was the man’s real motive in choosing wealth ?
viii) Write the synonym of the word jealousy from the passage.
Answers:
i) three
ii) No. Because the precious gift is still there.
iii) The word T refers to the man, the lead character in the story.
iv) with wealth
v) with the envy of his mockers and despisers.
vi) The fact is that wealth is power.
vii) to make his life worth the living and to make his mockers feel jealous of him.
viii) envy

7. The fairy came, bringing again four of the gifts, but Death was wanting. She said :
‘I gave it to a mother’s pet, a little child. It was innocent, but trusted me, asking me to choose for it. You did not ask me to choose.”
“Oh, miserable me ! What is left for me ?”
“What not even you have deserved: the wanton insult of Old Age.” [Revision Test – II]
Questions :
i) The fairy brought the gift, Death too. Say Yes or No.
ii) Who does the word I refer to ?
iii) What did the fairy give little child ?
iv) What is the epithet used to describe the little child ?
v) The man didn’t repose faith in the fairy. Write true or false.
vi) What was left for the man ?
vii) Write the word from the passage that means very unhappy.
viii) Write the antonymn of the word intelligent from the passage.
Answers:
i) No
ii) The word T refers to the fairy,
iii) Death
iv) a mother’s pet
v) true
vi) not even what he deserved; the wanton insult of old age
vii) miserable
viii) innocent

8. “The years have taught you wisdom-surely it must be so.
Three gifts remain. Only one of them has any worth-remember it, and choose warily.”
The man reflected long, then chose Fame; and the fairy, sighing, went her way.
Years went by and she came again, and stood behind the man where he sat solitary in the fading day, thinking. And she knew his thought. [Model Question Paper]
Questions :
i) Two of the remaining gifts are worthy. Write true or false.
ii) What did the man opt this time ?
iii) Was the fairy happy with his selection ?
iv) Which word in the passage indicates that the man was alone ?
v) Write the synonym of the word renown from the passage.
vi) Write the word from the passage which means disappearing gradually.
Answers:
i) false
ii) fame
iii) No, the fairy was not happy.
iv) solitary
v) fame
vi) fading

THE SHORT-SIGHTED BROTHERS

9. “Three elderly brothers, all very short-sighted, lived in a large house on the outskirts of a city, in China. One day the youngest brother suggested that he should take charge of the finances. “Elder brother’s sight is so bad, he cannot see how much money he’s receiving or giving,” he said, “and people take advantage of his disability.”

Questions :
i) Where did the three brothers live ?
ii) What did the youngest brother propose one day ?
iii) How did the youngest brother support his claim ?
iv) How would people take the eldest brother’s short-sightedness, according to the youngest brother ?
v) Was the youngest brother sincere in his suggestion ?
vi) Their sight problem was negligible. Is it true or false ?
vii) Give the synonym of edge from the passage.
viii) Write a set of antonyms you find in the passage as good is the antonym of bad’.
Answers:
i) in a large house on the outskirts of a city, in China.
ii) that he should take charge of the finances.
iii) That his eldest brother’s sight was so bad that he couldn’t see how much money he was receiving or giving
iv) People would take advantage of the eldest brother’s disability.
v) No. The youngest brother was not sincere.
vi) False. They were all very short-sighted, outskirts
viii) receiving × giving

10. ‘The tablet has a flowery border”. The second brother went away very pleased with himself. Hardly had he gone when the third brother arrived there. He too enquired about the inscription and on being told what it was, asked if there was any other writing on it.” “Only the donor’s name, Wang Lee, at the bottom,” said the monk.

Questions :
i) What did the tablet have for its decoration ?
ii) Who answered the question about its decoration ?
iii) Who was very happy to know about that decoration ?
iv) When did the third brother reach the monastery ?
v) What did the youngest brother want to know particularly ?
vi) Where was the donor’s name mentioned ?
vii) Why was the second brother happy with himself ?
viii) Write the antonym of departed from the passage.
Answers:
i) a flowery border
ii) the monk
iii) the second brother
iv) hardly when the second brother had gone away from the monastery
v) if there was any other writing on the tablet
vi) at the bottom of the tablet
vii) as he thought he alone learnt about the inscription and the flowery decoration
viii) arrived

11. The monk they had talked to the previous evening came out of the monastery just then and walked towards the short sighted brothers.
“Oh you’ve come to see the inscription,” he said. “So sorry. We couldn’t put it up yesterday evening. We are going to put it up today.”
The short-sighted brothers realised their follies.
Questions :
i) When did they all talk to the monk ?
ii) Did each brother know that the others also had talked to the monk ?
iii) Where did the monk go ?
iv) Why did the monk say sorry ?
v) What does the word it refer to?
vi) When were they going to put it up ?
vii) What did the brothers realise ?
viii) They didn’t put up the inscription as originally scheduled. How did it help the brothers ?
Answers:
i) The previous evening
ii) No. They did not know about the others’ visit.
iii) The monk went towards the short-sighted brothers.
iv) Because they couldn’t put up the tablet the previous evening, the monk, said sorry.
v) The word ‘it’ refers to ‘the tablet’.
vi) They were going to put up the tablet that day.
vii) They realised their follies.
viii) It helped the brothers realise their follies.

12. “I have my doubts about that,” said the eldest brother. “Let’s settle this once and for all. I’ve heard the monastery is putting up a tablet inscribed with a saying, above the main doorway, tonight. Let’s go there tomorrow and test our vision. Whoever can read the inscription with the least strain will get charge of our money. Agreed ?” [Revision Test – III]

Questions :
i) Name the story from which this passage is taken.
ii) Who does “I” in the paragraph refer to ?
iii) What did the speaker want to settle once and for all ?
iv) What did the speaker come to know of ?
v) Where should they go to get their vision tested, the following day ?
vi) Who would get the charge of their money, according to the proposal ?
vii) When was the monastery putting up a tablet above their doorway ?
viii) Write the word that is used in the passage that means a place where monks live.
Answers:
i) The Short-sighted Brothers.
ii) “I” in the paragraph refers to the eldest brother.
iii) as to who among the three brothers had a better sight.
iv) of the monastery putting up a tablet inscribed with a saying, above the main doorway, that night.
v) to the monastery
vi) the one who could read the inscription with the least strain
vii) on that night (when this discussion in the story was going on)
viii) monastery

SANGHALA PANTHULU

13. Thus, things were moving happily. But the farmers were perturbed. They observed the lives of people on the other side of the river Krishna ruled by the British and found that people were happy there. There was no drudgery, no penalties-no beatings either. But if the people of Ramasagaram were to migrate to that side leaving the households and assets earned by their ancestors and their caste trades as well, how would they live ?

Questions :
i) “Thus, things were moving happily.” ‘Happily’ to whom ?
ii) What did the farmers observe ?
iii) What did the farmers find out ?
iv) What was the reason for the vast difference in the lives of Ramasagaram people and that of those living on the other side of the river ?
v) Was it possible for the people of Ramasagaram to migrate to the other side of the river ?
vi) Give the word from the passage that means disturbed /worried Anxious .
vii) According to the passage, two groups of persons were happy. Name those two groups.
viii) Write the Noun form of the word migrate.
Answers:
i) moving happily to the police personnel
ii) observed the lives of people on the other side of river Krishna ruled by the Britishers.
iii) found that people on the other side of river Krishna were happy
iv) The reason : Ramasagaram was under the Nizam’s rule and the village on the other side of river Krishna was ruled by the Britishers.
v) No, it was not possible for them to migrate.
vi) ‘perturbed’
vii) the police; the people on the other side of river Krishna
viii) ‘migration’

14. In the evening, about five hundred people gathered under the peepul. Pahthulu explained about the nature of drudgery to all of them. He also taught them legal points. He insisted that nobody should agree to drudgery if wages were not paid. While the meeting was still going or, Ameen Saab arrived pompously on a horse along with eight jawans. “Panthulu, are you aware of the Nizam’s Act number 53 pertaining to patrolling ?!’ asked the Ameen.

Questions :
i) Where did the people assemble ?
ii) What did Panthulu explain to them ?
iii) The villagers meeting under the peepul tree already knew legal points. Say true or false
iv) What did Panthulu insist on ?
v) When did the Ameen Saab come there ?
vi) What did Act number 53 deal with ?
vii) Write the antonym of modestly from the passage.
viii) The passage pictures Panthulu as a man (fill in).

1. interested in legal practise
2. promoting violence
3. committed to the cause of the common man <
4. serving the purpose of the police

Answers:
i) under the peepul tree
ii) about the nature of drudgery
iii) false
iv) that nobody should agree to drudgery if wages were not paid
v) while the meeting was still going on
vi) with patrolling
vii) pompously
viii) (3) committed to the cause of the common man

15. After a week, the Mohathemeem arrived. The police didn’t divulge the disgrace they faced. But they recorded that Panthulu had instigated the villagers to revolt by trying to run a parallel government. They also mentioned that if the army was not sent, a great danger was looming large. The Mohathemeem summoned the farmers and inquired with them about the incident.

Questions :
i) Why did the Mohathemeem come to Ramasagaram ?
ii) What did the police hide from the Mohathemeem ?
iii) What did the police charge Panthulu with ?
iv) What did the police request for ?
v) Who did the Mohathemeem call to know more about the incident ?
vi) Write the idiomused in the passage that means something very frightening was certain to happen.
vii) Write the synonym from the passage of called.
viii) Name the part of speech of divulge.
Answers:
i) to inquire into the reported violence
ii) the disgrace the police faced
iii) that Panthulu instigated the villagers to revolt by trying to run a parallel government.
iv) for the army to be sent to the village
v) the farmers
vi) looming large
vii) summoned
viii) verb

16. The news about the arrived of the elderly man from city had spread in the village by morning. The villagers said he was a tall stout man. He helped form associations in villages. He sported a kerchief like lawyers do. He brought a leather suitcase which was full of books. He knew all the bigwigs in the city. He would do away with all our troubles. [Revision Test – IV]
Questions :
i) Name the story from which this passage is taken.
ii) Who does the phrase the elderly man refer to ?
iii) How would the elderly man help the villagers ?
iv) What did he bring with him ?
v) The villagers talked about the elderly man ………………… (fill in).

1. adversely
2. appreciatively
3. accusingly
4. arrogantly

vi) Find the Phrasal Verb used in the passage that means put an end to; eliminate.
vii) The word sported as used in this passage means ……………… (fill in).

1. game
2. athletics
3. wore

played
viii) Find the passage the Antonym of lean.
Answers:
i) Sanghala Panthulu
ii) Sanghala Panthulu
iii) He would help the villagers form associations and solve their problems.
iv) a leather suitcase, full of books
v) (2) appreciatively
vi) do(es) away with
vii) (3) wore
viii) stout

THE DINNER PARTY

17. A spirited discussion springs up between a young girl who insists that women have outgrown the jumping-on-a-chair-at-the-sight-of-a-mouse era and a colonel who says that they haven’t. “A woman’s unfailing reaction in any crisis,” the colonel says, “is to scream. And while a man may feel like it, he has that ounce more nerve control than a woman has. And that last ounce is what counts.”

The American does not join in the argument but watches the other guests. As he looks, he sees a strange expression come over the face of the hostess. She is staring straight ahead, her muscles contracting slightly. With a slight gesture, she summons the native boy standing behind her chair and whispers to him. The boy’s eyes widen: he quickly leaves the room. Of the guests, none except the American notices this or sees the boy place a bowl of milk on the veranda just outside the open doors.

Questions :
i) What are the young girl and the colonel arguing about ?
ii) The American joins the discussion. Say true or false.
iii) What does the American naturalist notice ?
iv) What does the hostess want the servant to do ?
v) Identify the synonym of calls from the passage.
vi) Find the antonym of familiar in the passage.
vii) “…………. he has that ounce more nerve control than a woman has. “What does the word nerve mean ?
viii) Pick out the word that fits the meaning of making narrower in the passage.
Answers:
i) about the nerve control of women
ii) false
iii) a strange expression on the face of the hostess
iv) to place a bowl of milk on the veranda just outside the open doors
v) summons
vi) strange
vii) emotions-anxiety
viii) contracting

18. “I want to know just what control everyone at this table has. I will count to three hundred- that’s five minutes-and not one of you is to move a muscle. Those who move will forfeit fifty rupees . Ready !”

The twenty people sit like stone images while he counts. He is saying. two hundred and eighty…” when, out of the corner of his eye, he sees the cobra emerge and make for the bowl of milk. Screams ring out as he jumps to slam the veranda doors safely shut.

‘You were right, Colonel!” the host exclaims. “A man has just shown us an example of perfect control.” “Just a minute,” the American says, turning to his hostess. “Mrs. Wynnes, how did you know that cobra was in the room ?” A faint smile lights up the woman’s face as she replies : “Because it was crawling across my foot.”

Questions :
i) What is the proposal from the American ?
ii) What does the American do to make the guests at the party stay stable ?
iii) Pick out the word which means lose as punishment from the passage.
iv) Pick out the word from the passage that means weak or dull.
v) Why does the American shut the doors ?
vi) How does the American react, when the host gives credit to him for having the most control?
vii) What does the hostess prove to her guests ?
viii) When does the cobra come out ?
Answers:
i) that he will count to three hundred and none of them is to move a muscle
ii) To make, the guests say stable and silent, the naturalist throws a challenge at them.
iii) forfeit
iv) faint
v) for safety, to prevent the snake from coming back again
vi) He asks them to wait and proves to them who really has the nerve control.
vii) The hostess proves to the guests that she has lots of nerve control.
viii) When the naturalist is saying two hundred and eighty.

19. The country is India. A colonial official and his wife are giving a large dinner party. They are seated with their guests-army officers and government attaches and their wives, and a visiting American naturalist-in their spacious dining room, which has a bare marble floor, open rafters and wide glass doors opening onto a veranda. [Revision Test – V]

Questions :
i) In which country is this story set ?
ii) Who is the host of the party ?
iii) Where is the party arranged ?
iv) What is synonym for the word porch in the passage ?
v) Who is the special guest in the party ?
vi) Describe the place where the dinner is hosted.
vii) Write the antonym of narrow from the passage.
viii) Identity the word from the passage that means touring.
Answers:
i) in India
ii) a colonial official and his wife
iii) in the spacious dining room of the hosts
iv) veranda
v) a visiting American naturalist
vi) a spacious dining room with a bare marble floor, open rafters and wide glass doors opening onto a veranda
vii) spacious
viii) visiting

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Transformation of Sentences Exercise Questions and Answers.

TS Inter 1st Year English Grammar Transformation of Sentences

Q. No. 13 (4 × 1 = 4)

A. VOICE

The term ‘voice’ in grammar refers to one aspect of the verb. If the verb group in a sentence has ‘be + pp of the verb’, that sentence is said to be in the passive voice. If any of or both the elements (be + V₃) are missing in the structure of the verb, the sentence is said to be in the active voice. (వాక్యంలోని verb groupలో ‘be + V3 ఉంటే ఆ వాక్యాన్ని passive voice అని, be + V₃ లలో ఏ ఒక్కటి లేకున్నా active voice అని అంటారు.)

If the verb is in the passive form, the subject of that sentence is just the ‘sufferer’ of the action indicated by the verb, (verb passive లో ఉంటే ఆ వాక్యంలోని subject, verb సూచించిన పని ఫలితాన్ని అనుభవిస్తుంది. అందుకనే ఈ రూపానికి ‘passive’ ‘సోమరిగా’, ‘పనిచేయకుండా’ అని పేరు.)

If the verb group is in the active voice, the subject of that sentence is the ‘doer’ of the action shown by the verb. (Verb Active voice లో ఉన్నప్పుడు, ఆ వాక్యం యొక్క subject, verb సూచించిన పనిని చేస్తుంది. అందుకే ఈ రూపానికి ‘Active చురుకుగా పనిచేస్తున్న’ అని పేరు)

If the doer of an action is either unimportant or unknown, the passive structure is natural. (పనిచేసినవారు ప్రధానం కాకున్నా, తెలియకున్నా, passive నిర్మాణాలు సహజంగా ఉంటాయి. )

Someone admitted him to the hospital immediately. (AV)
He was admitted to the hospital immediately. (PV)
He was killed in the war.
Rice is grown in many parts of Telangana and Andhra Pradesh.
English is spoken all over the world.
The roads have been swept The syllabus was changed last year.
The shop has been closed.
The doubts have been cleared by my teacher Agent

Transformation of sentences from one voice to the other involves five steps.

1. The object in the Active Voice sentence becomes the subject in the Passive Voice sentence.
Ex : They made (ten kites)Object (A.V)
(Ten kites) Subject were made by them. (P.V.)

If there are two objects in the A.V. sentence, either of them can be made the subject in the PV. sentence.

An interesting book was given to her brother by her.

Active :

1. I told them an interesting story.
2.  I told an interesting story to them.

Passive:

1. An interesting story was told to them.
2. They were told an interesting story.

2. A suitable ‘be’ form is to be introduced. This is the most important step. Selection of the right be’ form is based on two factors : (a) the number and person of the subject in the passive voice and (b) the tense of the verb in the Active Voice.
The following table helps one select the right be form.

S. No.	Subject in the Passive Voice	Tense of the verb in Active Voice	‘Be form to be used in the Passive Voice
1	I	Simple Present	am
2	he, she, it singlar nouns	Simple Present	is
3	We, you, they plural nouns	Simple Present	are
4	I, he, she, it singular nouns	Simple Past	was
5	We, you, they plural nouns	Simple Past	were
6	I	Present Continuous	am being
7	he, she, it singular nouns	Present Continuous	is being
8	We, you, they plural subjects	Present Continuous	are being
9	1, he, she, it singular nouns	Past Continuous	was being
10	We, you, they plural subjects	Past Continuous	were being
11 ‘	he, she, it Singular nouns	Present Perfect	has been
12	I, we, you, they plural nouns	Present Perfect	have been
13	Any subject	Past Perfect	had been
14	Any subject	will/shall/can/ may/would should/could/ might/ must, etc.	will, etc. +be
15	Any subject	will etc + have	will, etc. + have been

3. The main verb in the Active Voice sentence is to be changed into its past participle form. One must know the correct past participle forms of irregular verbs.

4. The preposition ‘by’ is used.

5. The subject in the Active voice sentence is made the object of the preposition ‘by’ in the passive voice sentence.
Ex : He broke the glass. (A.V.)
The glass was broken by him. (P.V.)
prep+object
If the subject in the A.V. sentence is either unimportant or a general one, ‘by + object’ may be dropped.
Ex : Someone removed the dead snake. (A.V.)
The dead snake was removed. (P-V.)
(by someone’ is dropped)
People call him ‘Babuji’. (A.V.)
He is called ‘Babuji’. (‘by people’ not necessary) (P.V.)

CHANGE IMPERATIVE SENTENCES INTO PASSIVE

We follow a slightly different method to change imperative sentences from the A.V. to the P.V. (Sentences with dropped subject and with V + object structure, conveying a request or an order are called imperative sentences.)
1. The passive voice sentence begins with ‘Let’.
Ex : Close the door. (A.V.) imperative
Let the door be closed. (P.V.)

2. The object in the A.V. sentence becomes the subject in the RV. sentence.
Answer this question. (A.V.)
Object
Let this question be answered. (P.V.)
Subject

3. The ‘be’ form ‘be’ is introduced.
(Whatever be the subject or the tense form, it is always ‘be’ in imperative sentences.)
Ex : Clean the room. (A.V.)
Let the room be cleaned. (P.V.)

4. The M.V. is changed into its Past Participle form.
Ex : Paint this chair. (A.V.)
M.V.
Let this chair be painted. (P.V.)
As the subject is not explicit in the imperative sentences, the need to use ‘by’ and its object doesn’t arise, while changing imperative sentences into the passive voice.

CHANGING QUESTIONS INTO PASSIVE

Study carefully the transformation of the interrogative sentences into the passive voice.
Ex : Who brought this book ? (A.V.)
By whom was this book brought ? (P.V.)
All the steps followed here are exactly like those we follow in changing the declarative sentences. But the word order is different.
By whom + be + Subject + M.V in PP + etc ?
Nbw, notice the transformation of Questions with other ‘wh’ words.
Ex : Where did you put my pen ? (A.V.)
Where was my pen put ? (P.V.)
‘Wh’ word + be + subject + MV in PP ?
Observe how ‘yes / no’ Questions are rewritten in the passive voice.
Have you solved the problem ? (A.V.)
Has the problem been solved ? (P.V.)
Helping Verb + Subject + be + MV in pp + etc. ?

EXAMPLES
1. Advertise the post. (A.V.)
Let the post be advertised. (P.V.)

2. America imports Indian tea. (A.V.)
Indian tea is imported by America. (P.V.)

3. The auditors are checking the accounts. (A.V.)
The accounts are being checked by the auditors. (P.V.)

4. They have sent the information. (A.V.)
The information has been sent. (P.V.)

5. Hurry can gain nothing. (A.V.)
Nothing can be gained by hurry. (P.V.)

6. Put the culprit in prison. (A.V)
Let the culprit be put in prison. (P.V.)

7. John teaches us English. (A.V.)
We are taught English by John. (P.V.)
English is taught to us by John. (P.V.)

8. The Manager sent a mail yesterday. (A.V)
A mail was sent by the manager yesterday. (P.V)

9. The conductor has issued tickets to all the passengers. (A.V.)
Tickets have been issued to all the passengers. (P.V.)
All the passengers have been issued tickets. (P.V.)

10. Narayana Murthy started Infosys. (A.V.)
Infosys was started by Narayana Murthy. (P.V.)
Look at the following sentence and observe the changes.

Tense	Active Voice	Passive Voice
Simple present	Floods cause a lot of damage.	am/are/is+v3(past participle) A lot of damage is caused by floods
Present continuous	The gardener is watering the plants.	am/are/is+being+v3 The plants are being watered by the gardner.
Present perfect	We have organized a special programme for children.	have/has+been+v3 A special programme has been organized for children (by us).
Simple past	Raghavendar Rao directed the film ‘Annamayya’.	was/were + v3 The film Annamayya’ was directed by Raghavendar Rao.
Past continuous	When they were shifting the patient to the ICU, he died.	Was / were + being v3 When the patient was being shifted to the ICU, he died.
Past perfect	The driver had already alerted the passengers before the robbers entered the bus.	had + been + v3 The passengers had already been alerted before the robbers entered the bus.
Simple future	I will conduct a spelling – contest tomorrow.	shall/will + be + v3 A spelling – contest will be conducted tomorrow.
Future perfect	They will have decorated the hall by evening.	shall / will + have + been + v3 The hall will have been decorated by evening.
The future of intention (be going to)	Keeravani is going to compose music this song.	is going to be + v3 Music is going to be composed by Keeravani for this song.
Modal veibs (should, must, ought to, can, etc.)	I will type this letter tomorrow.	should/would/must/tought/can + be + v3 ThisletowiBbetypedbyrnetorrKxrcw.
It is said (that) or subject+is said to be	Villagers say that there is a ghost in the old building.	It is said that there is a ghost in the old building.
Imperative	Check the spelling.	Let…. be + v3 Let the spelling be checked.

EXERCISES

I. Change the following sentences into the passive voice.

1. We practise yoga every day in the morning.
2. He will make all the arrangements.
3. The judge declared the verdict.
4. They had already announced the results before we entered the hall.
5. Many students sacrificed their precious lives for Telangana.
6. The students borrowed some books from the library.
7. Nobody can save him.
8. How much loan amount has the Bank sanctioned ?
9. One should wear a helmet while riding a two-wheeler.
10. Money alone cannot solve all problems.
11. Switch off the lights.
12. Please maintain silence in the prayer hall.
13. We have to undergo many formalities for getting a visa.
14. The workers called off the strike.
15. The teacher is explaining the lesson.
16. The postman will deliver the letters at noon.
Answers:
1. Yoga is practised by us every day in the morning.
2. All the arrangements will be made by him.
3. The verdict was declared by the judge.
4. The results had already been announced before we entered the hall.
5. Their precious lives were sacrificed by many students for Telangana.
6. Some books were borrowed by the students from the library.
7. He cannot be saved.
8. How much loan amount has been sanctioned by the Bank ?
9. Helmet should be worn while riding a two-wheeler.
10. All problems cannot be solved by money.
11. Let the lights be switched off.
12. Let silence be maintained in the prayer hall.
13. Many formalities have to be undergone for getting a visa.
14. The strike was called off by the workers.
15. The lesson is being explained by the teacher.
16. The letters will be delivered by the postman at noon.

II. Change the following sentences into the active voice.

1. The parcels will be delivered at any time (by the courier agents)
2. Surya was invited to tea by Chandra.
3. Traffic rules should be followed.
4. Vegetables are washed before cooking.
5. Let the following sentences be changed into the passive voice.
6. How many times were you reminded of the medicine ?
7. Let the dustbin be kept away from the eatables.
8. Hpve all your friends been invited to your birthday ?
9. Every sentence can’t be changed into the passive voice.
10. If the ointment isn’t applied to the wound, it will not heal.
11. My brother has never been beaten at chess by anyone in his school.
12. It is believed that Sammakka and Saralamma are the saviors of their lives in times of crisis by the villagers.
Answers:
1. The courier agents will deliver the parcels at any time.
2. Chandra invited Surya to tea.
3. One (We) should follow traffic rules.
4. We wash vegetables before cocking.
5. Change the following sentences into the passive voice.
6. How many times did I (we) remind you of the medicine ?
7. Keep the dustbin away from the eatables, (food)
8. Have you invited all your friends to your birthday party ?
9. One (We) can’t change every sentence into the passive voice.
10. If you don’t apply the ointment to the wound, it will not heal.
11. No one in his school has ever beaten my brother in chess.
12. The villagers believe that Sammakka and Saralamma are the saviours of their lives in times of crisis.

III. Change the following sentences into the passive voice.

1. Rainwater fills potholes on roads.
2. He is buying a TV set at the moment.
3. I have been growing plants since 1990.
4. They were reading the newspaper.
5. She had answered it already.
6. I will write an essay tonight.
7. You will have posted it by Monday.
8. Can she play the violin ?
9. They may not telecast it.
10. One must do one’s duty.
11. Gall in the doctor.
12. Close the door.
13. The Government has to do it.
14. Someone has already cast my vote.
15. Who could help him ?
Answers:
1. Potholes on roads are filled with (by) rainwater.
2. A TV set is being bought by him at the moment.
3. Plants have been being grown by me since 1990.
4. TKe newspaper was being read (రెడ్) by them.
5. It had already been answered by her.
6. An essay will be written by me tonight.
7. It will have been posted by you by Monday.
8. Can the violin be played by her ?
9. It may not be telecast, (by them)
10. One’s duty must be done, (by one)
11. Let the doctor be called in.
12. Let the door be closed.
13. It has to be done by the Government.
14. My vote has already been cast, (by someone)
15. By whom could he be helped ?

IV. Change the following into Active Voice.

1. He was seen crossing the road.
2. You are advised to be careful.
3. Let the picture be seen by me.
4. Her purchases were paid for by me.
5. There are no shops to be let.
6. She has been selected their monitor (by the class)
7. It is said that the earth is round.
8. The road had been repaired.
9. I am surprised at this news.
10. It is hoped that I shall win.
Answers:
1. We saw him crossing the road.
2. We advise you to be careful.
3. Let me see the picture.
4. I paid for her purchases.
5. There are no shops to let (out).
6. The class has selected her their monitor.
7. People say that the earth is round.
8. They had repaired the road.
9. This news surprises me.
10. I hope that I shall win.

V. Change the following sentences into the passive voice.

1. I have made a mistake.
2. Your students will respect you a great deal more for your frankness and honesty.
3. Call the attention of your near neighbour at the table to the excellence of the coffee.
4. Do you apply Pythagoras Theorem or Newton’s Law of Gravity ?
5. Rahul lost a quarter mark in English.
6. She planted trees; fenced, watered and guarded them.
7. Their hope and encouragement gave me greater strength.
8. Instantly remove that hatter.
Answers:
1. A mistake has been made by me.
2. You will be respected by your students a great deal more for your frankness and modesty.
3. Let the attention of your near neighbour at the table be called to the excellence of the coffee.
4. Is Pythagoras Theorem or Newtons Law of Gravity applied by you ?
5. A quarter mark in English was lost by Rahul.
6. Trees were planted, fenced, watered and guarded by her.
7. I was given greater strength by their hope and encouragement. (OR) Greater strength was given to me by their hope and encouragement.
8. Let that hatter be removed instantly.

B. DIRECT SPEECH AND INDIRECT SPEECH/REPORTED SPEECH

Sometimes it becomes necessary to report a person’s words. It can be done in two different ways. One -way is to reproduce the actual words of the speaker. The speaker’s actual words are shown in quotation marks in writing. This kind of reporting is called the ‘Direct Speech’.

The other kind is to express the idea of the speaker in the reporter’s words. This type is referred to as the ‘Indirect Speech’.
(ఒక వ్యక్తి అన్న మాటలకు అన్నట్లుగా తిరిగి చెబితే Direct Speech. కేవలం భావం మాత్రమే తిరిగి చెబితే Indirect Speech.)

The sentence in the Direct Speech has two parts. They are (a) the part outside the quotation marks (called the reporting part and (b) the part within the quatation marks (called the reported part).

The verb in the reporting part is called “the reporting verb.”
“Don’t make a noise,” said the teacher. (Reporting verb)
To change a sentence from the direct speech to the indirect speech, the following steps are to be followed.

1. The reporting part is to be written first. This is applicable only when the reporting part is either in the middle or at the end of the sentence.

Direct Speech :
“Ramu has solved all the problems.” said the teacher (Reporting part at the end)
Indirect Speech :
The teacher said that Ramu had solved all the problems. Reporting part begins the sentence.

2. The order of the words in the reporting part is arranged as “Subject + Verb.” This is applicable only when this order is not followed in the direct speech sentence.

3. The reporting verb is changed to words like ordered, requested, enquired, told, asked, ad-vised exclaimed, depending on the meaning of the reported part.

4. The quotation marks and the comma between the reporting and reported parts are removed in the indirect speech.
Ex : The boy said,” This plant has grown very tall.” (D.S.)
The boy said that plant had grown very tall. (I.S.)

5. A suitable connecting word is used to connect the reporting and reported parts. The selection of the right connecting word needs careful observation.
The following table will help you to select the right word.

SI.No.	Sentence in quotation marks	Connecting word
1.	Assertive or Exclamatory	that
2.	Imperative	to
3.	Yes/No type Question	if or whether
4.	‘Wh’ type Question	No connecting word

6. The pronouns in the reported part are to be changed suitably. This depends on the first speaker, the speaker’s listener, the reporter and the reporter’s listener. One has to ensure that the original speaker’s intention is correctly reported. (Indirect speech లోకి మార్చేటప్పుడు, ఎవరు ఎవరితో అన్న మాటలను తిరిగి ఎవరు ఎవరికి చెబుతున్నారు అనే దానిని బట్టి pronouns మారతాయి. మొదట మాట్లాడిన వ్యక్తి ఉద్దేశ్యం మారకుండా సరిగ్గా చెప్పడం అవసరం.)

Now notice the following examples carefully :
a) The teacher said to you, ‘You are late again.” (D.S.)
The teacher told you that you were late again. (I.S.) ‘you’ not changed.

b) The teacher said to me, “You have improved your performance.” (D.S.)
The teacher told me that I had improved my performance.
You → 1 & Your → my

c) The teacher said to Geetha, “You have to submit your assignments tomorrow.” (D.S.)
The teacher told Geetha that she had to submit her assignments the following day. (I.S.)
You → she and you → her

d) The teacher said to Prudhvi, “When will you finish your computer course” ? (D.S.)
The teacher asked Prudhvi when he would finish his computer course. (I.S.)
You → he and your → his

e) The teacher said to the students, “You must consult your parents.” (D.S.)
The teacher told the students that they should consult their parents. (I.S.)
You → they and Your → their
As the examples a, b, c, d and e show you, pronouns change in accordance with the speaker’s reference to a person/persons.

7. In changing sentences into the indirect speech, the tense form of the verbs in the reported part is to be changed. This is the most important part of the transformation. This is done in accor-dance with the concept popularly known as “the sequence of tenses.” (Reported part లోని verbs tenses మార్చడం అతి ప్రధాన భాగము దీనిని ‘Sequence of tenses’ అనే నియమానుసారంగా )
The following table clearly shows when and how to change the tense.

8. The next step in the transformation from the direct to the indirect is to change the adjectives or adverbs showing ‘nearness to those showing ‘distance as explained in the following table:

SI.	Ajjective / Adverb	Adjective or Adverb
No.	in ‘direct speech’	in ‘indirect speech’
1.	this	that
2.	these	those
3.	here	there
4.	now	then
5.	today/tonight	that day/that night
6.	yesterday	the previous day/ the day before
7.	tomorrow	the next day/the following day
8.	ago	before

9. The transformation from the direct to the indirect involves a change in the word order. This principle is applicable to interrogative and exclamatory sentences only. This is exemplified in the table given below.

SI. No.	Word order in Direct Speech with examples	Word order in Indirect Speech with examples
1	… Helping verb+subject+ Main verb … said … “are you coming”…	subject-1-Helping verb+Main verb asked … you were / coming
2	Helping verb 4- subject + Main verb said … do you like …	subject +…… + Main verb asked … he …. liked
3	‘wh’ word+adj/adv+ subject + verb said “How beautiful + the toy is …	subject + verb + intensifier adj/adv exclaimed – the toy was very n beautiful.

Note that the ‘wh’ word in exclamatory sentences doesn’t have the meaning of a question. It just emphasises the adjective or adverb. Hence this ‘wh’ word in the direct speech becomes the intensifier ‘very’ in the indirect speech.

10. The last step in this kind of transformation is to change the ‘Question mark’ or the ‘exclamatory mark’ into a full stop in the indirect speech.

You must note that all the Ten steps detailed above are not necessary to follow in the case of all kinds of sentences. You carefully check which steps are necessary and which steps are not necessary to follow in changing a given sentence.

EXAMPLES

1. Direct speech:
Reporter : Are you a vegetarian or non-vegetarian?
Bernard Shaw : I don’t want to make my stomach a burial ground for dead animais.
Reporter : How wonderful your answer is!
Bernard Shaw : Thank you for your compliments.

Indirect speech :
A reporter asked the famous writer Bernard Shaw whether he was a vegetarian or a non-vegetarian. Bernard Shaw replied that he didn’t want to make his stomach a burial ground for dead animals. The reporter responded that it was a wonderful answer. Bernard Shaw thanked him for his compliments.

3. Gandhi said, “I respect all religions.” (D.S)
Gandhi said that he respected all religions. (I.S)

4. He said to me, “Who is your favourite politician ?” (D.S)
He asked me who my favourite politician was (I.S)

5. An American said, “How hard-working Indians are !” (D.S)
An American exclaimed that Indians were hard-working. (I.S)

6. A customer said to the manager, “Can you do me a favour ?” (D.S)
A customer requested the manager to do him/her a favour. (I.S)

7. Abdul Kalam said, “I have come from a poor background.” (D.S)
Abdul Kalam said that he had come from a poor background. (I.S)

8. The teacher said to a student, “Are you confident ?” . (D.S)
The teacher asked a student whether he / she was confident. (I.S)

9. Amulya says, “I am learning music”. (D.S)
Amulya says that she is learning music. (I.S)

10. The teacher said,’The sun rises in the east.” (D.S)
The teacher said that the sun rises in the east. (I.S)

11. Dr. Rahul said, “1 will try my best to save the patient.” (D.S)
Dr. Rahul said that he would try his best to save the patient. (I.S)

12. Yasoda said to Krishna, “You are mischievous and trouble me a lot. (D.S)
Yasoda told Krishna that he was mischievous and troubled her a lot.” (I.S)

13. They said, “The Minister has at last unveiled the statue today. It has not been unveiled for so many months for reasons unknown.” (D.S)
They said that the Minister had at last unveiled the statue that day and added (that) it had not been unveiled for so many months for reasons unknown. (I.S)

14. A North Indian friend of mine said, “Unlike in Delhi, the climate in Hyderabad is moderate.” (D.S)
A North Indian friend of mine remarked that unlike in Delhi, the climate in Hyderabad was moderate. (I.S)

15. “How is your health ?” said Dr. Charan to a patient. (D.S)
Dr. Charan asked a patient how his/her health was. (I.S)

16. A stranger said to me, “Where is Golconda ?”. (D.S)
A stranger asked me where Golconda was. (I.S)

17. “Dheeraj said to his friend, “Are you interested in teaching ?” (D.S)
Dheeraj asked his friend if/whether he was interested in teaching. (I.S)

18. I said my daughter, “Do you want to do B.Tech. or B.Sc ?” (D.S)
I asked my daughter whether she wanted to do B.Tech or B.Sc. (I.S)

II. Study the following examples and observe how statements are changed into the indirect speech.

Direct Speech	Indirect Speech
1. Simple present He said, “I have many problems.”	1. Simple past He said that he had many problems.
2. Present continuous “I am reporting her words,” he said.	2. Past continuous He said that he was reporting her words.
3. Present perfect The cashier in the Bank said, “I have sent a report.”	3. Past perfect The cashier in die Bank said that he had sent a report.
4. Present perfect continuous A student said, “I have been trying to speak English for two years.”	4. Past perfect continuous A student said that he had been trying to speak English for two years.
5. Simple past ‘I forgot my hall ticket,” a candidate said.	5. Simple pastíPast perfect A candidate said that he / she had forgotten his hail ticket.
6. Past continuous “I was watering the plants in the garden,” she said.	6. Past continuous / Past per.cont. She said that she had been watering the plants in the garden.
7. Simple future “We will move to Hyderabad next year, “Rajitha said.	7. Rajitha said that they would move to Hyderabad the following year.
8. The English teacher said to the class, “I will tell you the difference between the two sentences.”	8. The English teacher told the class that he would tell them the difference between the two sentences.

III. Study the following examples and observe how questions are changed into indirect speech.

IV. All the verbs are changed into infinitives [to + verb (v₁)] while reporting imperative sentences.

Direct Speech	Indirect Speech
1. Order/Command The Site engineer said to his colleagues : “Don’t deviate from the plan.”	1. The Site engineer ordered his colleagues not to deviate from the plan.
2. Request A student said to the teacher : “Can you, please, repeat the question, Madam ?”	2. A student requested the teacher to repeat the question.
3. Advice Ahmad said to his son : “Don’t waste time and money.”	3. Ahmad advised his son not to waste time and money.
4. Instruction The invigilator said to the candidates in the examination hall:”Write your hall ticket number on the question paper.”	4. The invigilator instructed the candidates in the examination hall to write their hall ticket number on the question paper.
5. Threat A girl said to a boy : “I will complain to the police, if you tease me.”	5. A girl warned the boy that she would complain to the police if he teased her.
6. Offer A volunteer said to me : “Can I help you ?”	6. A volunteer offered to help me.
7. Invitation Praveena said to her friends : “Welcome to my home.”	7. Praveena invited her friends to her house.
8. Warning Mother said to her son : “Don’t swim in the turbulent river.”	8. Mother warned her son not to swim in the turbulent river.
9. Promise You said to your mother : Til be careful.”	9. You promised to your mother to be careful.
10. Apology Raheem said to you : “I am sorry I am unable to help you.”	10. Raheem apologized to you for not being able to help you.
Direct Speech	Indirect Speech
1. Keerthi said to us, “Let us have some snacks.”	1. Keerthi suggested (invited) to us that we should have some snacks.
2. Dev said to us, “Shall we visit the Thousand-Pillar temple today ?”	2. Dev proposed that we should visit the Thousand- Pillar temple that day.

Exclanations :

Direct Speech	Indirect Speech
1. “Oh ! They have lost the match,” he said.	1. He expressed regret that they had lost the match.
2. “Hurrah ! We have won the match,” said the boys.	2. The boys exdamed with delight that they had won the match.
3. My brothers said to me, “Better luck next time.”	3. My brothers wished me better luck next time.

Some More Examples :
1. He said to me, ‘You are lucky.” (D.S)
He told me (that) I was lucky. (I.S)

2. He said, “My father went to Chennai.” (D.S)
He said (that) his father had gone to Chennai. (I.S)

3. He said, “The sun rises in the east”. (D.S)
He said that the sun rises in the east. (I.S)

4. He said, “Kutub Minar is in Delhi.” (D.S)
He said that Kutub Minar is in Delhi. (I.S)

5. He said, “1 always go to bed early.” (D.S)
He said he always goes to bed earlier (I.S)

6. He said to me, “Do you want coffee ?” (D.S)
He asked me if I wanted coffee. (I.S)

7. He said to me, “Where did you go ?” (D.S)
He asked me where I had gone. (I.S)

8. He said to his son, “Go out and play.” (D.S)
He told his son to go out and play. (I.S)

9. He said, “Don’t disturb me.” (D.S)
He instructed him not to disturb him. (I.S)

10. He said, “What a terrible storm it is !” (D.S)
He exclaimed that it was a terrible storm. (I.S)

11. He said to her, “How foolish of you !” (D.S)
He exclaimed that she was very foolish. (I.S)

12. He said, “Alas ! What a tragedy.” (D.S)
He exclaimed with sorrow that it was a great tragedy. (I.S)

EXERCISES

I. Report the following statements.

1. Sunil said to his daughter, “I will take care of you.”
2. The M.L.A. said to villagers, ‘You have every right to question me.”
3. The Inspector said to the constable, “I am your boss.”
4. It is better for you to join M.PC.” said Bharath’s mother.
5. The principal said to the lecturers, You should maintain records.”
6. “I have been waiting here for you for one hour, “Vasundhara said to Vandana.
7. Kranthi said to the Inspector, “I met with an accident while taking a turn.”
8. He said, “I have lost my bag.”
9. The girl said, “I can change any given sentence into reported speech.”
10. “I am your fan,” said the boy to Allu Aijun.
Answers:
1. Sunil told his daughter that he would take care of her.
2. Tlje M.L.A told villagers that they had every right to question him.
3. The Inspector told the constable that he was his boss.
4. Bharath’s mother said that it would be better for him to join MPC.
5. The principal told the lecturers that they should maintain records.
6. Vasundhara told Vandana that she had been waiting there for her for an hour.
7. Kranthi informed the Inspector that he had met with an accident while taking a turn.
8. He said that he had lost his bag.
9. The girl said that she could change any given sentence into reported speech.
10. The boy told Allu Aijun that he was his fan.

II. Match the reported clause of Set n A with each reported statement of the direct speech in Set n B and then change the sentence into the reported speech.

Answers:
1. The palmist told a woman that she would become a good writer.
2. In a press meet, the Union Minister promised that the government would take all precautionary measures regarding cyclone.
3. Dr. Gopal told them that the operation was successful and the patient was out of danger.
4. The lawyer told the client that they could file an appeal in the High Court.
5. The boy came late to the class and told his teacher that his father had been ill for a few days.

III. Report the following questions in indirect speech.

1. A visitor said to me, “Are there any places worth seeing in Warangal ?”
2. The mother said to her son, “When will you have your breakfast ?”
3. The shopkeeper said to the customer, “Shall I show you the latest model ?”
4. I said to the shop assistant, “What is the price of this dress ?”
5. A classmate said to me, “Is your father a businessman ?”
6. Harika said to her friend, “Will you come to my home tomorrow ?”
7. The passenger said to the driver, “Does the bus stop at the crossroads ?”
8. A girl said to the principal, “Do I need to be a postgraduate to become an IAS officer ?”
9. The father said to his daughter, “Who teaches you English, Anitha ?”
10. The teacher said to Kavitha, “What does the word ‘corruption’ mean ?”
Answers:
1. A visitor asked me if there were any worth seeing places in Warangal.
2. The mother asked her son when he would have his breakfast.
3. The shopkeeper asked the customer if he could show him the latest model.
4. I asked the shop assistant what the price of that dress was.
5. A classmate asked me if my father was a businessman.
6. Harika asked her friend if she would come to her home the following day.
7. The passenger asked the driver if the bus would stop at the crossroads.
8. A girl asked the principal if she needed to be a postgraduate to become an IAS officer.
9. The father asked his daughter Anita who would teach them English.
10. The teacher asked Kavitha what the word corruption meant.

IV. Change the following imperatives into the indirect speech.

1. Hima said, “Get out from here.”
2. Neha said, “Mom, please give me your mobile.”
3. Hardik said to Annu, “Go and study.”
4. Nani said to me, “Exercise daily.”
5. Father said to Swetha, “Switch off the fan.
Answers:
1. Hima ordered him/them to get out from there.
2. Neha requested her mother to give her mobile.
3. Hardik asked Annu to go and study.
4. Nani advised me to exercise daily.
5. Father asked Swetha to switch off the fan

V. Change the following exclamations into the indirect speech.

1. Nivya said to her sister, “How interesting the serial is !”
2. My friend said to me, “What a wonderful opportunity it is !”
3. “Oh ! he is dead,” the doctor said.
4. “Thank goodness ! I’ve passed my exams,” my son said.
5. “Hurray ! I’ve got first rank in the entrance examination !” my friend said.
6. “How awful! She has missed the chance,” Mahesh said.
7. A visitor said, “What sultry weather !”
8. “What a pity ! Many passengers died in the accident,” said an eye witness.
9. Akshay said to his partner, “Bad luck, never mind.”
10. “Oh ! What a beautiful place it is !” he said.
Answers:
1. Nivya exclaimed to her sister that the serial was very interesting.
2. My friend exclaimed to me that it was a wonderful opportunity.
3. The doctor declared sadly that he was dead.
4. My son happily said that he had passed his exams.
5. My friend cheerfully declared that he had got the first rank in the entrance examination.
6. Mahesh rather sadly exclaimed that she had missed the chance.
7. A visitor exclaimed that it was very sultry weather.
8. An eye witness exclaimed sadly that many passengers had died in the accident.
9. Akshay told his partner that it was their bad luck and said better not mind that.
10. He exclaimed that it was a very beautiful place.

VI. Report the following sentences.

1. A father said to his son. “Ramu, concentrate on your studies now.”
2. “What a wonderful poem it is !” said the teacher.
3. The principal said to the student, “Vinay, can you spell this word ?”
4. “Is there any train to Mumbai now ?” I asked the enquiry clerk.
5. “How much time does a ray of the sun take to reach the earth ?” the quiz master asked the team.
6. “Alas ! India has lost a famous scientist,” said the Prime Minister on the death of Abdul Kalam.
7. “How would you help develop the company ?” the interview board member said to the can-didate.
8. “If I get a job, I will arrange a grand party,” said Spandana.
9. “Stand where you are,” the officer said to the cadets.
10. “Don’t make friends with bad boys,” said the mother to her son.
11. “Hearty welcome to our village !” Radha said to her friends.
12. “Hurrah ! We have defeated Pakistan in T 20 too,” said Kohli.
13. “Please be seated. My father is sleeping,” said the girl to the visitors.
14. “I am a pure vegetarian,” Gandhi said.
15. “How exciting it is to see Telangana as a separate state !” said a hundred-year old man.
16. “Remember, Man is mortal,” said the Swamiji.
17. He said, “We need not wait here for the bus.”
18. “While I was going to see Deepthi, it started raining,” Kiran said.
19. The doctor said, “Sorry, I cannot help it.”
20. “Nothing is in our hands,” said the priest.
Answers:
1. A father advised his son Ramu to concentrate on his studies then.
2. The teacher exclaimed that it was a very wonderful poem.
3. The principal asked Vinay if he could spell that word.
4. I asked the enquiry clerk if there was any train to Mumbai then.
5. The quiz master asked the team how much time a ray of the sun took to reach the earth.
6. The Prime Minister said on the death of Abdul Kalam that India had lost a famous scientist.
7. The interview board member asked the candidate how he would help the company develop.
8. Spandana said that if she got a job she would arrange a grand party.
9. The officer ordered the cadets to stand where they were.
10. The mother advised her son not to make friends with bad boys.
11. Radha extended hearty welcome to her friends to her village.
12. Kohli gladly said that they had defeated Pakistan in T20 too.
13. The girl requested the visitors to be seated and informed them that her father was sleeping.
14. Gandhi said that he was a pure vegetarian.
15. A hundred – year old man exclaimed that it was very exciting to see Telangana as a separate state.’
16. The Swamiji said emphatically that man is mortal.
17. He said that they needed not wait there for the bus.
18. Kiran said that while he was going to see Deepthi, it had started raining.
19. The doctor apologetically said that he couldnot help it.
20. The priest said that nothing was in their hands.

VII. Change the following sentences into the indirect speech.

1. I said to her, “I had already applied for a job.”
2. You said to me, “She loves you.” * ‘
3. We said to him, “Can we use your phone ?”
4. The teacher said to her, ‘Why did you fail the exam ?”
5. She will say to me, “How do you solve the problem ?”
6. I said to them, “I don’t ever waste my time.”
7. You said to me, “I have not yet met them.”
8. The minister said, “The problem will surely be looked into.”
9. They said to rile, “Have you been working here since 2015 ?”
10. We said to them, ‘We were watching a movie.”
11. She said, “I went to the market yesterday.”
12. They said to us, “We will be waiting for you.”
13. John said to her, “1 will call a doctor for you.”
14. I said to her, “Alas, I am undone !”
15. She said to him, “Please complete the job.”
16. He shouted at them, “Shut up!”
17. The officer said to him, “Don’t repeat this mistake in the future.”
18. The teacher said, “Akbar died in 1605 AD.”
19. She said to her, “Knowledge is power.”
20. A soft voice said, “What a cold day !”
Answers:
1. I told her that I had already applied for a job.
2. You told me that she loved me.
3. We asked him if we could use his phone.
4. The teacher asked her why she had failed the exam.
5. She will ask me how I will solve the problem.
6. I told them that I never wasted my time. (OR) I told them that I never waste my time.
7. You told me that you had not yet met them.
8. The minister said that the problem would surely be looked into.
9. They asked me if I had been working there since 2015.
10. We told them that we had been watching a movie.
11. She said that she had gone to the market the previous day.
12. They told us that they would be waiting for us.
13. John told her that he would call a doctor for her.
14. I exclaimed with sorrow to her that I was undone.
15. She requested him politely to complete the job.
16. He ordered them loudly to shut up.
17. The officer warned him not to repeat that mistake in the future.
18. The teacher said that Akbar had died in 1605 AD.
19. She told her that knowledge is power.
20. A soft voice exclaimed that it was a very cold day.

VIII. Change the following sentences into the other speech.

1. A good teacher will say frankly and clearly, “I don’t know, 1 cannot answer that question.”
2. I asked my Biology teacher what I should do to save it.
3. I asked my grandmother how she got to be so wise.
4. Thimmakka concludes, “Even one sapling each could make a better place for our children.”
5. Box : Stop ! Can you inform me who the individual is that I invariably encounter going downstairs when I’m coming up, and coming upstairs when I’m going down ?
Answers:
1. A good teacher will say frankly and clearly that he/she doesn’t know. He/She will add that he/ she cannot answer that question. (I.S)
2. I said to my Biology teacher, “What shuld I do to save it ?” (D.S)
3. I said to my grandmother, “How did you get to be so wise” ? (D.S)
4. Thimmakka concludes that even one sapling each could make a better place for our children. (I.S)
5. Box asked (her) to stop. He further asked her if she could inform him who that individual was that he invariably encountered going downstairs when he was coming up and coming upstairs when he was going down. (I.S)

C. DEGREES OF COMPARISON

Man craves for variety. He knows that variety is the spice of life. Hence he tries to express the same idea in various different ways. The result is the tradition of describing the quality, quantity size etc., of some thing or person in three different ways. Most adjectives and some adverbs have three different forms. They are positive, comparative and superlative. They are known as the three degrees of comparision.

(ఒక వ్యక్తి యొక్క వస్తున్న యొక్క గుణ, పరిమాణాలను మూడు భిన్న రూపాలలో వ్యక్తీకరించడం ఆనవాయితీ. ఈ రూపాలనే Degrees of comparison అని పిలుస్తారు)

When the comparison is among a minimum of three persons/things, that can be ex-pressed in all the three degrees. If the compared persons are two only, the superlative degree is not possible. When we talk of the quality, quantity of a single item, only positive degree is possible.

Interchange of Degrees of comparison :
Transforming one sentence from the Superlative Degree to Comparative Degree involves the following steps.

1. First, remove ‘the’ that is used before the superlative adjective, while changing it into the comparative.
2. Write the comparative form of the adjective.
3. Then, add ‘than any other’ after the adjective.
Ex : Praveen is the youngest boy in the class. (S.D.)
Praveen is younger than any other? boy in the class.

To change that sentence into the positive :

1. Begin the sentence with ‘No other’.
2. Secondly, place the phrase after the superlative adjective after ‘No other’.
3. Use ‘so + adj in positive Degree + as’ after the verb.
4. Place the subject in the S.D after ‘as’ in RD.
Ex : He is the richest man in our town. (S.D.)

Transforming one sentence from the superlative Degree to comparative or positive Degree involves the following steps.

Sentences with ‘one of phrase in the Superlative degree follow a slightly different wording. This is as explained here under.

Some sentences show the comparison between only two things/persons. In such cases, the superlative is not possible. The interchange is made as shown here.

EXAMPLES

Change the following sentences into other degrees.

Question 1.
Bengaluru is as cool as Ooty. (P.D)
Answer:
Ooty is not cooler than Bengaluru. (C.D)

Question 2.
Australia is the biggest island in the world. (S.D)
Answer:
Australia is bigger than any other island in the world. (C.D)
No other island in the world is so big as Australia. (PD)

Question 3.
The moon is brighter than the stars. (C.D)
Answer:
The stars are not so bright as the moon. (P.D)

Question 4.
There is no vice so bad as drink. (P.D)
Answer:
Drink is worse than any other vice. (C.D)

Question 5.
Kashmir is the most beautiful place in India. (S.D)
Answer:
No other place in India is so beautiful as Kashmir. (P.D)
Kashmir is more beautiful than any other place in India. (C.D)

Question 6.
Silver is not so precious as gold. (P.D)
Answer:
Gold is more precious than silver. (C.D)

Question 7.
Kumble is one of the ablest bowlers. (S.D)
Answer:
Very few bowlers are so able as Kumble. (P.D)
Kumble is abler them most other bowlers. (C.D)

Question 8.
The lion is more ferocious than any other animal. (C.D)
Answer:
The lion is the most ferocious of all the animals. (S.D)
No other animal is so ferocious as the lion. (P.D)

Question 9.
The sword is not so mighty as the pen. (P.D)
Answer:
The pen is mightier than the sword. (C.D).

Question 10.
Copper is more useful than most other metals. (C.D)
Answer:
Copper is one of the most useful metals. (S.D)
Very few metals are so useful as copper. (P.D)

Question 11.
Mount Everest i.s higher than any other peak in the world. (C.D)
Answer:
Mount Everest is the highest of all the peaks in the world. (S.D)
No other peak in the world is so high as Mount Everest. (P.D)

Question 12.
I like you better than him. (C.D)
Answer:
I didn’t like him as well as I like you. (P.D)

Question 13.
A deer runs faster than a horse. (C.D)
Answer:
A horse does not run so fast as a deer. (P.D)

Question 14.
Sangeetha is the tallest girl in the class. (S.D)
Answer:
No other girl in the class is so tall as Sangeetha. (P.D)

Question 15.
This is one of the most powerful earthquakes that occurred. (S.D)
Ans. This is more powerful than most other earthquakes that occurred. (C.D)
Very few earthquakes that occurred are as powerful as this earthquake. (P.D)

More Examples :

1. For me, dancing is easier to learn than singing. (C.D)
For me, singing is not as easy to learn as dancing. (P.D)

2. Singing is the best of all art forms. (S.D)
Singing is better than any other art forms. (C.D)
No other art form is as good as singing. (P.D)

3. The sentence is more impressive than the earlier one. (C.D)
The earlier sentence is not so impressive as this one. (P.D)

4. A car, of course, is costlier than a bike. (C.D)
A bike, of course, is not as costly as a car. (P.D)

5. Uttar Pradesh is the largest state in India. (S.D)
Uttar Pradesh is larger than any other state in India. (C.D)
No other state in India is as large as Uttar Pradesh. (P.D)

6. Bill Gates is the wealthiest man in the world. (S.D)
Bill Gates is wealthier than any other man in the world. (C.D)
No other man in the world is as wealthy as Bill Gates. (P.D)

7. English is the most widely used language in the world. (S.D)
English is more widely used than any other language in the world. (C.D)
No other language in the world is as widely used as English. (P.D)

Types of transformation :

EXERCISES

I. Change the following sentences into other Degrees.

1. LIC is one of the most popular insurance companies in India.
2. The custard apple is better for health than the apple.
3. A computer works faster than the human brain.
4. A Governor is sometimes more powerful than a Chief Minister.
5. The teaching profession is the best of all professions.
6. Laxmi Mittal is one of the most popular industrialists.
7: No other bank in India has as many branches as SBI.
8. Virus infects a person faster than bacteria.
9. Cancer is one of the most dangerous diseases.
10. No other boy in the class is as active as Surya Teja.
11. The Amazon is one of the longest rivers in the world.
12. No other animal lives as long as the turtle.
13. Jupiter is bigger than any other planet.
14. A rainbow is one of the most beautiful sights in nature.
15. Very few English poets are as great as John Keats.
16. The lotus is the most beautiful flower.

17. Mathematics is more difficult than most other subjects.
18. Shimla is cooler than Ooty.
19. He can’t run as fast as I.
20. Vinay is not as mischievous as some other boys in the college.
21. There are some vegetarian food as healthy as eggs.
22. Of all the Telugu singers S.R Balasubramanyam has the most melodious voice.
23. Health is more important than wealth.
24. I cannot speak as fast as you.
25. Very few TV channels are as popular as E TV.
Answers:
1. Very few insurance companies in India are as popular as LIC. (P.D)
LIC is more popular than many other insurance companies in India. (C.D)

2. The apple is not as good for health as the custard apple. (P.D)

3. The human brain does not work as fast as a computer. (P.D)

4. Sometimes a Chief Minister is not as powerful as a Governor. (P.D)

5. The teaching profession is better than any other profession. (C.D)
No other profession is as good as the teaching profession. (P.D)

6. Very few industrialists are as popular as Laxmi Mittal. (P.D)
Laxmi Mittal is more popular than many other industrialists. (C.D)

7. SBI has the most number of branches. (S.D)
SBI has more branches than any other bank in India. (C.D)

8. Bacteria does not infect a person as fast as virus. (P.D)

9. Cancer is more dangerous than many other diseases. (C.D)
Very few diseases are as dangerous as cancer. (P.D)

10. Surya Teja is more active than any other boy in the class. (C.D)
Surya Teja is the most active boy in the class. (S.D)

11. The Amazon is longer than many other rivers in the world. (C.D)
Very few rivers in the world are as long as the Amazon. (P.D)

12. The turtle lives longer than any other animal. (C.D)
The turtle lives the longest of all animals. (S.D)

13. No other planet is as big as Jupiter. (P.D)
Jupiter is the biggest planet. (S.D)

14. A rainbow is more beautiful than many other sights in nature. (C.D)
Very few sights in nature are as beautiful as a rainbow. (P.D)

15. John Keats is one of the greatest poets of English. (S.D)
John Keats is greater than many other poets of English. (C.D)

16. The lotus is more beautiful than any other flower. (C.D)
No other flower is as beautiful as the lotus. (P.D)

17. Mathematics is one of the most difficult subjects. (S.D)
Very few subjects are as difficult as Mathematics. (P.D)

18. Ooty is not as cool as Shimla. (P.D)

19. I can run faster than he. (C.D)

20. Vinay is one of the most mischievous boys in the college. (S.D)
Vinay is more mischievous than many other boys in the college. (C.D)

21. Eggs are not healthier than some vegetarian foods. (C.D)

22. No other Telugu singer has a voice as melodious as S.R Balu’s. (P.D)
S.R Balu’s voice is more melodious than that of any other Telugu singer. (C.D)

23. Wealth is not as important as health. (P.D)

24. You can speak faster than I. (C.D)

25. E TV is more popular than many other TV channels. (C.D)
E TV is one of the most popular TV channels (S.D)

II. Rewrite the following comparisons as directed.

1. The taste of Pizza is more pleasing than that of Burger. (into the other degree)
2. Sheela is getting smarter and smarter than Neela. (into the other degree)
3. Raj is one of the bravest fighters. (into comparative)
4. Radha speaks more fluently than Sudha. (into the other degree)
5. ’ Riding a horse is not as easy as riding a motorbike. (into the other degree)
6. ‘Silence’ is the most potent weapon to win an argument. (into positive)
7. Rachana’s sister is taller than yours. (into the other degree)
8. Dogs don’t look as cute as rabbits. (into the other degree)
9. He is not the worst student in the class. (into comparative)
10. Very few heroes are as great as Gandhiji in the world history. (into superlative)
Answers:
1. The taste of Burger is not so pleasing as that of Pizza. (P.D)
2. Neela is not getting as smart as Sheela. (P.D)

3. Raj is braver than any other fighter. (C.D)
No other fighter is as brave as Raj. (P.D)

4. Sudha does not speak as fluently as Radha. (P.D)

5. Riding a motorbike is easier than riding a horse. (C.D)

6. No other weapon is so potent as silence to win an argument. (P.D)
Silence is more potent than any other weapon to win an argument. (C.D)

7. Your sister is not so tall as Rachana’s sister. (P.D)
8. Rabbits look cuter than dogs. (C.D)

9. He is not worse than most students in the class. (C.D)
Many students in the class are as bad as he. (P.D)

10. Gandhi is one of the greatest heroes in the world history. (S.D)
Gandhi is greater than many other heroes in the world history. (C.D)

III. Change the following sentences into other degrees of comparison.

1. Bus journey is not as comfortable as train journey.
2. Radhakrishnan is more highly respected than any other teacher.
3. Robert Frost is one of the best American poets.
4. No other road in the world is as long as the Pan American Highway.
5. Kashmir is one of the coolest places in India.
6. A foolish friend can be more dangerous than a wise enemy.
7. Money is not as important as character.
8. Modern culture is not as stable as Traditional culture.
9. For many Indians, cricket gives greater pleasure than football.
10. Natural flowers appeal more to our senses than artificial flowers.
Answers:
1. Train journey is more comfortable than bus journey. (C.D)

2. Radhakrishnan is the most highly respected teacher. (S.D)
No other teacher is so highly respected as Radhakrishnan. (P.D)

3. Robert Frost is better than many other American poets. (C.D)
Very few American poets are as good as Robert Frost. (P.D)

4. Pan American Highway is longer than any other road in the world. (C.D)
Pan American highway is the longest road in the world. (S.D)

5. Kashmir is cooler than many other places in India. (C.D)
Very few places in India are so cool as Kashmir. (P.D)

6. A wise enemy cannot be so dangerous as a foolish friend. (P.D)
7. Character is more important than money. (C.D)
8. Traditional culture is more stable than modern culture. (C.D)
9. For many Indians, football doesn’t give as much pleasure as cricket. (P.D)
10. Artificial flowers do not appeal to our senses as much as natural flowers. (P.D)

D. QUESTION TAGS

Question tags are short questions added to statements. They seek confirmation or agreement. (ఏకీభావాన్ని లేదా ద్రువీకరణను ఆశిస్తూ, ఒక వాక్యానికి అనుబంధంగా చేర్చబడే చిన్న ప్రశ్నలే Question tags.)

Question tags have a fixed structure. It is represented as : Helping verb + (n’t) + pronoun ? ((n’t) in brackets means it is optioned.)
e.g. She loves her children, doesn’t she?
(Q.T.) does (H.V.) n’t she (pronoun)

1. While constructing a tag, selecting the suitable helping verb is the first and the most important step. If the given sentence has a helping verb, then that helping verb is used in the tag too.
E.g : She has answered all the questions, hasn’t she?
has → Helping Verb in the sentence used in the tag too.

When the sentence doesn’t have a helping verb, the verb is either in the simple present Or in the simple past and in the affirmative. In case of the simple present, the suitable helping verb is “do/does”. If the verb in the sentence is in the simple past, then we must use ‘did’ in the tag.
Eg : a) She hates smokers, doesn’t she ? Simple present, hence ‘does’
b) They grow vegetables, do n’t they ? Simple present
c) He sold his car, didn’t he ? Simple past; hence ‘did’.

2. The second step is to decide whether to use ‘n’t’ or not. The guiding principle is quite simple. If the given sentence doesn’t have ‘rio/not’, use ‘n’t’ is the tag. If the given sentence has ‘no/not’ then the tag doesn’t have ‘n’t’.
Eg : a) He was sleeping in the class, was n’t he ?
‘not’ is not used n’t in the tag
b) They don’t grow paddy, do they?
‘n’t’ in sentence ‘n’t’ not used in tag
Remember to use the short form ‘n’t’ in the tag.

3. The last part of the tag as a pronoun. This corresponds to the subject of the given sentence. If the subject of the sentence is a pronoun, the same pronoun is used in the tag. If the subject is a noun, select a suitable pronoun as shown below.

SI.No.	Subject in the sentence	Pronoun used in the tag
1.	Name of a man	he
2.	Name of a woman	she
3.	Name of a single thing	it
4.	Name referring to, a group of persons, places, things	they

Eg : a) Gourav is a good cricketer, isn’t he? ?
b) Sania Mirza played well, didn’t she ?
c) Ooty is a beautiful hill station, isn’t it ?
d) Mangoes aren’t good this season, are they ?

4. Note that ‘amn’t’ is not used. Instead, aren’t is used.
Eg : lam rather slow, aren’t I ?

5. Imperative sentences take ‘will’ as the helping verb in the tags.
Eg : a) Close the door, Won’t you ? (Will + n’t = won’t)
b) Don’t leave the place, will you ?

6. Sentences expressing proposals with the help of “Let us ……………..” have for their tags “Shall we?”
Eg : Let us walk fast, shall we ?

EXAMPLES

1. Srinivasa Ramanujan is a famous mathematician, isn’t he ?
2. Normally players have a lot of practice, don’t they ?
3. Sania Mirza is not a cricketer, is she ?
4. She can speak English, can’t she ?
5. He should believe us, shouldn’t he ?
6. We must all learn English, mustn’t we ?
7. The Governor administers the oath of office to ministers, doesn’t she / he ?
8. He does not support any one, does he ?
9. Most of the villagers depend on agriculture, don’t they ?
10. They do not seem to lead a happy life, do they ?
11. Children love to play with toys, don’t they ?
12. On their picnic, the children did not have a chance to play in the garden, did they ?
13. The children are not yet back home, are they ?
14. Do me a favour, Raju, can you ?
15. Take a right decision, won’t you ?
16. Don’t waste time, will you ?
17. Let’s understand their problems, shall we ?
18. No one complained against us, did they ?
19. Everyone appreciated his performance, didn’t they ?
20. Someone should take the initiative, shouldn’t they ?
21. Nothing is impossible, is it ?
22. There will be problems in that case, won’t there ?
23. One can achieve anything by faith, can’t one ?
24. I am not disturbing you, am I ?
25. It is used for present actions, isn’t it ?
26. You understand the point, don’t you ?
27. You are attending a job interview tomorrow, aren’t you ?
28. It is clear to you now, isn’t it ?
29. They are ready, aren’t they ?
30. They are not ready, are they ?
31. He did not attend the class, did he ?
32. Oh, you got two prizes, did you ? (Special case. Positive statement, positive tag. Express reaction to a surprising news.)
33. I am going home, aren’t I ? (Note that ‘aren’t is used instead of ‘amn’t.)

EXERCISES

I. Add an appropriate question tag to each one of the following statements.

1. Sandeep has attended all classes, ……………….. ?
2. We are lucky to be born in India, ………………..?
3. English is an interesting language, ……………….. ?
4. He was very busy yesterday, ……………….. ?
5. I am very happy now, ……………….. ?
6. I can face challenges, ……………….. ?
7. Ravi always thinks positively, ……………….. ?
8. He does not criticize, ……………….. ?
9. Some people always depend on others, ……………….. ?
10. Discipline must be maintained at any cost, ……………….. ?
11. Let us keep to the pavement, ……………….. ?
12. Don’t blame others for everything, ……………….. ?
13. One can do wonders with knowledge, ……………….. ?
14. Nothing is permanent except change, ……………….. ?
15. Students are our best judges, ……………….. ?
Answers:
1. hasn’t he ?
2. aren’t we ?
3. isn’t it ?
4. wasn’t he ?
5. aren’t I ?
6. can’t I ?
7. doesn’t he ?
8. does he ?
9. Don’t they
10. mustn’t it be ?
11. shall we ?
12. will you ?
13. can’t one ?
14. is it ?
15. aren’t they ?

II. Add an appropriate question tag to each one of the following statements.

1. You don’t lime me, ……………….. ?
2. It is not raining, ……………….. ?
3. You have done your homework, ……………….. ?
4. I am not late, ……………….. ?
5. I am invited to your party, ………………..?
6. You like fast food, ……………….. ?
7. You will come to my party, ……………….. ?
8. You remembered to feed the cat, ……………….. ?
9. Let’s play tennis, ……………….. ?
10. There’s a problem here, ……………….. ?
11. He never says a word, ……………….. ?
12. Nobody came to your party, ……………….. ?
13. Don’t forget, ……………….. ?
14. You think you’re clever, ……………….. ?
15. So you think you’re clever, ……………….. ?
16. We don’t have to go to the party, ……………….. ?
17. It stopped raining, ……………….. ?
18. Have a seat, ……………….. ?
19. Help yourself to some cake, ……………….. ?
20. Children, be quiet, ……………….. ?
Answers:
1. do you
2. is it
3. haven’t you
4. am I (Note . possible tag-am ; negative tag are in the place of ami
5. aren’t I
6. don’t you
7. won’t you
8. didn’t you
9. shall we
10. isn’t there
11. does he (never = not; negative statement; hence positive tag)
12. did they
13. will you
14. don’t you (you think is the main clause)
15. do you (expressing sudden reaction)
16. do we
17. didn’t it
18. won’t you
19. will you
20. will you

III. Add an appropriate question tag to each one of the following statements.

1. I am unable to answer your question, ………………..?
2. Rahul’s first rank is at stake, ……………….. ?
3. The noise in my ears was that of the faithful Oxford crowd, ……………….. ?
4. The stop-watches held the answer, ……………….. ?
5. It belongs to both of you, ……………….. ?
Answers:
1. am I (unable = not able)
2. isn’t it
3. wasn’t it
4. didn’t they
5. doesn’t it ?

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Correction of Errors in Sentences Exercise Questions and Answers.

TS Inter 1st Year English Grammar Correction of Errors in Sentences

Q.No. 14 (4 × 1 = 4 Marks)

Exercises

I. Correct the underlined parts of the following sentences :

1. Your informations are wrong.
2. He has bought expensive furnitures.
3. Children should learn the alphabets with joy.
4. He has deep knowledges of various fields.
5. The sceneries of Darjeeling are very beautiful.
6. Keep your surrounding clean.
7. We must express thank to those who help us.
8. Economics are an interesting subject.
9. Athletics are an interesting sport.
10. The news of the earthquake have spread like wildfire.
11. Measles are an infectious disease.
12. Ocean sands are not used for construction.
13. C.V. Raman’s knowledges of all branches of physics are amazing.
14. There is a scarcity of man-servants nowadays.
15. Many passer-bys observed the accident.
16. Mouses have spoiled the crop.
17. There are five womans in the team.
18. My cousin brother is a doctor.
19. We should wash our foots before coming into the house.
20. He has many sheeps.

21. One of my classmates are an army officer.
22. One should respect her teachers.
23. Every men are responsible for this situation.
24. His both sons are lawyers.
25. The principal and the chairman has attended the meeting.
26. Each of them were given a gift.
27. Student’s must avail the opportunities.
28. The two players blamed one another for their defeat.
29. All Indians must respect each other.
30. Yourself are responsible for your future.
31. There are no less than ten employed persons in their village.
32. He and me are brothers.
33. As there are only few students, I can interact with them easily.
34. My all friends are very active.
35. Gandhi is more truthful than any political leader.
36. This is taller than many buildings in Hyderabad.
37. Raghu is my older brother.
38. The streets of Hyderabad are wider than Warangal.
39. Suma is an popular anchor.
40. I waited for a hour.
41. Sun rises in east.
42. Onions cost Rs. 20 kilogram.
43. They are staying in the same flat for the last many years.
44. How long are you waiting here ?
45. He is interested to do a job.
46. They have moved to the new house last week.
47. He is having many imported clothes.
48. As soon as I opened the doors, the birds flu away.
49. If I will stand on my own legs, my parents will feel happy.

50. If you consult me, I would have advised you what to do.
51. He is visiting the library daily.
52. He walks very fastly.
53. We don’t , hardly believe it.
54. They don’t do anything careful.
55. She scarce attends classes.
56. He is walking very slow.
57. The property was divided between the four brothers.
58. I prefer fruits than sweets.
59. He is afraid with darkness.
60. We entered into the hall to watch the play.
61. Beside being a poet, Tagore is also a short story writer.
62. The shops will be open between 10 a.m. to 8 p.m.
63. Either you must take up a job or start a business.
64. Neither she drinks tea nor coffee.
65. Sheela is as proud like a peacock.
66. They asked me that where SBI was.
67. They asked me what was my name.
68. This article is made with cotton.
69. We look forward to meet the minister.
70. I know them for the last many years.
71. Why is she hating classical music ?
72. It is raining since yesterday.
73. I am good in English.
74. If cleanliness will be maintained, we will be healthy.
75. We had seen them two weeks ago.
76. The river has overflown its bank in many places.
77. He has hanged his coat on a nail.
78. All banks are open from 10.30 a.m. and 4 p.m.
79. He is not an expert in grammar, isn’t it ?
80. Children are fond for chocolates.
Answer:
1. Your informations is wrong.
2. He has bought expensive furniture.
3. Children should learn the alphabet with joy.
4. He has deep knowledge of various fields.
5. The scenery of Darjeeling are very beautiful.
6. Keep your surroundings clean.
7. We must express thanks to those who help us.
8. Economics is an interesting subject.
9. Athletics is an interesting sport.
10. The news of the earthquake has spread like wildfire.

11. Measles is an infectious disease.
12. Ocean sands is not used for construction.
13. C.V. Raman’s knowledge of all branches of physics are amazing.
14. There is a scarcity of men-servants nowadays.
15. Many passer-by observed the accident.
16. Mice have spoiled the crop.
17. There are five womens in the team.
18. My cousin is a doctor.
19. We should wash our feet before coming into the house.
20. He has many sheep.
21. One of my classmates is an army officer.
22. One should respect one’s teachers.
23. Every men is responsible for this situation.
24. Both his sons are lawyers.
25. The principal and the chairman have attended the meeting.
26. Each of them was given a gift.
27. Student’s must avail themselves of the opportunities.
28. The two players blamed each another for their defeat.
29. All Indians must respect one other.
30. You are responsible for your future.
31. There are no fewer than ten employed persons in their village.
32. He and I are brothers.
33. As there are only a few students, I can interact with them easily.
34. All my friends are very active.
35. Gandhi is more truthful than any other political leader.
36. This is taller than many other buildings in Hyderabad.
37. Raghu is my elder brother.
38. The streets of Hyderabad are wider than those of Warangal.

39. Suma is a popular anchor.
40. I waited for an hour.
41. The sun rises in east.
42. Onions cost Rs. 20 a kilogram.
43. They have been staying in the same flat for the last many years.
44. How long have you been waiting here ?
45. He is interested in doing a job.
46. They moved to the new house last week.
47. He has many imported clothes.
48. As soon as I opened the doors, the birds flew away.
49. If I stand on my own legs, my parents will feel happy.
50. If you had consulted me, I would have advised you what to do.
51. He visits the library daily.
52. He walks very fast.
53. We don’t beleive it/. We hardly believe it.
54. They don’t do anything careful.
55. She scarcely attends classes.
56. He is walking very slowly.
57. The property was divided among the four brothers.
58. I prefer fruits to sweets.
59. He is afraid of darkness.
60. We entered the the hall to watch the play.
61. Besides being a poet, Tagore is also a short story writer.
62. The shops will be open between 10 a.m. and 8 p.m.
63. You must either take up a job or start a business.
64. She drinks neither tea nor coffee.

65. Sheela is as proud as a peacock.
66. They asked me where SBI was.
67. They asked me what my name was.
68. This article is made of cotton.
69. We look forward to meetimg the minister.
70. I have know them for the last many years.
71. Why does she hate classical music ?
72. It has been raining since yesterday.
73. I am good at English.
74. If cleanliness is maintained, we will be healthy.
75. We saw them two weeks ago.
76. The river has overflowed its bank in many places.
77. He has hung his coat on a nail.
78. All banks are open from 10.30 a.m. to 4 p.m.
79. He is not an expert in grammar, is he ?
80. Children are fond of chocolates.

II. Correct the following sentences :

1. Very good morning.
2. What is your good name ?
3. Why you are late today ?
4. The staff meeting has been pre-poned.
5. 1 will report it to the concerned teacher.
6. We go home by walk.
7. Where you come from ?
8. 1 do not know what is your name.
9. He went to the back-side of the house.
10. Please bring your luggages here.
11. This road is more shorter than that.
12. We often chit-chat with our friends.
13. Please shut the TV.
14. I and my wife went to a movie.
15. Mohan and myself will come.
16. I’m having a scooter.
17. This costed me a lot.

18. I doubt that he will succeed.
19. We are living in Bengaluru since 2015.
20. We have a lot of furnitures in our house.
Answer:
1. Good morning.
2. What is your name ?
3. Why are late today ?
4. The staff meeting has been advanced.
5. I will report it to the teacher concerned.
6. We go home on foot.
7. Where are you coming from ?
8. 1 do not know what your name is.
9. He went to the back of the house.
10. Please bring your luggage here.
11. This road is shorter than that.
12. We often chat with our friends.
13. Please switch of the TV.
14. My wife and I went to a movie.
15. Mohan and I will come.
16. 1 have a scooter.
17. This cost me a lot.
18. I doubt if he will succeed.
19. We have been living in Bengaluru since 2015.
20. We have a lot of furniture in our house.

III. Rewrite the following sentences after correcting the errors :

1. Can you read the Urdu alphabets ?
2. The police is investigating the case.
3. Here is your glasses.
4. Look at the man in the blue jean.
5. Have you seen the table of content of this textbook ?
6. Mumps are a kind of disease.
7. There were a series of programmes to mark the occasion.
8. She bought a toothpaste.
9. This box has twelve dozens of apples.
10. The athlete ran a four-miles race.

11. I received five thousands rupees from the manager.
12. Most male deers have horns.
13. Eight hundred rupees are too much for a pen.
14. A black and white dog are following me.
15. A black and a white dog is following me.
16. Many a man were cheated in this way.
17. I lost my all belongings.
18. Neither he nor his wife have arrived yet.
19. More than one person are working here.
20. The first inning is over.
Answer:
1. Can you read the Urdu alphabet ?
2. The police are investigating the case.
3. Here are your glasses.
4. Look at the man in the blue jeans.
5. Have you seen the table of contents of this textbook ?
6. Mumps is a kind of disease.
7. There was a series of programmes to mark the occasion.
8. She bought a tube of toothpaste.
9. This box has twelve dozen of apples.
10. The athlete ran a four-mile race.
11. I received five thousand rupees from the manager.
12. Most male deer have horns.
13. Eight hundred rupees is too much for a pen.
14. A black and white dog is following me.
15. A black and a white dog are following me.
16. Many a man was cheated in this way.
17. I lost all my belongings.
18. Neither he nor his wife has arrived yet.
19. More than one person is working here.
20. The first innings is over.

IV. Rewrite the following sentences after correcting the errors :

1. It is myself.
2. She and me are twins.
3. Each of these girls have wide eyes.
4. Neither of these cars are worth the money.
5. Both didn’t attend the meeting.
6. We all didn’t go.
7. Sharma plays cricket better than me.
8. Any one of these two boys are sent for training.
9. I haven’t got some pens.
10. If I were him. I wouldn’t have played the game.
11. Only they who have admit passes will be allowed in.
12. Drink water which tastes better.
13. Every woman raised their voice.
14. Sheila and Nancy like one another.
15. My neighbour that works in a bank has gone to Mumbai.
16. My three sisters like each other.
17. Divide the pieces of bread between him and me.
18. Shrika is senior than Srihan.
19. These all mangoes are ripe.

20. There are no less than ten thousand books in this library..
21. She is worst than her cousin.
22. This picture is the best of the two.
Answer:
1. It is I.
2. She and I are twins.
3. Each of these girls has wide eyes.
4. Neither of these cars is worth the money.
5. Neither attend the meeting.
6. None of us went.
7. Sharma plays cricket better than I.
8. Either of these two boys is sent for training.
9. I haven’t got any pens.
10. If I were he, I wouldn’t have played the game.
11. Only those who have admit passes will be allowed in.
12. Drink water that tastes better.
13. Every woman raised her voice.
14. Sheila and Nancy like each other.
15. My neighbour who works in a bank has gone to Mumbai.
16. My three sisters like one other.
17. Divide the pieces of bread between he and I.
18. Shrika is senior to Srihan.
19. All these mangoes are ripe.
20. There are no fewer than ten thousand books in this library.
21. She is worse than her cousin.
22. This picture is the better of the two.

V. Rewrite the following sentences after correcting the errors.

1. I need an advice from an expert.
2. He is more better than she.
3. This is the most best book I have read.
4. Virat Kohli is better than any cricketer in India.
5. This is strongest of any other metals.
6. She has been helping the poors.
7. The two first chapters are very interesting.
8. I am hopeless about his success.
9. She was not smart early.
10. This is a worth reading novel.
11. He hasn’t written much stories.
12. Open your book at six page.
13. This is a best book.
14. Ooty is cool than Thirupathi.
15. I paid him rupees hundred.
Answer:
1. I need a piece of advice from an expert.
2. He is better than she.
3. This is the best book I have read.
4. Virat Kohli is better than any other cricketer in India.
5. This the strongest of all metals.
6. She has been helping the poor.
7. The first two chapters are very interesting.
8. I have no hope of his success.
9. She was not smart earlier.
10. This is a novel worth reading.
11. He hasn’t written many stories.
12. Open your book at page six.
13. This is a very good book.
14. Ooty is cooler than Thirupathi.
15. I paid him a hundred rupees.

VI. Rewrite the following sentences after correcting the errors.

1. Sania Mirza is more popular than any tennis player.
2. Of the two this is the best article.
3. Our apartment is on third floor.
4. When Rome was burning, Nero is playing on the flute.
5. She enjoys to sing patriotic songs.
6. Why you were absent yesterday ?
7. She scarce blames others.
8. This is different to others.
9. She resembles to her mother.
10. I was prevented to do my work.
11. Would you mind to open the door ?
Answer:
1. Sania Mirza is more popular than any other tennis player.
2. Of the two this is the better article.
3. Our apartment is on the third floor.
4. When Rome was burning, Nero was playing on the flute.
5. She enjoys singing patriotic songs.
6. Why were you absent yesterday ?
7. She scarcely blames others.
8. This is different from others.
9. She resembles her mother.
10. I was prevented from doing my work.
11. Would you mind opening the door ?

VII. Rewrite the following sentences after correcting the errors.

1. My brother is working in tribal area.
2. I have an urgent business.
3. Kaleshwaram Project is built on Godavari.
4. Moon is very bright today.
5. I shall be visiting UK next month.
6. Himalayas form the northern boundary of India.
7. We should love the nature.
8. The Calcutta is a big city.
9. I have read Ramayana thrice.
10. David goes to the church every Sunday.
11. The gold is a precious metal.
12. He can play sitar well.
13. Lion is the king of beasts.
14. I had the lunch at noon.
15. I met my friend at college.
Answer:
1. My brother is working in a tribal area.
2. I have (some) urgent business.
3. Kaleshwaram Project is built on the Godavari.
4. The Moon is very bright today.
5. I shall be visiting the UK next month.
6. The Himalayas form the northern boundary of India.
7. We should love nature.
8. Calcutta is a big city.
9. I have read the Ramayana thrice.
10. David goes to church every Sunday.
11. Gold is a precious metal.
12. He can play the sitar well.
13. The Lion is the king of beasts.
14. I had lunch at noon.
15. I met my friend at the college.

VIII. Rewrite the following sentences after correcting the errors.

1. One of my friends have gone to the UAE.
2. Each of the girls were given the medals.
3. Everyone of the workers have stayed away from work.
4. Neither of the participants were able to win the match.
5. Age and experience bring wisdom to man.
6. Bread and butter are my favourite breakfast.
7. Where you are going ?
8. When vou will come here again ?
9. He asked me if I am going to Dubai.
10. I asked her if she is learning dance.
11. You are married, isn’t it ?
12. I like skating, do I ?
13. As I was ill so I could not go.
14. Both Hari as well as Krishna came to see me.
15. She said that she saw him last night.
16. The train left before I arrived.
17. I had been to Vijayawada yesterday.
18. I took my supper.
19. He knows to swim.

20. She cut her pencil.
21. I said to him to go.
22. He gave a speech.
23. He made a lecture.
24. My tooth is painting.
25. He made a goal.
Answers:
1. One of my friend has gone to the UAE.
2. Each of the girls was given the medals.
3. Everyone of the workers has stayed away from work.
4. Neither of the participants was able to win the match.
5. Age and experience brings wisdom to man.
6. Bread and butter is my favourite breakfast.
7. Where are you going ?
8. When will you come here again ?
9. He asked me if I was going to Dubai.
10. I asked her if she was learning dance.
11. You are married, aren’t you ?
12. I like skating, don’t I ?
13. As I was ill I could not go. (OR)
I was ill so I could not go.
14. Both Hari and Krishna came to see me. (OR)
Hari as well as Krishna came to see me.
15. She said that she had seen him last night.
16. The train had left before I arrived.
17. I went to Vijayawada yesterday.
18. I had my supper.
19. He knows how to swim.
20. She sharpened her pencil.

21. I told him to go.
22. He made a speech.
23. He gave a lecture.
24. My tooth is aching.
25. He scored a goal.

IX. Rewrite the following sentences after correcting the errors.

1. I am too glad to see you.
2. This coffee is very hot to drink.
3. She is very much sorry.
4. You are leaving back your bag.
5. You can lift it by and by.
6. We scarcely see a lion.
7. He behaved cowardly.
8. Sarojini Naidu was called as the Nightingale of India.
9. She works very hardly.
10. He speaks English very good.
11. His version is somewhat true.
12. She speaks the truth always.
13. He comes to me often.
14. Most likely the shops will remain closed tomorrow.
15. The whole India is proud of his achievement.
Answer:
1. I am very glad to see you.
2. This coffee is too hot to drink.
3. She is very sorry.
4. You are leaving behind your bag.
5. You can lift it little and little.
6. We rarely see a lion.
7. He behaved in a cowardly manner.
8. Sarojini Naidu was called the Nightingale of India.
9. She works very hard.
10. He speaks English very well.
11. His version is partly true.
12. She always speaks the truth.
13. He often comes to me.
14. Most probably the shops will remain closed tomorrow.
15. The whole of India is proud of his achievement.

X. Rewrite the following sentences after correcting the errors.

1. I have ordered for three dishes.
2. He entered into the hall.
3. She has married with her cousin.
4. We are awaiting for the train.
5. He has been suffering with Corona.
6. She covered her head by a shawl.
7. They waited three hours.
8. I enquired her where her husband was.
9. He went for riding.
10. She is called with different names.
11. Sion the document with ink.
12. This is a comfortable house to live.
13. I filled water in the bucket.
14. I shall explain them this.
15. I travelled by my Principal’s car.
16. She was accused for stealing a gold ring.
17. He congratulated me for my success.
18. She got down the bus at Mancherial.
19. I don’t agree for your proposal.
20. This is the road to go.

21. Though he is fat, but he runs fast.
22. Scarcely had he gone than the phone rang.
23. No sooner had he gone when the officer came.
24. She is not only beautiful but intelligent.
25. I doubt that he will get through the exam.
26. He was handsome but everyone admired him.
27. You are not right nor wrong.
28. Wait while I come.
29. Unless you do not try, you will never succeed.
30. City life is tense and village life is relaxed.
Answers:
1. I have ordered three dishes.
2. He entered the hall.
3. She has married her cousin.
4. We are awaiting the train./We are waiting for the train.
5. He has been suffering from Corona.
6. She covered her head with a shawl.
7. They waited for three hours.
8. I enquired of her where her husband was.
9. He went riding. OR He went for a ride.
10. She is called by different names.
11. Sign the document in ink.
12. This is a comfortable house to live in.
13. I filled the bucket with water.
14. I shall explain this to them.
15. I travelled in my Principal’s car.
16. She was accused of stealing a gold ring.
17. He congratulated me on my success.
18. She got down from the bus at Mancherial.
19. I don’t agree to your proposal.
20. This is the road to go by.

21. Though he is fat, he runs fast.
22. Scarcely had he gone when the phone rang.
23. No sooner had he gone than the officer came.
24. She is not only beautiful but also intelligent.
25. I doubt whether he will get through the exam.
26. He was handsome and everyone admired him.
27. You are neither right nor wrong.
28. Wait till I come.
29. Unless you try, you will never succeed.
30. City life is tense but village life is relaxed.

XI. Rewrite the following sentences after correcting the errors.

1. A bouquet of yellow roses lend colour and fragrance.
2. Here is the keys.
3. Two miles are too far to walk.
4. If I was a bird, I would fly.
5. I wish it was Sunday.
6. All the staff are here.
7. The captain with his men were killed.
8. It is a two days programme.
9. Neither he came nor he wrote.
10. One of my friend is here.
11. He enioyed at the party.
12. He is in class tenth.
13. He gave his examination.
14. Columbus invented America.
15. Edison discovered electric bulb.
16. Never I have seen such a mess.
17. Should I cut this word ?
18. He asked a help.
19. They insisted to pay.
20. Choose the best of the two options.

21. He went to foreign.
22. I asked a question to him.
23. We are doing Yoga every day.
24. All the roads are covered by snow.
25. He does not care for my words.
Answers:
1. A bouquet of yellow roses lends colour and fragrance.
2. Here are the keys.
3. Two miles is too far to walk.
4. If I were a bird, I would fly.
5. I wish it were Sunday.
6. All these staff is here.
7. The captain with his men was killed.
8. It is a two day programme.
9. Neither did he come nor did he write.
10. One of my friends is here.
11. He enjoyed himself at the party.
12. He is in class ten.
13. He took his examination.
14. Columbus discovered America.
15. Edison invented electric bulb.
16. Never have I seen such a mess.
17. Should I erase this word ?
18. He asked for help.
19. They insisted on paying.
20. Choose, the better of the two options.
21. He went abroad.
22. I, asked him a question.
23. We do Yoga every day.
24. All the roads are covered with snow.
25. He pays no attention to what I say.

XII. Rewrite the following sentences after correcting the errors.

1. We elected Ram as President.
2. Did you went to school yesterday ?
3. She explained me the matter.
4. Are you really interested with English grammar ?
5. Bring me half glass water.
6. Every Sunday we go to the church.
7. He plays tennis, isn’t he ?
8. I returned back from London.
9. A bend in the road is not an end of the road.
10. I must change my cloths at once.
Answers:
1. We elected Ram President.
2. Did you go to school yesterday ?
3. She explained the matter to me.
4. Are you really interested in English Grammar ?
5. Bring me half a glass water.
6. Every Sunday we go to church.
7. He plays tennis, doesn’t he ?
8. I returned from London.
9. A bend in the road is not the end of the road.
10. I must change my clothes at once.

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Product of Vectors Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Product of Vectors Important Questions Long Answer Type

Question 1.
By the vector method, prove that altitudes of a triangle are concurrent.
Answer:
Let A, B, C be the vertices of a triangle.
Let \(\overline{\mathrm{OA}}\) = a̅, \(\overline{\mathrm{OB}}\) = b̅, \(\overline{\mathrm{OC}}\) = c̅
Let the altitudes through A, B meet at H and \(\overline{\mathrm{OH}}\) = r̅ (say)

Now \(\overline{\mathrm{AH}} \perp \overline{\mathrm{BC}} \Rightarrow \overline{\mathrm{AH}} \cdot \overline{\mathrm{BC}}=0 \Rightarrow(\overline{\mathrm{OH}}-\overline{\mathrm{OA}}) \cdot(\overline{\mathrm{OC}}-\overline{\mathrm{OB}})\) = 0
⇒ (r̅ – a̅). (c̅ – b̅) = 0
⇒ r̅ . (c̅ – b̅) – a̅. (c̅ – b̅) = 0 ……………(1)

Also \(\overline{\mathrm{BH}} \perp \overline{\mathrm{AC}} \Rightarrow \overline{\mathrm{BH}} \cdot \overline{\mathrm{AC}}=0 \Rightarrow(\overline{\mathrm{OH}}-\overline{\mathrm{OB}}) \cdot(\overline{\mathrm{OC}}-\overline{\mathrm{OA}})\) = 0
⇒ (r̅ – b̅). (c̅ – a̅) = 0
⇒ r̅ .(c – a̅) – b̅.(c̅ – a̅) = 0 ………….(2)

(1) – (2) ⇒ r̅ . (c̅ – b̅) – a̅. (c̅ – b̅) – r . (c̅ – a̅) + b̅. (c̅ – a̅) = 0

⇒ r̅ . (a̅ – b̅) – c̅ (a̅ – b̅) = 0
⇒ (r̅ – c̅) . (a̅ – b̅) = 0
⇒ \((\overline{\mathrm{OH}}-\overline{\mathrm{OC}}) \cdot(\overline{\mathrm{OA}}-\overline{\mathrm{OB}})\) = 0
⇒ \((\overline{\mathrm{OH}}-\overline{\mathrm{OC}}) \cdot(\overline{\mathrm{OA}}-\overline{\mathrm{OB}})=0 \Rightarrow \overline{\mathrm{CH}} \cdot \overline{\mathrm{BA}}=0 \Rightarrow \overline{\mathrm{CH}} \perp \overline{\mathrm{BA}}\)
∴ The altitude through C also passes through H.
∴ The altitudes of a Ale are concurrent.

Question 2.
By vector method, prove that the perpendicular bisectors of the sides of a triangle are concurrent. [Mar. ’92]
Answer:
Let A, B, C be the vertices of a triangle. Let \(\overline{\mathrm{OA}}\) = a, \(\overline{\mathrm{OB}}\) = b, \(\overline{\mathrm{OC}}\) = c

Let D, E, F are the midpoints of BC, CA, AB respectively

∴ The bisector of AB passes through ‘O’.
∴ The perpendicular bisectors of a triangle are concurrent.

Question 3.
If a̅, b̅, c̅ be three vectors. Then show that (a̅ × b̅) × c̅ = (a̅. c̅) b̅ – (b̅ . c̅) a̅.
Answer:
Let a̅ = a₁ i̅ + a₂ j̅ + a₃k̅, b̅ = b₁ i̅ + b₂ j̅ + b₃k̅ and c̅ = c₁ i̅ + c₂ j̅ + c₃k̅
= i̅ (a₂b₃ – a₃b₂) – j̅ (a₁b₃ – a₁b₃) + k̅ (a₁b₂ – a₂b₁)

(a̅ × b̅) × c̅ = \(\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
a_2 b_3-a_3 b_2 & a_3 b_1-a_1 b_3 & a_1 b_2-a_2 b_1 \\
c_1 & c_2 & c_3
\end{array}\right|\)
= i̅(a₃b₁c₃ – a₁b₃c₃ – a₁b₂c₂ + a₂b₁c₂) – j̅(a₂b₃c₃ – a₃b₂c₃ – a₁b₂c₁ + a₂b₁c₁) + k̅ (a₂b₃c₂ – a₃b₂c₂ – a₃b₁c₁ + a₁b₃c₁)
c̅.a̅ = (c₁i̅ + c₂j̅ + c₃k̅). (a₁i̅ + a₂j̅ + a₃k̅) = c₁a₁ + c₂a₂ + c₃a₃
c̅.b̅ = (c₁i̅ + c₂j̅ + c₃k̅). (b₁̅ + b₂j̅ + b₃k̅) = c₁b₁ + c₂b₂ + c₃b₃

Now (c̅.a̅)b̅ – (c̅.b̅)a̅ = (c₁a₁ + c₂a₂ + c₃a₃) (b₁ i + b₂ j + b₃k) – (c₁b₁ + c₂b₂ + c₂b₃) (a₁ i̅ + a₂j̅ + a₃k̅)
= a₁b₁c₁ i̅ + a₁b₂c₁ j̅ + a₁b₃c₁ k̅ + a₂b₁c₂ i̅ + a₂b₂c₂j̅ + a₂b₃c₂ k̅ + a₃b₁c₃ i̅ + a₃b₂c₃ i̅ + a₃b₂c₃ j̅ + a₃b₃c₃ k̅ – a₁b₁c₁i̅ – a₂b₁c₁j̅ – a₃b₂₁c₁ k̅ – a₁b₂c₂ i̅ – a₁b₂c₂ j̅ – a₃b₂c₂k̅ – a₁b₃c₃i̅ – a₂b₃c₃j̅ – a₃b₃c₃k̅
= i̅(a₂b₁c₂ + a₃b₁c₃ – a₁b₂c₂ – a₁b₃c₃) – j̅(-a₁b₂c₁ – a₃b₂c₃ + a₂b₁c₁ + a2₂b₃c₃) + k̅(a₁b₃c₁ + a₂b₃c₂ – a₃b₁c₁ – a₃b₂c₂)
∴ (a̅ × b̅) × c̅ = (a̅. c̅) b̅ – (b̅ . c̅) a̅

Question 4.
Find the equation of the plane passing through the points A = (2, 3, -1), B = (4, 5, 2) and C = (3, 6, 5).
Answer:
Let the position vectors of the points A, B and C with respect to the origin ‘O’ are
\(\overline{\mathrm{OA}}\) = 2 i̅ + 3 j̅ – k̅; \(\overline{\mathrm{OB}}\) = 4 i̅ + 5 j̅ + 2k̅; \(\overline{\mathrm{OC}}\) = 3 i̅ + 6 j̅ + 5k̅
Let r̅ = OP = x i̅ + y j̅ + zk̅ be the position vector of any point P in the plane of ΔABC.
AP = OP-OA = xi̅ + yj̅ + zk̅ – 2i̅ – 3j̅ + k̅ = (x – 2)i̅ + (y – 3)j̅ +(z + 1)k̅
AB = OB – OA = 4i̅ + 5j̅ + 2k̅ – 2i̅ – 3j̅ + k̅ = 2i̅ + 2j̅ + 3k̅
AC = OC -OA = 3i̅ + 6j̅ + 5k̅ – 2i̅ – 3j̅ + k̅ = i̅ + 3j̅ + 6k̅
The equation of the plane passing through the points A, B and C is \([\overline{\mathrm{AP}} \overline{\mathrm{AB}} \overline{\mathrm{AC}}]\) = 0
⇒ \(\left|\begin{array}{ccc}
x-2 & y-3 & z+1 \\
2 & 2 & 3 \\
1 & 3 & 6
\end{array}\right|\) = 0
⇒ (x – 2) [12 – 9] – (y – 3) [12 – 3] + (z + 1) [6 – 2] = 0
⇒ (x – 2) (3) – (y – 3) (9) + (z + 1) (4) = 0
⇒ 3x – 6 – 9y + 27 + 4z + 4 = 0
⇒ 3x – 9y + 4z + 25 = 0.

Question 5.
Find the equation of the plane passing through the point A = (3, -2, -1) and parallel to the vectors b̅ = i̅ – 2j̅ + 4k̅ and c̅ = 3i̅ + 2j̅ – 5k̅. [May ’01]
Answer:
The position vector of the point with respect to origin O’ is OA = 3 i̅ – 2 j̅ – k̅
Given b = i̅ – 2j̅ + 4k̅, c = 3i̅ + 2j̅ – 5k̅
Let r̅ = \(\overline{\mathrm{OP}}\) = xi̅ + yj̅ + zk̅ be the position vectqr of any point P in the plane of AABC.
\(\overline{\mathrm{AP}}=\overline{\mathrm{OP}}-\overline{\mathrm{OA}}\) = xi̅ +yj̅ + zk̅ – 3i̅ +2j̅ + k̅ = (x – 3)i̅ +(y + 2)j̅ +(z + 1)k̅

The equation of the plane passing through the point A parallel to the vectors b̅ & c̅ is [\(\overline{\mathrm{AP}} b̅ c̅ ] = 0
⇒ \left|\begin{array}{ccc}
x-3 & y+2 & z+1 \\
1 & -2 & 4 \\
3 & 2 & -5
\end{array}\right|\) = 0
⇒ (x – 3)[10 – 8] – (y + 2)[-5 – 12] + (z + 1)[2 + 6] = 0
⇒ (x + 3)(2) – (y + 2)(-17) + (z + 1)(8) = 0
⇒ 2x – 6 + 17y + 34 + 8z + 8 = 0
⇒ 2x + 17y + 8z + 36 = 0.

Question 6.
Find the shortest distance between the skew lines r̅ = (6i̅ + 2j̅ + 2k̅) + t (i̅ – 2j̅ + 2k̅) and r̅ = (-4i̅ – k̅) + s (3i̅ – 2j̅ – 2k̅).
Answer:
Given equations of the straight lines are r̅ = (6 i̅ + 2j̅ +2k̅) + t(i̅ – 2j̅ + 2k̅)

Comparing this equation with r̅ = a̅ + tb̅
a̅ = 6 i̅ + 2 j̅ + 2k̅, b̅ = i̅ – 2j̅ + 2k̅
r̅ = (- 4 i̅ – k̅) + s (3 i̅ – 2 j̅ – 2k̅)

Comparing this equation with r̅ = c̅ + sd̅
c̅ = -4i̅ – k̅, d̅ = 3i̅ – 2j̅ – 2k̅
a̅ – c̅ = 6i̅ + 2j̅ + 2k̅ + 4i̅ + k ̅= 10i̅ + 2j̅ + 3k̅
[a̅ – c̅ b̅d̅] = \(\left|\begin{array}{ccc}
10 & 2 & 3 \\
1 & -2 & 2 \\
3 & -2 & -2
\end{array}\right|\) = 10(4 + 4) – 2(-2-6) + 3(-2 + 6)
= 10(8) – 2(-8) + 3(4) = 80 + 16 + 12 = 108

b̅ × d̅ =
= i̅(4 + 4) – j̅(-2-6) + k̅(-2+6) = 8i̅ + 8j̅ + 4k̅

|b̅ × d̅| = \(\sqrt{8^2+8^2+4^2}\)
= \(\sqrt{64+64+16}=\sqrt{144}\)
= 12

The shortest distance between the two skew lines is
PQ = \(\frac{\mid\left[\begin{array}{ll}
\overline{\mathrm{a}}-\overline{\mathrm{c}} & \overline{\mathrm{b}} \overline{\mathrm{d}}]
\end{array} \mid\right.}{|\overline{\mathrm{b}} \times \overline{\mathrm{d}}|}=\frac{|108|}{12}=\frac{108}{12}\) = 9

Question 7.
If A = (1, – 2, – 1), B = (4,0, – 3), C = (1, 2, – 1) and D = (2, – 4, – 5), find the distance between \(\overline{\mathrm{A B}}\) and \(\overline{\mathrm{C D}}\).
Answer:
Given points are A = (1, – 2, – 1), B = (4, 0, – 3), C = (1, 2, – 1), D = (2, -4,-5)

The vector equation of the straight line passing through the points A and B is
r = (1 – t)a̅ + tb̅, t ∈ R = (1 – t) (i̅ – 2j̅ – k̅) + t(4 i̅ – 3k̅) = i̅ – 2j̅ – k̅ – ti̅ + 2t j̅ + tk̅ + 4ti̅ – 3tk̅
= (i̅ – 2j̅ – k̅) + 3ti̅ +2tj̅ – 2tk̅
r = (i̅ – 2j̅ – k̅) + t(3i̅ + 2j̅ – 2k̅)

Comparing this equation with r̅ = a̅ + tb̅
We get a̅ = i̅ – 2 j̅ – k̅, b̅ = 3 i̅ + 2 j̅ – 2k̅

The vector equation of a straight line passing through the points C and D is
r̅ = (1 – s) c̅ + sd̅, s ∈ R = (1 – s) (i̅ + 2j̅ – k̅) + s(2i̅ – 4j̅ – 5k̅)
= i̅ + 2j̅ – k̅ – si̅ — 2s j̅ + sk̅ + 2si̅ – 4s j̅ – 5sk̅
= i̅ + 2j̅ – k̅ + si̅ – 6sj̅ – 4sk̅
r = (i̅ – 2 j̅ – k̅) + s(i̅ – 6j̅ – 4k̅)

Comparing this equation with r = c̅ + sd̅, we get c̅ = i̅ + 2j̅ – k̅, d̅ = i̅ – 6j̅ – 4k̅
a̅ – c̅ = i̅ – 2j̅ – k̅ – i̅ – 2j̅ + k̅ = – 4j̅
[a̅ – c̅ b̅d̅] = \(\left|\begin{array}{rrr}
0 & -4 & 0 \\
3 & 2 & -2 \\
1 & -6 & -4
\end{array}\right|\)
= 0(-8 – 12) + 4(-12 + 2) + 0(-18 – 2) = 0 + 4(-10) + 0 = -40

b̅ × d̅ = \(\left|\begin{array}{ccc}
\overline{\mathbf{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
3 & 2 & -2 \\
1 & -6 & -4
\end{array}\right|\)
= i̅(-8 – 12) – j̅(-12 + 2) + k̅(-18 – 2) = 20i̅ + 10j̅ – 20k̅

b̅ × d̅ = \(\sqrt{(-20)^2+(10)^2+(-20)^2}\)
= \(\sqrt{400+100+400}=\sqrt{900}\)
= 30

The shortest distance between the two skew lines is PQ = \(\frac{\left|\left[\begin{array}{ll}
\bar{a}-\bar{c} & \bar{b} \bar{d}
\end{array}\right]\right|}{|\bar{b} \times \bar{d}|}=\frac{|-40|}{30}=\frac{4}{3}\)

Question 8.
If a̅ = i̅ – 2j̅ + k̅, b̅ = 2i̅ + j̅ + k̅, c̅ = i̅ + 2j̅ – k̅ find a̅ × (b̅ × c̅) and |(a̅ × b̅) × c̅|.
Answer:
Given vectors are a̅ = i̅ – 2j̅ + k̅, b̅ = 2i̅ + j̅ + k̅, c̅ = i̅ + 2j̅ – k̅

Question 9.
If a̅ = i̅ – 2j̅ – 3k̅, b̅ = 2i̅ + j̅ – k̅ and c̅ = i̅ + 3j̅ – 2k̅, verify that a̅ × (b̅ × c̅) = (a̅ × b̅) × c̅. [Mar ’11, ’98; Mar. ’08, ’04; B.P]
Answer:
Given vectors a̅ = i̅ – 2j̅ – 3k̅, b̅ = 2i̅ + j̅ – k̅, c̅ = i̅ + 3j̅ – 2k̅

Question 10.
If a̅ = 2i̅ + j̅ – 3k̅, b̅ = i̅ – 2j̅ + k̅, c̅ = – i̅ + j̅ – 4k̅ and d̅ = i̅ + j̅ + k̅, then compute |(a̅ × b̅) × (c̅ × d̅)|. [Mar. ’15(TS); Mar. ’03]
Answer:
Given vectors are a̅ = 2i̅ + j̅ – 3k̅, b̅ = i̅ – 2j̅ + k̅, c̅ = – i̅ + j̅ – 4k̅, d̅ = i̅ + j̅ + k̅

Question 11.
If a̅ = i̅ – 2j̅ + 3k̅, b̅ = 2i̅ + j̅ + k̅, c̅ = i̅ + j̅ + 2k̅, then find |(a̅ × b̅) × c̅| and |a̅ × (b̅ × c̅)|. [Mar. ’18(TS); Mar ’13]
Answer:
Given vectors are a̅ = i̅ – 2j̅ + 3k̅, b̅ = 2i̅ + j̅ + k̅, c̅ = i̅ + j̅ + 2k̅
Now (a̅ × b̅) × c̅ = (a̅. c̅)b̅ – (b̅. c̅) a̅
= [(i̅ – 2j̅ + 3k̅). (i̅ + j̅ + 2k̅)] b̅ – [(2i̅ + j̅ + k̅). (i̅ + j̅ + 2k̅)]a̅
= (1 – 2 + 6)b̅ – (2 + 1 + 2)a̅ = 5b̅ – 5a̅
= 5 [2i̅ + j̅ + k̅ – i̅ + 2j̅ – 3k̅] = 5 [i̅ + 3j̅ – 2k̅]
|(a̅ × b̅) × c̅| = 57l + 9 + 4 = 57l4

a̅ × (b̅ × c̅) = (a̅. c̅)b̅ – (a̅. b̅) c̅
= [(i̅ – 2j̅ + 3k̅). (i̅ + j̅ + 2k̅)] b̅ – [(i̅ – 2j̅ + 3k̅). (2i̅ + j̅ + k̅)]c̅
= (1 – 2 + 6) b̅ – (2 – 2 + 3) c̅ = 5b̅ – 3c̅
= 5 (2 i̅ + j̅ + k̅) — 3(i̅ + j̅ + 2k̅) = 7 i̅ + 2 j̅ – k̅
Hence |a̅ × (b̅ × c̅) | = |7i̅ + 2j̅ – k̅|
= \(\sqrt{49+4+1}=\sqrt{54}\) = 3√6

Question 12.
Let a̅ = 2i + j – 2k, b̅ = i + j. If c isavec- tor such that a̅.c̅ = |c̅|,|c̅-a̅| = 2√2 and the angle between a̅ × b̅ and c is 30°, then find the value of |(a̅ × b̅) × c̅|. [May ’15(TS)]
Answer:
Given that a̅ = 2i̅ + j̅ – 2k̅, b̅ = i̅ + j̅
⇒ |a̅| = \(\sqrt{4+1+4}\) = √9 =3, |b̅| = \(\sqrt{1+1}\) = √2
also, a̅ . c̅ = |c̅| and |c̅ – a̅| = 2√2
⇒ (c̅ – a̅)² = 8 ⇒ |c̅|² + |a̅|² – 2c̅.a̅ = 8
⇒ |c| + 9 – 2|c̅| = 8 ⇒ |c̅| — 2|c̅| + 1 = 0
⇒ (|c̅| – 1) = 0 ⇒ |c̅| = 1

Now a̅ × b̅ = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
2 & 1 & -2 \\
1 & 1 & 0
\end{array}\right|\)
= i̅(0 + 2) – j̅(0 + 2) + k̅ (2 – 1) = 2i̅ – 2j̅ + k̅
|a̅ × b̅| = ^/4 + 4 + l =3
|(a̅ × b̅) × c̅| = |a̅ × b̅| |c̅| sin (a̅ × b̅, c̅)
= |a̅ × b̅| |c̅| sin 30° = 3.1.\(\frac{1}{2}=\frac{3}{2}\)

Some More Maths 1A Product of Vectors Important Questions

Question 1.
If a̅ = 6i̅ + 2j̅ + 3k̅ and b̅ = 2i̅ – 9j̅ + 6k̅ then find a̅. b̅ and the angle between a̅ and b̅.
Answer:
Given a̅ = 6i̅ + 2j̅ + 3k̅ and b̅ = 2i̅ – 9j̅ + 6k̅
Now a̅.b̅ = (6i̅ + 2j̅ + 3k̅).(2i̅ – 9j̅ + 6k̅) = 12 – 18 + 8 = 12

Let ‘θ’ be the angle between a̅ & b̅ then cos θ = \(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}||\overline{\mathrm{b}}|}=\frac{12}{7.11}=\frac{12}{77}\)
∴ θ = cos^-1\(\left(\frac{12}{77}\right)\)

Question 2.
For what values of λ, the vectors i̅ – λj̅ + 2k̅ and 8i̅ + 6 j̅ – k̅ are at right angles ?
Answer:
Let a̅ = i̅ – λj̅ + 2k̅ and b̅ = 8i̅ + 6j̅ – k̅
Since a,b form a right angle i.e., the vectors a and b are perpendicular then a̅ b̅ =0
(i̅ – λj̅ + 2k̅). (8i̅ + 6j̅ – k̅) = 0
⇒ 8 – 6λ – 2 = 0
⇒ 6 – 6λ = 0
⇒ 6λ = 6
⇒ λ = 1.

Question 3.
If the vectors λi̅ – 3j̅ + 5k̅ and 2λi̅ – A,j̅ – k̅ are perpendicular to each other then find X.
Answer:
Let a̅ = λi̅ – 3j̅ + 5k̅ and b̅ = 2λi̅ – λj̅ – k̅
Since the vector a̅ = λi̅ – 3j̅ + 5k̅ and b̅ = 2λi̅ – λj̅ – k̅
(λi̅ – 3j̅ + 5k̅). (2λ.i̅ – λ j̅ -k̅) = 0
2λ² + 3λ – 5 = 0
2λ² + 5λ – 2λ – 5 = 0

λ(2λ + 5) – 1(2λ + 5) = 0
(2λ + 5) (λ – 1) = 0
2λ + 5 = 0, λ – 1 = 0
λ = \(\frac{-5}{2}\), λ = 1
∴ λ = \(\frac{-5}{2}\) (or) 1.

Question 4.
If a̅ = i̅ – j̅ – k̅ and b̅ = 2i̅ – 3j̅ + k̅, then find the projection vector of b on a and its magnitude.
Answer:
Given vectors are a̅ = i̅ – j̅ – k̅ and b̅ = 2 i̅ – 3 j̅ + k̅
The projection vector of b̅ on a̅

Question 5.
Find the area of the parallelogram for which the vectors a̅ = 2i̅ – 3j̅ and b̅ =3i̅ – k̅ are adjacent sides.
Answer:
Given vectors are = 2j̅ – 3j̅ and b̅ = 3i̅ – k̅
The area of the parallelogram for which the vectors and 6 are adjacent sides
= a̅ × b̅ = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
2 & -3 & 0 \\
3 & 0 & -1
\end{array}\right|\)
= i̅(3 – o) – j̅(-2 – o) + k̅(0 + 9) = 3i̅ + 2j̅ + 9k̅
∴ Area of the parallelogram = |a̅ × b̅| = \(\sqrt{(3)^2+(2)^2+(9)^2}=\sqrt{9+4+81}=\sqrt{94}\) sq.units.

Question 6.
If a̅ = 2i̅ – 3j̅ + 5k̅, b̅ = -i̅ + 4j̅ + 2k̅ then find a̅ × b̅ and unit vector perpendicular to both a̅ and b̅.
Answer:
Given a̅ = 2i̅ – 3j̅ + 5k̅, b̅ = -i̅ + 4j̅ + 2k̅
a̅ × b̅ =
= i̅(-6 – 20) – j̅(4 + 5) + k(8 – 3) = 26i̅ – 9j̅ + 5k̅
|a̅ × b̅| = \(\sqrt{(-26)^2+(-9)^2+(5)^2}=\sqrt{676+81+25}\)
= \(\sqrt{782}\)
∴ The unit vector perpendicular to both a̅ & b̅ is = \(\frac{(\overline{\mathrm{a}} \times \overline{\mathrm{b}})}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}=\pm \frac{(-26 \overline{\mathrm{i}}-9 \overline{\mathrm{j}}+5 \overline{\mathrm{k}})}{\sqrt{782}}\)

Question 7.
Find unit vector perpendicular to both i̅ + j̅ + k̅ and 2i̅ + j̅ + 3k̅.
Answer:
Let a̅ = i̅ + j̅ + k̅, b̅ = 2i̅ + j̅ + 3k̅
a̅ × b̅ = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
1 & 1 & 1 \\
2 & 1 & 3
\end{array}\right|\)
= i̅(3 – 1) – j̅(3 – 2) + k̅(1 – 2)
= 2i̅ – j̅ – k̅
|a̅ × b̅| = \(\sqrt{(2)^2+(-1)^2+(-1)^2}=\sqrt{4+1+1}\) = √6
∴ The unit vector perpendicular to both a̅ & b̅ is = \(\pm \frac{(\overline{\mathrm{a}} \times \overline{\mathrm{b}})}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}=\pm \frac{(2 \overline{\mathrm{i}}-\overline{\mathrm{j}}-\overline{\mathrm{k}})}{\sqrt{6}}\)

Question 8.
If θ is the angle between the vector i̅ + j̅ and j̅ + k̅, then find sin θ.
Answer:
Let a̅ = i̅ + j̅ , b̅ = j̅ + k̅
a̅ × b̅ = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
1 & 1 & 0 \\
0 & 1 & 1
\end{array}\right|\)
= i̅(1 – 0) – j̅(1 – 0) + k̅(1 – 0) = i̅ – j̅ + k̅

If ‘θ’ is the angle between the vector a̅ & b̅ then sin θ = \(\frac{(\overline{\mathrm{a}} \times \overline{\mathrm{b}})}{|\overline{\mathrm{a}}| \overline{\mathrm{b}} \mid}=\frac{\sqrt{3}}{\sqrt{2} \cdot \sqrt{2}}=\frac{\sqrt{3}}{2}\)

Question 9.
If a̅ = (1, – 1, -6), b̅ = (1, -3, 4) and c̅ = (2, -5, 3), then compute a̅.(b̅ × c̅).
Answer:
Given a̅ = (1, – 1, -6), b̅ = (1, -3, 4), c̅ = (2, -5, 3)
Now a̅.(b̅ × c̅) = [a̅ b̅ c̅]= 1 —3 4
= 1(-9 + 20) + 1(3 – 8) – 6(-5 + 6)
= 11 – 5 – 6 = 0

Question 10.
Find the volume of the parallelopiped whose coterminus edges are represented by the vectors 2i̅ – 3j̅ + k̅, i̅ -j̅ + 2k̅ and 2i̅ + j̅ – k̅ .
Answer:
Let a̅ = 2i̅ – 3j̅ + k̅, b̅ = i̅ -j̅ + 2k̅, c̅ = 2i̅ + j̅ – k̅
Volume of the parallelopiped
V = |[a̅ b̅ c̅]|
[a̅ b̅ c̅] = \(\left|\begin{array}{rrr}
2 & -3 & 1 \\
1 & -1 & 2 \\
2 & 1 & -1
\end{array}\right|\) = 2(1 – 2) + 3(-1-4) + 1(1 + 2)
= -2 – 15 + 3 = -14

∴ |[a̅ b̅ c̅]| = 14 cubic units

Question 11.
Find ‘t’ for which the vectors 2i̅ – 3j̅ + k̅, i̅ + 2j̅ – 3k̅ and j̅ – tk̅ are coplanar.
Answer:
Let a̅ = 2i̅ – 3j̅ + k̅, b̅ = i̅ + 2j̅ – 3k̅, c̅ = j̅ – tk̅
Since the vectors are coplanar then [a̅ b̅ c̅] = 0
⇒ \(\) = 0
⇒ 2(-2t + 3) + 3(-t – 0) + 1(1 – 0) = 0
⇒ -4t + 6 – 3t + 1 = 0
⇒ -7t + 7 = 0
⇒ t = 1

Question 12.
If the vectors a̅ = 2i̅ – j̅ + k̅, b̅ = i̅ + 2j̅ – 3k̅ and c̅ = 3i̅ + pj̅ + 5k̅ are coplanar then find p.
Answer:
Given a̅ = 2i̅ – j̅ + k̅, b̅ = i̅ + 2j̅ – 3k̅, c̅ = 3i̅ + pj̅ + 5k̅
Since the vectors are coplanar then [a̅ b̅ c̅] = 0
⇒ \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & -3 \\
3 & \mathrm{p} & 5
\end{array}\right|\) = 0
= 2(10 + 3p) + 1(5 + 9) + 1(p – 6) = 0
⇒ 20 + 6p + 14 + p – 6 = 0
⇒ 7p + 28 = 0
⇒ p = -4

Question 13.
In ΔABC if \(\overline{\mathrm{B C}}\) = a̅, \(\overline{\mathrm{C A}}\) = b̅ and \(\overline{\mathrm{A B}}\) = c̅ then show that a̅ × b̅ = b̅ × c̅ = c̅ × a̅.
Answer:
a̅ + b̅ + c̅ = \(\overline{\mathrm{BC}}+\overline{\mathrm{CA}}+\overline{\mathrm{AB}}=\overline{\mathrm{BA}}+\overline{\mathrm{AB}}=\overline{0}\) (by vector addition)
∴ a̅ + b̅ = -c̅

⇒ a̅ × b̅ = c̅ × a̅ ………….(1) (∵ a̅ × a̅ = 0)
Again a̅ + b̅ = -c̅
(a̅ + b̅) × b̅ = -c̅ × b̅ ⇒ a̅ × b̅ + b̅ × b̅ = b̅ × c̅
⇒ a̅ × b̅ = b̅ × c̅ ……………..(2) (∵ b̅ × b̅ = 0)
∴ From (1) and (2); a̅ × b̅ = b̅ × c̅ = c̅ × a̅

Question 14.
Find a unit vector perpendicular to the plane determined by the points P(1, – 1, 2), Q(2, 0, – 1) and R (0, 2, 1).
Answer:
Let O be the origin.
\(\overline{\mathrm{OP}}\) = i̅ – j̅ + 2k̅, \(\overline{\mathrm{OQ}}\) = 2i̅ – k̅, \(\overline{\mathrm{OR}}\) = 2j̅ + k̅
\(\overline{\mathrm{PQ}}=\overline{\mathrm{OQ}}-\overline{\mathrm{OP}}\) = (2i̅ – k̅)-(i̅ – j̅ + 2k̅) = i̅ + j̅ – 3k̅
\(\overline{\mathrm{PR}}=\overline{\mathrm{OR}}-\overline{\mathrm{OP}}\) = (2j̅ + k̅) – (i̅ – j̅ + 2k̅) =-i̅ + 3j̅ – k̅

Now \(\overline{\mathrm{PQ}} \times \overline{\mathrm{PR}}\) = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
1 & 1 & -3 \\
-1 & 3 & -1
\end{array}\right|\)
= i̅(-1 + 9) – j̅(-1 – 3) + k̅(3 + 1)
= 8i̅ + 4j̅ + 4k̅
= 4(2i̅ + j̅ + k̅)

\(|\overline{\mathrm{PQ}} \times \overline{\mathrm{PR}}|=4 \sqrt{4+1+1}\) = 4√6
∴ Unit vector perpendicular to the plane determined by the points P, Q and R is
= \(\pm \frac{(\overline{\mathrm{PQ}} \times \overline{\mathrm{PR}})}{|\overline{\mathrm{PQ}} \times \overline{\mathrm{PR}}|}=\pm \frac{4(2 \overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}})}{4 \sqrt{6}}=\pm\left(\frac{2 \overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}}{\sqrt{6}}\right)\)

Question 15.
If a̅, b̅ and c̅ are non coplanar vectors, then prove that the four points with position vectors 2a̅ + 3b̅ – c̅, a̅ – 2b̅ + 3c̅, 3a̅ + 4b̅ – 2c̅ and a̅ – 6b̅ + 6c̅ are coplanar.
Answer:
Suppose A, B, C, D are the given points with respect to a fixed origin ‘O’ and given that
\(\overline{\mathrm{OA}}\) = 2a̅ + 3b̅ – c̅,
\(\overline{\mathrm{OB}}\) = a̅ – 2b̅ + 3c̅ ,
\(\overline{\mathrm{OC}}\) = 3a̅ + 4b̅ – 2c̅ and
\(\overline{\mathrm{OD}}\) = a̅ – 6b̅ + 6c̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = (a̅ – 2b̅ + 3c̅) – (2a̅ + 3b̅ – c̅) = – a̅ – 5b̅ + 4c̅
\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = (3a̅ + 4b̅ – 2c̅) – (2a̅ + 3b̅ – c̅) = a̅ + b̅ – c̅
\(\overline{\mathrm{AD}}=\overline{\mathrm{OD}}-\overline{\mathrm{OA}}\) =(a̅ – 6b̅ + 6c̅) – (2a̅ + 3b̅ – c̅) = -a̅ – 9b̅ + 7c̅
∴ \(\left[\begin{array}{lll}
\overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}}
\end{array}\right]=\left|\begin{array}{rrr}
-1 & -5 & 4 \\
1 & 1 & -1 \\
-1 & -9 & 7
\end{array}\right|\) [a̅ b̅ c̅]
=[-1(7 – 9)+ 5(7 – 1)+ 4 (-9 + 1)] [a̅ b̅ c̅]
= [-1(- 2) + 5 (6) + 4 (- 8)] [a b c]
= (2 + 30 – 32) [a̅ b̅ c̅] = 0
Hence the given points A, B, C, D are coplanar.

Question 16.
If a, b, c are three vectors, then show that a̅ × (b̅ × c̅) = (a̅ . c̅) b̅ – (b̅. c̅) a̅.
Answer:
Case – (i) : If a and b are parallel, then the result is true.
Case – (ii) : If a and b are not parallel, then suppose \(\overline{\mathrm{OA}}\) = a̅ and \(\overline{\mathrm{OB}}\) = b̅ with respect to an origin O. Since a and b are not parallel we have O, A, B are non collinear and hence determine a plane.
Let i̅ be a unit vector along OA and j̅ be a unit vector perpendicular to i̅ in XY plane, k̅ is a unit vector perpendicular to XZ plane such that i̅, j̅, k̅ form a right handed system and k̅ = i̅ × j̅

∴ a̅ = a₁i̅, b̅ = b₁i̅ + b₂j̅, c̅ = c₁i̅ + c₂j̅ + c₃k̅ for a₁, b₁, c₁, c₂, c₃, b₂ being scalars.
Since a̅ and b̅ are non parallel,
a̅ × b̅ = a₁i̅ × (b₁j̅ i̅ + b₂j̅) = a₁b₂k̅
∴ (a̅ × b̅) × c̅ = (a₁b₂k) × (c_1i̅  + c₂j + c₃k) = a₁b₂c₁j̅ – a₁b₂c₂i̅ ……………….(1)
(a̅.c̅)b̅ —(b̅.c̅)a̅ = a₁c₁ (b₁i̅ + b₂j̅) – (b₁c₁ + b₂c₂) a₁i̅ = a₁c₁b₂j̅ – a₁b₂c₂i̅ ……………. (2)
From (1) and (2),
∴ (a̅ × b̅) × c̅ = (a̅.c̅)b̅ – (b̅.c̅)a̅

Question 17.
Find the cartesian equation of the plane through the point A (2, – 1, – 4) and parallel to the plane 4x – 12y – 3z – 7 = 0.
Answer:
The equation of the parallel plane is 4x – 12y – 3z = p
It passes through A (2, – 1, – 4) then 4 (2) – 12 (- 1) – 3 (- 4) = p
⇒ 8 + 12 + 12 = p
⇒ p = 32
∴ The equation of the required plane is 4x – 12y – 3z = 32

Question 18.
Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4.
Answer:
Equation of the plane is 2x – 3y – 6z = 5 ⇒ r̅ . (2 i̅ – 3 j̅ – 6k̅) = 5 …………….(1)
Similarly 6x + 2y- 9z = 4 ⇒ r̅ . (6i̅ + 2 j̅ – 9k̅) = 4 ………………(2)
If the angle between the planes r̅. n̅₁ = d₁ and r̅.n̅₂ = d₂ is θ then
cos θ = \(=\frac{\overline{\mathrm{n}}_1 \cdot \overline{\mathrm{n}}_2}{\left|\overline{\mathrm{n}}_1\right|\left|\overline{\mathrm{n}}_2\right|}\)
= \(\frac{(2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}-6 \overline{\mathrm{k}}) \cdot(6 \overline{\mathrm{i}}+2 \overline{\mathrm{j}}-9 \overline{\mathrm{k}})}{\sqrt{4+9+36} \sqrt{36+4+81}}=\frac{12-6+54}{7(11)}\)
= \(\frac{60}{77}\)
∴ θ = cos^-1\(\left(\frac{60}{77}\right)\)

Question 19.
a̅ = 2i̅ – j̅ + k̅, b̅ = i̅ – 3 j̅ – 5k̅. Find the vector c̅ such that a̅, b̅ and c̅ form the sides of a triangle.
Answer:
a̅ = 2i̅ – j̅ + k̅, b̅ = i̅ – 3j̅ – 5k̅

∵ a̅, b̅, c̅ form the sides of a triangle a̅ + b̅ + c̅ = 0 (∵ \(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CA}}\) = 0)
∴ c̅ = -a̅ – b̅ = -(2i̅ – j̅ + k̅) – (i̅ – 3j̅ – 5k̅) = -3i̅ + 4j̅ + 4k̅

Question 20.
Find the angle between the planes r̅ (2i̅ – j̅ + 2k̅) = 3 and r̅ . (3i̅ + 6j̅ + k̅) = 4. [Mar. ’15(TS)]
Answer:
If the angle between planes r̅.n̅₁ = p₁ and r̅.n̅₂ = p₂ is θ
then cos θ = \(\frac{\bar{n}_1 \cdot \bar{n}_2}{\left|\bar{n}_1\right| \cdot\left|\bar{n}_2\right|}=\frac{(2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+2 \overline{\mathrm{k}}) \cdot(3 \overline{\mathrm{i}}+6 \overline{\mathrm{j}}+\overline{\mathrm{k}})}{\sqrt{4+1+4} \sqrt{9+36+1}}\)
= \(\frac{6-6+2}{3 \sqrt{46}}=\frac{2}{3 \sqrt{46}}\)
∴ θ = cos^-1\(\left(\frac{2}{3 \sqrt{46}}\right)\)

Question 21.
If a̅ + b̅ + c̅ = 0, |a̅| = 3, |b̅| =5 and |c̅| = 7, then find the angle between a̅ and b̅.
Answer:
Given a̅ + b̅ + c̅ = 0 ⇒ c̅ =-(a̅ + b̅)
⇒ |c̅|² = (a̅ + b̅) = a̅² + b̅² + 2(a̅ b̅)
⇒ 49 = 9 + 25 + 2(a̅ b̅)
⇒ 2(a̅.b̅) = 49 – 34 = 15
⇒ a̅ . b̅ = \(\frac{15}{2}\)

If θ is the angle between a and b then cos θ = \(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}||\overline{\mathrm{b}}|}=\frac{\frac{15}{2}}{3(5)}\)
= \(\frac{15}{2} \cdot \frac{1}{15}=\frac{1}{2}\)
∴ θ = cos^-1\(\left(\frac{1}{2}\right)=\frac{\pi}{3}\)
∴ Angle between a̅ and b̅ is \(\frac{\pi}{3}\)

Question 22.
If |a̅| = 2, |b̅| =3 and |c̅| = 4 and each of a̅, b̅, c̅ is perpendicular to the sum of the other two vectors, then find the magnitude of a̅ + b̅ + c̅.
Answer:
Given |a̅| = 2, |b̅| = 3 and |c̅| = 4
Since each of a̅, b̅, c̅ is perpendicular to the sum of other two vectors i.e., a̅ is perpendicular to b̅ + c̅.
.\a̅ – (b̅ + c̅) = 0
⇒ a̅ – b̅ + a̅ – c̅= 0
Similarly b̅ (c̅ + a̅) = 0
⇒ b̅ c̅ + b̅ a̅ = 0 and c̅.(a̅ + b̅) = 0
⇒ c̅.a̅ + c̅.b̅ = 0
Adding we get 2 [(a̅. b̅) + (b̅. c̅) + (c̅ . a̅)] = 0 ……………..(1)
Also (a̅ + b̅ + c̅) =|a̅|2 + |b̅|2 + |c̅|2 + 2(a̅b̅ + b̅c̅ + c̅a̅)
=4 + 9 + 16 + 2(a̅b̅ + b̅c̅ + c̅a̅)
= 4 + 9 + 16 + 2 (0) = 29
∴ |a̅ + b̅ + c̅| = \(\sqrt{29}\)

Question 23.
For any two vectors a̅ and b̅ show that |a̅ × b̅|2 = \(\overline{\mathrm{a}^2} \overline{\mathrm{b}^2}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2\)
Answer:
We know that |a̅ × b̅| = |a̅| |b̅| sin θ, where ‘θ’ is the angle between a̅ and b̅.
Squaring on both sides |a̅ × b̅| = |a̅|² |b̅|² sin² θ = |a̅| |b̅| (1 – cos² θ)
= |a̅| |b̅̅| – |a̅|² |b̅|² cos² θ
= \(\overline{\mathrm{a}^2} \overline{\mathrm{b}^2}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2\)

Question 24.
Find the area of the triangle formed by the two sides a̅ = i̅ + 2j̅ + 3k̅ and b̅ = 3i̅ + 5j̅ – k̅.
Answer:
Given a̅ = i̅ + 2 j̅ + 3k̅ and b̅ = 3 i̅ + 5 j̅ – k̅
Then a̅ × b̅ = \(\left|\begin{array}{rrr}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
1 & 2 & 3 \\
3 & 5 & -1
\end{array}\right|\)
= i̅(-2 – 15) – j̅(-1 – 9) + k̅(5 – 6)
= -17i̅ + 10j̅ – k̅
Area of the triangle = \(\frac{1}{2}\)|a̅ × b̅| = \(\frac{1}{2} \sqrt{289+100+1}=\frac{1}{2} \sqrt{390}\)

Question 25.
For any vector g show that |a̅ × i̅|² + |a̅ × j̅|² +|a̅ × k̅|² = 2|a̅|².
Answer:
Let a̅ = xi̅ + yj̅ + zk̅, then a̅ × i̅ = -yk̅ + zj̅
|a̅ × i̅|² = y² + z²
Similarly |a̅ × j̅|² = z² + x² and |a̅ × k̅| = x²+ y²
∴ |a̅ × i̅|² + |a̅ × j̅|² + |a̅ × k̅|² = 2(x² + y² + z²) = 2|a̅|²

Question 26.
Find the area of the triangle having 3i̅ + 4j̅ and – 5i̅ + 7j̅ as two of its sides.
Answer:
Area of the triangle = \(\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|\)
\(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}\) = (3i̅ + 4 j̅) × (- 5i̅ + 7j̅) =-15(i̅ × i̅) – 20(j̅ × j̅) + 21(i̅ × j̅) + 28(j̅ × j̅)
= 20k̅ + 21k̅ = 41k̅
∴ Area of the ΔABC = \(\frac{1}{2}\)(41) = 20.5 sq. units

Question 27.
If a̅ = 2i̅ + j̅ – k̅, b̅ = -i̅ + 2j̅ – 4k̅ and c̅ = i̅ + j̅ + k̅, then find (a̅ × b̅) . (b̅ × c̅).
Answer:
Given a̅ = 2i̅ + j̅ – k̅, b̅ = -i̅ + 2j̅ – 4k̅ and c̅ = i̅ + j̅ + k̅
Then a̅ × b̅ = \(\left|\begin{array}{rrr}
\bar{i} & \bar{j} & \bar{k} \\
2 & 1 & -1 \\
-1 & 2 & -4
\end{array}\right|\)
= i̅(-4 + 2) – j̅(-8 – 1) + k̅(4 + 1) = -2i̅ + 9j̅ + 5k̅

b̅ × c̅ = \(\left|\begin{array}{rrr}
\overline{\mathrm{j}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
-1 & 2 & -4 \\
1 & 1 & 1
\end{array}\right|\)
= i̅(2 + 4) – j̅(-1 + 4) + k̅(-1 – 2) = 6i̅ – 3j̅ – 3k̅
(a̅ × b̅).(b̅ × c̅) = (-2i̅ + 9j̅ + 5k̅).(6i̅ – 3j̅ – 3k̅) = -12 – 27 – 15 = -54

Question 28.
Find the vector area and the area of the parallelogram having a̅ = i̅ + 2j̅ – k̅ and b̅ = 2i̅ – j̅ + 2k̅ as adjacent sides.
Answer:
Given a̅ = i̅ + 2 j̅ – k̅ and b̅ = 2 i̅ – j̅ + 2k̅
Then the vector area of parallelogram = a̅ × b̅
∴ (a̅ × b̅) = \(\left|\begin{array}{rrr}
\bar{i} & \bar{j} & \bar{k} \\
1 & 2 & -1 \\
2 & -1 & 2
\end{array}\right|\)
= i̅(4 – 1) – j̅(2 + 2) + k̅(-1 – 4) = 3i̅ – 4j̅ – 5k̅
Magnitude of the area = \(\sqrt{9+16+25}=\sqrt{50}\) = 5√2 sq. units

Question 29.
If a̅ × b̅ = b̅ × c̅ ≠ 0, then show that a̅ + c̅ = pb̅, where p is some scalar.
Answer:
Consider (a̅ + c̅ – pb̅) × b̅ = (a̅ × b̅) + (c̅ × b̅) – p(b̅ × b̅) = (b̅ × c̅) – (b̅ × c̅) – p(0) = 0
∴ a̅ + c̅ – pb̅ = 0
⇒ a̅ + c̅ = pb̅

Question 30.
Let a and b be vectors, satisfying |a̅|=|b̅| = 5 and (a̅, b̅) = 45°. Find the area of the triangle having a̅ – 2b̅ and 3a̅ + 2b̅ as two of its sides. [May ’15(AP); Mar. ’07] [Mar ’18(AP)]
Answer:
Area of the triangle = \(\frac{1}{2}\)|(a̅ – 2b̅) × (3a̅ + 2b̅)| ……………(1)
Now |(a̅ – 2b̅) × (3a̅ + 2b̅)| = |3(a̅ × a̅) – 2(b̅ × a̅) – 6(b̅ × a̅) – 4(b̅ × b̅)|
= |2(a̅ × b̅) + 6(a̅ × b̅)| = |8(a̅ × b̅)| = 8|a||b| sin 45°
= 8(5)(5)\(\frac{1}{\sqrt{2}}\)
= 100√2

Question 31.
Let a̅ and b̅ be vectors, satisfying |a̅| = |b̅| = 5 and (a̅, b̅) = 45°. Find the area of the triangle having a̅ – 2b̅ and 3a̅ + 2b̅ as two of Its sides. [May ’15(AP); Mar. ’07] [Mar ’18(AP)]
Answer:
Area of the triangle = \(\frac{1}{2}\)|(a̅ – 2b̅) × (3a̅ + 2b̅)| ………….(1)
Now |(a̅ – 2b̅) × (3a̅ + 2b̅)| = |3(a̅ × a̅) – 2(b̅ × a̅) – 6(b̅ × a̅) – 4(b̅ × b̅)|
= |2(a̅ × b̅) + 6(a̅ × b̅)| = |8(a̅ × b̅)| = 8|a̅||b̅| sin 45
= 8(5)(5)\(\frac{1}{\sqrt{2}}\)
= 100√2

From (1), Area of the triangle = \(\frac{1}{2}\)(100)√2 = 50√2 sq.units

Question 32.
Find the volume of the tetrahedron having the edges i̅ + j̅ + k̅; i̅ – j̅ and i̅ + 2j̅ + k̅.
Answer:
Denoting the edges by a̅ ,b̅, c̅ of tetrahedron, then its volume is
= \(\frac{1}{6}\)[a̅ b̅ c̅] = \(\frac{1}{6}\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 0 \\
1 & 2 & 1
\end{array}\right|\) = \(\frac{1}{6}\) [1 (- 1) – 1 (1) + 1 (2 + 1)]
= \(\frac{1}{6}\) [-1 – 1 + 3] = \(\frac{1}{6}\) cubic units.

Question 33.
If a̅, b̅, c̅ are mutually perpendicular unit vectors, then find the value of [a̅ b̅ c̅]².
Answer:
Given a̅, b̅, c̅ are mutually perpendicular unit vectors.
We have |a̅| = |b̅| = |c̅| = 1 and taking a̅ = i̅, b̅ = j̅, c̅ = k̅
We have [a̅ b̅ c̅] = [i̅ j̅ k̅] = i̅ . (j̅ × k̅) = i̅ . i̅ = 1
∴ [a̅ b̅ c̅]² =1

Question 34.
Show that (a̅ + b̅) – (b̅ + c̅) × (c̅ + a̅) = 2 [a̅ b̅ c̅]
Answer:
(a̅ + b̅) . (b̅ + c̅) × (c̅ + a̅) = \(\left|\begin{array}{lll}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1
\end{array}\right|\) [a̅ b̅ c̅]
= [1(1) – 1(-1)][a̅ b̅ c̅]
= 2[a̅ b̅ c̅]

Question 35.
Prove that a̅ × [a̅ × (a̅ × b̅)] = (a̅ . a̅) (b̅ × a̅)
Answer:
L.H.S = a̅ × [a̅ × (a̅ × b̅)] = a̅ × [(a̅ . b̅)a̅ – (a̅. a̅)b̅]
= (a̅ . b̅) (a̅ × a̅) – (a̅ . a̅) (a̅ × b̅)
= (a̅ . b̅) (0) – (a̅ a̅) (a̅ × b̅)
= (a̅.a̅)(b̅ × a̅) = R.H.S.

Question 36.
If a, b, c and d are coplanar vectors, then show that (a̅ × b̅) × (c̅ × d̅) = 0.
Answer:
Given a,b,c,d are coplanar vectors ⇒ a̅ × b̅ is perpendicular to the plane S.
In the similar way c̅ × d̅ is perpendicular to the plane S.
a̅ × b̅ and c̅ × d̅ are parallel vectors.
⇒ (a̅ × b̅) × (c̅ × d̅) = 0 (or) (a̅ × b̅) × (c̅ × d̅) = [a̅ c̅ d̅]b̅ – [b̅ c̅ d̅]a̅
= 0b̅ – 0a̅ = 0 (∵ a̅, b̅, c̅, d̅ are coplanar)

Question 37.
If a, b, c are non coplanar, then show that the vectors a̅ – b̅, b̅ + c̅, c̅ + a̅ are coplanar.
Answer:
Given that a̅, b̅, c̅ are non coplanar we have [a̅ b̅ c̅] ≠ 0
[a̅ – b̅ b̅ + c̅ c̅ + a̅] = [a̅ b̅ c̅] = [1 + 1(-1)][a̅ b̅ c̅] = 0
[a̅ b̅ c̅] = 0
∴ Vectors a̅ – b̅, b̅ + c̅, c̅ + a̅ are coplanar.

Question 38.
Let a̅, b̅ and c̅ be non coplanar vectors. If [2a̅ – b̅ + 3c̅, a̅ + b̅ – 2c̅, a̅ + b̅ – 3c̅] = λ [a̅ b̅ c̅] then find the value of λ.
Answer:
Given a̅, b̅, c̅ are non coplanar vectors
We have [a̅ b̅ c̅] ≠ 0
1 1 -2 [a̅ b̅ c̅] = [2 (- 3 + 2) + 1 (- 3 + 2) + 3 (1 – 1)] [a̅ b̅ c̅]
= [-2 – 1] [a̅ b̅ c̅] = -3[a̅ b̅ c̅]
Given [2a̅ – b̅ + 3c̅, a̅ + b̅ – 2c̅, a̅ + b̅ – 3c̅] = λ [a̅ b̅ c̅]
We have – 3 [a̅ b̅ c̅] = λ [a̅ b̅ c̅]
⇒ λ = -3

Question 39.
If a̅ = (1, – 1, – 6), b̅ = (1, – 3, 4) and c̅ = (2, – 5, 3), then compute a̅ × (b̅ × c̅)
Answer:
Given a̅ = (1, – 1, – 6), b̅ = (1, – 3, 4), c̅ = (2, – 5, 3)
Now b̅ × c̅ = \(\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
1 & -3 & 4 \\
2 & -5 & 3
\end{array}\right|\)
= i̅(-9 + 20) – j̅(3 – 8) + k̅(-5 + 6)
= i̅(11) – 7(-5) + k(1) = 11i̅ + 5j̅ + k̅

a̅ × (b̅ × c̅) = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
1 & -1 & -6 \\
11 & 5 & 1
\end{array}\right|\) = i̅(-1 + 30) – j̅ (1 + 66) + k̅(5 + 11)
= i̅(29) – j̅(67) + k̅(16)
= 29i̅ – 67j̅ + 16k̅

Question 40.
If |a̅| = 11, |b̅| = 23 and |a̅ – b̅| = 30, then find the angle between the vectors a̅, b̅ and also find |a̅ + b̅|.
Answer:
|a̅ – b̅| = 30 ⇒|a̅ – b̅|² = 900
⇒ |a̅|² + |b̅|² – 2(a̅.b̅) = 900
⇒ 121 + 529 – 2(a̅. b̅) = 900
⇒ 650 -2(a̅ . b̅) = 900
⇒ -2(a-b) = 250
⇒ -2 |a̅| |b̅| cos θ = 250
⇒ (-2)(253)cos θ = 25O
⇒ cos θ = \(\frac{125}{253}\)
⇒ θ = π – cos^-1\(\left(\frac{125}{253}\right)\)

Also |a̅ + b̅|² = |a̅|² + |b̅|² + 2(a̅.b̅) = 112 + 232 + 2|a̅||b̅|cos θ
= 121 + 529 + 2(11)(23)\(\left(\frac{-125}{253}\right)\) = 400
∴ |a̅ + b̅| = 20

Question 41.
Find λ in order that the four points A(3, 2, λ), B(4, λ, 5), C(4, 2, -2) and D(6, 5, – λ) be coplanar.
Answer:
\(\overline{\mathrm{OA}}\) = 3i̅ + 2j̅ + k̅; \(\overline{\mathrm{OB}}\) = 4i̅ + λj̅ + 5k̅; \(\overline{\mathrm{OC}}\) = 4i̅ + 2j̅ – 2k̅ and \(\overline{\mathrm{OD}}\) = 6i̅ + 5j̅ – k̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = i̅ + (A – 2) j̅ + 4k̅;
\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = i̅ – 3k̅;
\(\overline{\mathrm{AD}}=\overline{\mathrm{OD}}-\overline{\mathrm{OA}}\) = 3i̅ + 3j̅ – 2k̅

Given A, B, C, D are coplanar
⇒ \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}\) are coplanar
⇒ \(\left[\begin{array}{lll}
\overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}}
\end{array}\right]\) = 0
⇒ \(\left|\begin{array}{ccc}
1 & \lambda-2 & 4 \\
1 & 0 & -3 \\
3 & 3 & -2
\end{array}\right|\) = 0
⇒ 1 (9) – (λ – 2) (- 2 + 9) + 4(3) = 0
⇒ 21 – (λ – 2) (7) = 0
⇒ λ – 2 = 3
⇒ λ = 5

Question 42.
For any two vectors a̅ and b̅, show that (1 + |a̅|)² (1 + |b̅|)² = |1 – a̅. b̅|² + |a̅ + b̅ + a̅ × b̅|²
Answer:
Let |a̅| = a̅, |b̅| = b̅, (a̅, b̅) = θ.
LHS = (1 + a²) (1 + b²)
= 1 + a² + b² + a² b² …………..(1)
a̅. b̅ = |a̅||b̅| cos 9 = ab cos θ
|a̅ × b̅| = |a̅| |b̅| sin θ = ab sin θ
RHS = (1 – a . b)² + (a̅ + b̅ + a̅ × b̅)²
= 1 + (a̅ b̅)² – 2a̅ b̅ + a̅² + b̅² + |a̅ × b̅| + 2 a̅ . b̅ + 2(b̅ . a̅ × b̅) + 2(a̅ × b̅ . a̅)
= 1 + a̅²b̅² cos2 θ + a̅² + b̅² + a̅²b̅² sin2 θ + 0 + 0
= 1 + a̅²b̅² (cos2 θ + sin2 θ) + a̅² + b̅²
= 1 + a̅²b̅² (1) + a̅² +b̅²
= 1 + a̅²b̅² + a̅² + b̅²
= 1 + a̅² + b̅² + a̅²b̅²
= RHS.

Question 43.
If a̅ = 2i̅ – 3j̅ + k̅ and b̅ = i̅ + 4j̅ – 2k̅, then find (a̅ + b̅) × (a̅ – b̅)
Answer:
Given a̅ = 2i̅ – 3j̅ + k̅
b̅ = i̅ + 4j̅ – 2k̅
Now
a̅ + b̅ = 2i̅ – 3j̅ + k̅ + i̅ + 4j̅ – 2k̅ = 3i̅ + j̅ – k̅
a̅ – b̅ = 2i̅ – 3j̅ + k̅ – i̅ – 4j̅ + 2k̅ = i̅ – 7j̅ + 3k̅
(a̅ + b̅) x (a̅ – b̅) = \(\left|\begin{array}{rrr}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
3 & 1 & -1 \\
1 & -7 & 3
\end{array}\right|\)
= i̅(3 – 7) – j̅(9 + 1) + k̅(-21 – 1)
= -4i̅ – 10j̅ – 22k̅

Question 44.
Show that the volume of a tetrahedron with a̅, b̅ and c̅ as coterminous edges is \(\frac{1}{6}\)|[a̅ b̅ c̅]|
Answer:
Let OABC be a tetrahedron and
\(\overline{\mathrm{OA}}\) = a̅, \(\overline{\mathrm{OB}}\) = b̅, \(\overline{\mathrm{OC}}\) = c̅
Let V be the volume of the tetrahedron OABC. By definition the volume V is given by V = \(\frac{1}{3}\) (area of the base ΔOAB)
(Length of the perpendicular from C to the base ΔOAB)

CN is the perpendicular from C to AOAB and CM is the perpendicular from C on to the supporting line of a̅ × b̅ so that CN = OM = Length of the projection of c onto a̅ × b̅.

Question 45.
Find the equation of the plane passing through the point (3, -2,1) and perpendicular to the vector (4, 7, -4).
Answer:
The vector equation of the plane is
[r̅ – (3 i̅ – 2 j̅ + k̅)j̅ . (4 i̅ + 7 j̅ – 4k̅) = 0
⇒ r̅ . (4i̅ + 7j̅ – 4k̅) – (12 – 14 – 4) = 0
⇒ r̅ . (4 i̅ + 7 j̅ – 4k̅) +6 = 0
The Cartesian equation is 4x + 7y – 4z + 6 = 0

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Addition of Vectors Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions Short Answer Type

Question 1.
Let ABCDEF be a regular hexagon with centre ’O’. Show that [Mar. 16 (AP), 15 (AP), 03]
\(\overline{\mathrm{AB}}+\overline{\mathrm{AC}}+\overline{\mathrm{AD}}+\overline{\mathrm{AE}}+\overline{\mathrm{AF}}=3 \overline{\mathrm{AD}}=6 \overline{\mathrm{AO}}\)
Answer:
ABCDEF is a regular hexagon with centre ‘O’.

Question 2.
In the two dimensional plane, prove by using vector methods, the equation of the line whose intercepts on the axes are ‘a’ and ‘b’ is \(\frac{x}{a}+\frac{y}{b}\) = 1. [May 05, 01]
Answer:

Let A = (a, 0) and B = (0, b)
∴ A = aī, B = bj̄
The equation of the straight line AB is r̄ = (1 – t) ā + tb̄, t ∈ R
r̄ = (1 – t)aī + tbj̄
Let r̄ = xī + yj̄
⇒ xī + yj̄ = (1 – t)aī + tbj̄
Equating the corresponding coefficients of ī, j̄ we have
(1 -1) a = x
\(\frac{x}{a}\) = 1 – t ⇒ \(\frac{x}{a}\) + t = 1 …………… (1)
tb = y ⇒ t = \(\frac{y}{b}\)
From (a) ⇒ \(\frac{x}{a}+\frac{y}{b}\) = 1
∴ The equation of the lines is \(\frac{x}{a}+\frac{y}{b}\) = 1.

Question 3.
Find the point of intersection of the line r̄ = 2ā + b̄ + t(b̄ – c̄) and the plane r̄ = ā + x(b̄ + c̄) + y(ā + 2b̄ – c̄), where ā, b̄ and c̄ are non-coplanar vectors. [May 13]
Answer:
Given r̄ = 2ā + b̄ + t(b̄ – c̄), r̄ = ā + x(b̄ + c̄) + y(ā + 2b̄ – c̄)
Let r̄ be the position vector of the point P the intersection of the line and the plane.
∴ 2ā + b̄ + t(b̄ – c̄) = ā + x(b̄ + c̄) + y(ā + 2b̄ – c̄) (∵ ā, b̄, c̄ are non-coplanar)
Equating the coefficients of ā, b̄, c̄
2 = y + 1 ⇒ y = 1 ⇒ 1 + t = x + 2y ⇒ 1 + t = x + 2 ⇒ t – x = 1 ……………… (1)
– t = x – y ⇒ x + t = y ⇒ x + t = 1 …………….. (2)
Solving (1) and (2) equations t = 1, x = 0.
∴ Point of intersection is r̄ = 2ā + b̄ + b̄ – c̄ = 2ā + 2b̄ – c̄
Hence the position vector of the point of intersection is 2ā + 2b̄ – c̄.

Question 4.
If ā, b̄, c̄ are non coplanar, find the point of intersection of the line passing through the points 2ā + 3b̄ – c̄, 3ā + 4b̄ – 2c̄ with the line joining the points ā – 2b̄ + 3c̄, ā – 6b̄ + 6c̄.
Answer:
Let P = 2ā + 3b̄ – c̄, Q = 3ā + 4b̄ – 2c̄, R = ā – 2b̄ + 3c̄, S = ā – 6b̄ + 6c̄
The equation of the straight line passing through P(2ā + 3b̄ – c̄), Q(3ā + 4b̄ – 2c̄) is
r̄ = (1 – t) (ā + tb̄), t ∈ R
r̄ = (1 – t) (2ā + 3b̄ – c̄) + t(3ā + 4b̄ – 2c̄)
= 2ā + 3b̄ – c̄ – 2tā – 3tab̄ + tc̄ + 3tā + 4tb̄ – 2tc̄
= 2ā + 3b̄ – c̄ + tā + tb̄ – tc̄
r̄ = (2 + t) ā + (3 + t)b̄ + (- 1 – t)c̄ ………………… (1)
The equation of the straight line passing through
R(ā – 2b̄ + 3c̄) and S(ā – 6b̄ + 6c̄) is r̄ = (1 – s)ā + sb̄, s ∈ R
r̄ =(1 – s) (ā – 2b̄ + 3c̄) + s(ā – 6b̄ + 6c̄)
= ā – 2b̄ + 3c̄ – sā + 2sb̄ – 3sc̄ + sā – 6sb̄ + 6sc̄ = ā – 2b̄ + 3c̄ – 4sb̄ + 3sc̄
r̄ = ā + (- 2 – 4s)b̄ + (3 + 3s)c̄ …………….. (2)
From (1) and (2), (2 + t)ā + (3 + t)b̄ + (-1 – t)c̄ = a + (- 2 – 4s)b̄ + (3 + 3s)c̄
Equating the corresponding coefficients of ā, b̄ and c̄ we have

2 + t = 1
t = – 1

3 + t = – 2 – 4s
3 – 1 = – 2 – 4s
2 = – 2 – 4s
4 = – 4s
s = – 1

– 1 – t = 3 + 3s
– 4 = 3s + t
– 4 = 3 (- 1) – 1
– 4 = – 3 – 1
– 4 = – 4

∴ The lines (1) and (2) are intersect each other.
∴ Substituting the value of t = – 1 in (1) (or) s = – 1 in (2)
The point of intersection of the lines is r̄ = (2 – 1)ā + (3 – 1)b̄ + (- 1 + 1)c̄ ⇒ r̄ = ā + 2b̄

DTP. Show that the line joining the pair of points 6a – 4b + 4c, – 4c and the line joining the
pair of points – a – 2b – 3c, a + 2b – 5c intersect at the point -4c when a, b, c are non- coplanar vectors.

Question 5.
Find the vector equation of the plane which passes through the points 2ī + 4j̄ + 2k̄, 2ī + 3j̄ + 5k̄ and parallel to the vector 3ī – 2j̄ + k̄. Also find the point where this plane meets the line joining the points 2ī + j̄̄ + 3k̄ and 4ī – 2j̄+ 3k̄. [Mar. 12]
Answer:
The vector equation of the plane passing through 2ī + 4j̄ + 2k̄, 2ī + 3j̄ + 5k̄ and parallel to the vector 3ī – 2j̄ + k̄ is
r̄ = (1 – t)ā + tb̄ + sc̄; t, s ∈ R
= (1 – t) [2ī + 4j̄ + 2k̄] + t [2ī + 3j̄ + 5k̄] + s(3ī – 2j̄ + k̄)
= 2ī + 4j̄ + 2k̄ – 2tī – 4tj̄ – 2tk̄ + 2tī + 3tj̄ + 5tk̄ + 3sī – 2sj̄ + sk̄
= (2 – 2t + 2t + 3s)ī + (4 – 4t + 3t – 2s)j̄ + (2 – 2t + 5t + s)k̄
r̄ = (2 + 3s)ī + (4 – t – 2s)j̄ + (2 + 3t + s)k̄; s, t ∈ R ……………… (1)
The vector equation of a line passing through the points 2ī + j̄ + 3k̄ and 4ī – 2j̄ +3k̄ is
r̄ = (1 – x)ā + xb̄, x ∈ R
r̄ = (1 – x)(2ī + j̄ + 3k̄) + x(4ī – 2j̄ + 3k̄)
= 2ī + j̄ + 3k̄ – 2xī + xj̄ + 3xk̄ + 4xī – 2xj̄ + 3xk̄
= 2ī + j̄ + 3k̄ – 2xī – 3xj̄
r̄ = (2 + 2x)ī + (1 – 3x)j̄ + 3k̄, x ∈ R …………… (2)
The point of Intersect ion of (1) & (2) Is P. Let the position vector of P Is ÕP = y from (1) & (2)
(2 + 3s)ī + (4 – t – 2s)j̄ + (2 + 3t + s)k̄ = (2 + 2x)ī + (1 – 3x)j̄ + 3k̄
Equation the corresponding coefficients if ī, j̄ and k̄
2 + 3s = 2 + 2x
2x – 3s = 0 ………… (3)
4 – t – 2s = 1 – 3x
t + 2s – 3x = 3 …………… (4)
2 + 3t + s = 3
3t + s = 1 …………… (5)
3 × (4) – (5)
3t + 6s – 9x = 9
3t + s = 1
5s – 9x = 8
9x – 5s = – 8 ……………… (6)
Solve (3) and (6)

∴ The plane and line Intersect each other.
∴ Substituting the value of s, t in (1) (or) x in (2)
∴ The point of intersection is

Question 6.
Find the vector equation of the plane passing through the points 4ī – 3j̄ – k̄ , 3ī + 7j̄ – 10k̄ and 2ī + 5j̄ – 7k̄ and show that the point ī + 2j̄ – 3k̄ lies in the plane.
Answer:
Vector equation of the plane passing through A(4ī – 3j̄ – k̄); B(3ī + 7j̄ – 10k̄) and C(2ī + 5j̄ – 7k̄) is
r̄ = (1 – s – t) (4ī – 3j̄ – k̄) + s(3ī + 7j̄ – 10k̄) + t(2ī + 5j̄ – 7k̄)
Let D(ī + 2j̄ – 3k̄) lies on the plane, then
(ī + 2j̄ – 3k̄) = (1 – s – t) (4ī – 3j̄ – k̄) + s(3ī + 7j̄ – 10k̄) + t(2ī + 5j̄ – 7k̄)
Since i, j, k are non coplanar, equating co-efficients of ī, j̄, k̄ on both sides
4(1 – s – t) + 3s + 2t = 1 ⇒ 4 – 4s – 4t + 3s + 2t = 1 ⇒ s + 2t = 3 ……………. (1)
– 3(1 – s -1) + 7s + 5t = 2 ⇒ – 3 + 3s + 3t + 7s + 5t = 2 ⇒ 10s + 8t = 5 …………… (2)
Also – (1 – s -1) – 10s – 7t = – 3 ⇒ – 1 + s + t – 10s – 7t = – 3 ⇒ 9s + 6t = 2 …………… (3)
From (1), 3s + 6t = 9
Solving (1) and (3) equations 6s = – 7 ⇒ s = – \(\frac{7}{6}\)

∴ s = \(\frac{-7}{6}\), t = \(\frac{25}{12}\) satisfy (1), (2), (3) and D lies on the plane passing through A, B, C.

Question 7.
Let ā = ī + 2j̄ + 3k̄ and b̄ = 3ī + j̄. Find the unit vector in the direction of a + b.
Answer:
Given ā = ī + 2j̄ + 3k̄, b̄ = 3ī + j̄
Now ā + b̄ = ī + 2j̄ + 3k̄ + 3ī + j̄ = 4ī + 3j̄ + 3k̄
|ā + b̄| = \(\sqrt{4^2+3^2+3^2}\) = \(\sqrt{16+9+9}\) = √34
∴ The unit vector in the direction of
ā + b̄ = \(\frac{\bar{a}+\bar{b}}{|\bar{a}+\bar{b}|}\) = \(\frac{4 \bar{i}+3 \bar{j}+3 \bar{k}}{\sqrt{34}}\) = \(\frac{1}{\sqrt{34}}\) (4ī + 3j̄ + 3k̄)

Question 8.
ā = 2ī + 5j̄ + k̄ and b̄ = 4ī + mj̄ + nk̄ are collinear vectors, then find m and n.
Answer:
Given vectors are ā = 2ī + 5j̄ + k̄, b̄ = 4ī + mj̄ + nk̄
If a₁ī + b₁j̄ + c₁k̄ and a₂ī + b₂j̄ + c₂k̄ vectors are collinear, then \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
∴ Since ā, b̄ are collinear, then \(\frac{2}{4}=\frac{5}{m}=\frac{1}{n}\)
⇒ \(\frac{1}{2}=\frac{5}{m}=\frac{1}{n}\) ⇒ \(\frac{1}{2}=\frac{5}{m}, \frac{1}{2}=\frac{1}{n}\)
∴ m = 10 and n = 2

Question 9.
Show that the points whose position vectors are ā – 2b̄ + 3c̄, 2ā + 3b̄ – 4c̄, – 7b̄ + 10c̄ are collinear.
Answer:
Let A, B, C be the given points respectively. Then

Question 10.
Find the vector equation of the plane passing through the points (0, 0, 0), (0, 5, 0) and (2, 0, 1).
Answer:
Let ā = 0̄; b̄ = 5j̄; c̄ = 2ī + k̄
The vector equation of the plane passing through the points 0̄, 5j̄ and 2ī + k̄ is
r̄ = (1 – s – t)ā + tb̄; t, s ∈ R = (1 – t – s)(0̄) + t(5j̄) + s(2ī + k̄); t, s ∈ R
r̄ = (5t)j̄ + 5(2ī + k̄); t, s ∈ R

Question 11.
Show that the line joining the pair of points 6ā – 4b̄ + 4c̄, – 4c̄ and the line joining the pair of points – ā – 2b̄ – 3c̄, ā + 2b̄ – 5c̄ intersect at the point – 4c̄ when ā, b̄, c̄ are non- coplanar vectors.
Answer:
Let P = 6ā – 4b̄ + 4c̄; Q = – 4c̄; R = – ā – 2b̄ – 3c̄; S = ā + 2b̄ – 5c̄
The equation of the straight line passing through P(6ā – 4b̄ + 4c̄) and Q(- 4c̄) is
r̄ = (1 – t)ā + tb̄, t ∈ R
r̄ = (1 – t) (6ā – 4b̄ + 4c̄) + t(- 4c̄) =6ā – 4b̄ + 4c̄ – 6tā – 4tb̄ + 4tc̄ – 4tc̄
= 6ā – 4b̄ + 4c̄ – 6tā + 4tb̄ – 8tc̄
r̄ = (6 – 6t)ā + (- 4 + 4t)b̄ + (4 – 8t)c̄ ……………….. (1)

The equation of the straight line passing through R(- ā – 2b̄ – 3c̄) and s(ā + 2b̄ – 5c̄) is
r̄ = (1 – s)ā + sb̄, s ∈ R
= (1 – s)(- ā – 2b̄ – 3c̄) + s(ā + 2b̄ – 5c̄) = – ā – 2b̄ – 3c̄ + sā + 2sb̄ + 3sc̄ + sā + 2sb̄ – 5sc̄
= – ā – 2b̄ – 3c̄ + 2sā + 4sb̄ – 2sc̄
⇒ r̄ = (- 1 + 2s)ā + (- 2 + 4s)b̄ + (- 3 – 2s)c̄ …………….. (2)
From (1) & (2)
(6 – 6t)ā + (- 4 + 4t)b̄ + (4 – 8t)c̄
= (- 1 + 2s)ā + (- 2 + 4s)b̄ + (-3 – 2s)c̄
Equating the corresponding coefficients of ā, b̄ & c̄ we have
6 – 6t = – 1 + 2s ⇒ 6t + 2s = 7 ………. (3)
– 4 + 4t = – 2 + 4s ⇒ – 4t + 4s = – 2
2t – 2s = 1 ……… (4)
4 – 8t = – 3 – 2s = 8t – 2s = 7 ……………. (5)
Solve (3) & (4)

Substitute the values of s, t in equation (5)
8(1) – 2\(\left(\frac{1}{2}\right)\) = 7 ⇒ 8 – 1 = 7 ⇒ 7 = 7
∴ The equations (1) & (2) intersect each other.
Substituting the value of t = 1 in (1)
or s = \(\frac{1}{2}\) in (2)
The point of intersection of the lines is
r̄ = (6 – 6.1)ā + (- 4 + 4.1)b̄ + (4 – 8.1)c̄
r̄ = – 4c̄

Some More Maths 1A Addition of Vectors Important Questions

Question 1.
Find a vector in the direction of vector ā = ī – 2j̄ that has magnitude 7 units.
Answer:
Given ā = ī – 2j̄
|ā| = \(\sqrt{(1)^2+(-2)^2}\) = \(\sqrt{1+4}\) = √5
The unit vector in the direction of the vector a is \(\frac{\Lambda}{\mathbf{a}}\) = \(\frac{\bar{a}}{|\bar{a}|}=\frac{\bar{i}-2 \bar{j}}{\sqrt{5}}\)
∴ The vector having magnitude equal to 7 and in the direction of ā is
\(\frac{\Lambda}{\mathbf{a}}\) = \(7\left(\frac{\overline{\mathrm{i}}-2 \overline{\mathrm{j}}}{\sqrt{5}}\right)=\frac{7}{\sqrt{5}} \overline{\mathrm{i}}-\frac{14}{\sqrt{5}} \overline{\mathrm{j}}\)

Question 2.
Consider the two points P and Q with position vectors \(\overline{\mathrm{OP}}\) = 3ā – 2b̄ and \(\overline{\mathrm{OQ}}\) = ā + b̄. Find the position vector of a point R which divides the line joining P and Q in the ratio 2 : 1 (i) internally and (ii) externally.
Answer:
(i) The position vector of the point R dividing the joining of P and Q internally in the ratio 2: 1 is
\(\overline{\mathrm{OR}}\) = \(\frac{2(\overline{\mathrm{a}}+\overline{\mathrm{b}})+(3 \overline{\mathrm{a}}-2 \overline{\mathrm{b}})}{2+1}=\frac{5 \overline{\mathrm{a}}}{3}\)

(ii) The position vector of the point R dividing the joining of P and Q externally in the ratio 2: 1 is
\(\overline{\mathrm{OR}}\) = \(\frac{2(\overline{\mathrm{a}}+\overline{\mathrm{b}})-(3 \overline{\mathrm{a}}-2 \overline{\mathrm{b}})}{2-1}\) = 4b̄ – ā

Question 3.
Show that the points A(2ī – j̄ + k̄), B(ī – 3j̄ – 5k̄), C(3ī – 4j̄ – 4k̄) are the vertices of a right angled triangle.
Answer:
Given position vectors

Clearly, AB = AC and AB² = BC² + AC²
∴ ∆ABC is right angled.

Question 4.
In ∆ABC, if ‘O’ is the circumcentre and H is the orthocentre, then show that
(i) \(\overline{\mathrm{OA}}+\overline{\mathrm{OB}}+\overline{\mathrm{OC}}=\overline{\mathrm{OH}}\)
(ii) \(\overline{\mathrm{HA}}+\overline{\mathrm{HB}}+\overline{\mathrm{HC}}=2 \overline{\mathrm{HO}}\)
Answer:
Take ‘O’ as origin the position vectors of vertices A, B and C with respect to the origin ‘O’ are
\(\overline{\mathrm{OA}}=\overline{\mathrm{a}}\), \(\overline{\mathrm{OB}}=\overline{\mathrm{b}}\) and \(\overline{\mathrm{OC}}=\overline{\mathrm{c}}\)
Let D be the midpoint of BC then the position vector of D is

Question 5.
If the points whose position vectors are 3ī – 2j̄ – k̄, 2ī + 3j̄ – 4k̄, – ī + j̄ + 2k̄ and 4ī + 5j̄ + λk̄ are coplanar, then show that λ = \(\frac{-146}{17}\). [May 15 (TS)]
Answer:
Let A, B, C, D be the given points respectively.

Since A, B, C and D are coplanar, then [\(\overline{\mathrm{AB}}\) \(\overline{\mathrm{AC}}\) \(\overline{\mathrm{AD}}\)] = 0
⇒ \(\left|\begin{array}{ccc}
-1 & 5 & -3 \\
-4 & 3 & 3 \\
1 & 7 & \lambda+1
\end{array}\right|\) = 0 ⇒ – 1(3λ + 3 – 21) – 5(- 4λ – 4 – 3) – 3(- 28 – 3) = 0
⇒ – 1 (3λ – 18) – 5(- 4λ – 7) – 3(- 31) = 0
⇒ – 3λ + 18 + 20λ + 35 + 93 = 0 -146
⇒ 17λ + 146 = 0
⇒ λ = \(\frac{-146}{17}\)

Question 6.
ABCDE is a pentagon. If the sum of the vectors \(\overline{\mathbf{A B}}, \overline{\mathbf{A E}}, \overline{\mathbf{B C}}, \overline{\mathbf{D C}}, \overline{\mathbf{E D}}\) and \(\) is λ \(\overline{\mathbf{A C}}\), then find the value of λ.
Answer:
ABCDE is a pentagon. Given that

Question 7.
Is the triangle formed by the vectors 3ī + 5j̄ + 2k̄, 2ī – 3j̄ – 5k̄ and – 5ī – 2j̄ + 3k̄ equilateral ?
Answer:
Let ABC be the triangle with

∴ The given vectors formed on equilateral triangle.

Question 8.
ā, b̄, c̄ are non-coplanar vectors. Prove that the four points – ā + 4b̄ – 3c̄, 3ā + 2b̄ – 5c̄, – 3ā + 8b̄ – 5c̄, – 3ā + 2b̄ + c̄ are coplanar.
Answer:

= [4 (16 – 4) + 2 (- 8 – 4) – 2 (4 + 8)] [ā b̄ c̄]
= [4(12) + 2 (-12) – 2 (12)] [ā b̄ c̄]
= [48 – 24 – 24] [ā b̄ c̄] = 0 [ā b̄ c̄] = 0
∴ The given points are coplanar.

Question 9.
a, b, c are non-coplanar vectors. Prove that the four points 6ā + 2b̄ – c̄, 2ā – b̄ + 3c̄, – ā + 2b̄ – 4c̄, – 12ā – b̄ – 3c̄ are coplanar.
Answer:
Given that a, b, c are non-coplanar ⇒ [ā b̄ c̄] ≠ 0

Question 10.
If ī, j̄, k̄ are unit vectors along the positive directions of the coordinate axes, then show that the four points 4ī + 5j̄ + k̄, – j̄ – k̄, 3ī + 9j̄ + 4k̄ and – 4ī + 4j̄ + 4k̄ are coplanar. [Mar. 14]
Answer:

∴ The given vectors are coplanar.

Question 11.
Show that the points whose position vectors are 2ā + 5b̄ – 4c̄ , ā + 4b̄ – 3c̄, 4ā + 7b̄ – 6c̄ are collinear when a, b, c are non-coplanar vectors.
Answer:
Let A, B, C be the given points respectively.

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Addition of Vectors Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions Very Short Answer Type

Question 1.
Find the unit vector in- the direction of the sum of the vectors ā = 2ī + 2j̄ – 5k̄ and b̄ = 2ī + j̄ + 3k̄. [May. 13]
Answer:
Given ā = 2ī + 2j̄ – 5k̄, b̄ = 2ī + j̄ + 3k̄
Now ā + b̄ = 2ī + 2j̄ – 5k̄ + 2ī + kj̄ + 3k̄ = 4ī + 3j̄ – 2k̄
|ā + b̄| = |4ī + 3j̄ – 2k̄| = \(\sqrt{(4)^2+(3)^2+(-2)^2}\) = \(\sqrt{16+9+4}\) = √29
The unit vector in the direction of ā + b̄ is \(\frac{\bar{a}+\bar{b}}{|\bar{a}+\bar{b}|}\)
= \(\frac{4 \overline{\mathrm{i}}+3 \overline{\mathrm{j}}-2 \overline{\mathrm{k}}}{\sqrt{29}}\) = \(\frac{1}{\sqrt{29}}\) (4ī + 3j̄ – 2k̄).

Let ā = ī + 2j̄ + 3k̄ and b̄ = 3ī + j̄. Find the unit vector in the direction of ā + b̄ [Mar. 16 (TS)]
Answer:
\(\frac{1}{\sqrt{34}}\)(4ī + 3j̄ + 3k̄).

Question 2.
Find the unit vector in the direction of vector ā = 2ī + 3j̄ + k̄ . [Mar. 14]
Answer:
Given ā = 2ī + 3j̄ + k̄
Now |ā| = \(\sqrt{(2)^2+(3)^2+(1)^2}\) = \(\sqrt{4+9+1}\) = √14
∴ The unit vector in the direction of ā vector

Question 3.
If the vectors – 3ī + 4j̄ + λk̄ and μī + 8j̄ + 6k̄ are collinear vectors, then find λ andμ. [Mar. 18 (AP); May 14, 12; 10; Mar. 14]
Answer:
If a₁ī + b₁j̄ + c₁k̄, a₂ī + b₂j̄ + c₂k̄ are collinear vectors, then \(\).
Let ā = – 3ī + 4j̄ + λk̄, b = μī + 8j̄ + 6k̄
Since ā, b̄ are collinear vectors, then \(\frac{-3}{\mu}=\frac{4}{8}=\frac{\lambda}{6}\) ⇒ \(\frac{-3}{\mu}=\frac{1}{2}=\frac{\lambda}{6}\) ⇒ \(\frac{-3}{\mu}=\frac{1}{2}, \frac{1}{2}=\frac{\lambda}{6}\)
μ = – 6, λ = 3
∴ λ = 3, μ = – 6

ā = 2ī + 5j̄ + k̄ and b̄ = 4ī + mj̄ + nk̄ are collinear vectors, then find m and n.
Answer:
10, 2.

Question 4.
Let ā, b̄ be non – collinear vectors. If ᾱ = (x + 4y) ā + (2x + y + 1) b̄ and β̄ = (y – 2x + 2) ā + (2x – 3y – 1) b̄ are such that 3ᾱ = 2β̄ , then find x and y.
Answer:
Given vectors are ᾱ = (x + 4y) ā + (2x + y + 1) b̄, β̄ = (y – 2x + 2) ā + (2x – 3y – 1) b̄
Given 3ᾱ = 2β̄ ⇒ 3[(x + 4y) ā + (2x + y + 1) b̄] = 2[(y – 2x + 2) ā + (2x – 3y -1) b̄]
⇒ 3(x + 4y) ā + 3 (2x + y +1) b̄ = 2(y – 2x + 2) ā + 2(2x – 3y -1) b̄
Since ā, b̄ are non-collinear then
3(x + 4y) = 2(y – 2x + 2)
3x + 12y = 2y – 4x + 4
7x + 10y – 4 = 0 ……………. (1)
3(2x + y + 1) = 2(2x – 3y – 1)
6x + 3y + 3 = 4x – 6y – 2
2x + 9y + 5 = 0 …………….. (2)
Solve (1) and (2)

Question 5.
Show that the points whose position vectors are – 2ā + 3b̄ + 5c̄, ā + 2b̄ + 3c̄, 7ā – c̄ are collinear when ā, b̄, c̄ are non-coplanar vectors. [May 05, 92; Mar. 02]
Answer:
Let A, B, C be the given points.
The position vectors of A, B, C with respect to the origin ‘O’ are

∴ A, B and C are collinear.

Question 6.
If the position vectors of the points A, B and C are – 2ī + j̄ – k̄, – 4ī + 2j̄ + 2k̄ and 6ī – 3j̄ – 13k̄ respectively and \(\overline{\mathbf{A B}}\) = λ.\(\overline{\mathbf{A C}}\), then find the value of A.
Answer:
The position vectors of the points A, B and C with respect to origin ‘O’ are

Question 7.
If \(\overrightarrow{\mathbf{O A}}\) = ī + j̄ + k̄, \(\overline{\mathrm{AB}}\) = 3ī – 2j̄ + k̄, \(\overline{\mathrm{BC}}\) = ī + 2j̄ – 2k̄ and \(\overline{\mathrm{CD}}\) = 2ī + j̄ + 3k̄, then find the vector OD. [Mar. 19 (TS); Mar. 13; May 96]
Answer:

Question 8.
Let ā = 2ī + 4j̄ – 5k̄, b̄ = ī + j̄ + k̄ and c̄ = j̄ + 2k̄. Find the unit vector in the opposite direction of ā + b̄ + c̄. [Mar.19 (TS); May 15 (AP); Mar.19, 15 (AP); Mar.12, 10, 09, 04, B.P.]
Answer:
Given vectors are ā = 2ī + 4j̄ – 5k̄, b̄ = ī + j̄ + k̄ and c̄ = j̄ + 2k̄
Now, ā + b̄ + c̄ = 2ī + 4j̄ – 5k̄ + ī + j̄ + k̄ + j̄ + 2k̄ = 3ī + 6j̄ – 2k̄
|a + b + c| ⇒ \(\sqrt{3^2+6^2+(-2)^2}\) = \(\sqrt{9+36+4}\) = √49 = 7
∴ The unit vector in the opposite direction of
ā + b̄ + c̄ = \(\frac{-(\bar{a}+\bar{b}+\bar{c})}{|\bar{a}+\bar{b}+\bar{c}|}\) = \(\frac{-(3 \bar{i}+6 \bar{j}-2 \bar{k})}{7}\)

Question 9.
Find the vector equation of the line passing through the point 2ī + 3j̄ + k̄ and parallel to the vector 4ī – 2j̄ + 3k̄. [Mar. 17 (TS), 15 (AP); May 10, 07, 01; Mar. 07; 1. 92]
Answer:
Let ā = 2ī + 3j̄ + k̄, b̄ = 4ī – 2j̄ + 3k̄
The vector equation of the line passing through the point 2ī + 3j̄ + k̄ and parallel to the vector
4ī – 2j̄ + 3k̄ is r̄ = ā + tb̄, t ∈ R = 2ī + 3j̄ + k̄ + t(4ī – 2j̄ + 3k̄), t ∈ R
= 2ī + 3j̄ + k̄ + 4tī – 2tj̄ + 3tk̄
∴ r̄ = (2 + 4t)ī + (3 – 2t)j̄ +(1 + 3t)k̄, t ∈ R

Question 10.
OABC is a parallelogram. If \(\overline{\mathbf{O A}}\) = ā and \(\overline{\mathbf{O C}}\) = c̄, find the vector equation of the side \(\overline{\mathbf{B C}}\).
Answer:

OABC is a parallelogram.
\(\overline{\mathrm{OA}}=\overline{\mathrm{a}}, \overline{\mathrm{OC}}=\overline{\mathrm{c}}=\overline{\mathrm{AB}}\)
\(\overline{\mathrm{OB}}=\overline{\mathrm{OA}}+\overline{\mathrm{AB}}\)
\(\overline{\mathrm{OB}}\) = ā + c̄
The vector equation of a side BC
i.e., the vector equation of a line passing through C(c̄) and B(ā + c̄) is r̄ = (1 – t)ā + tb̄, t ∈ R
r̄ = (1 – t)c̄ + t(ā + c̄) = c̄ – tc̄ + tā + tc̄
⇒ r̄ – c̄ + tā, t ∈ R

Question 11.
If ā, b̄, c̄ are the position vectors of the vertices A, B and C respectively of ∆ABC, then find the vector equation of the median through the vertex A. [Mar. 13]
Answer:
The position vectors of the vertices A, B and C with respect to the origin are
\(\overline{\mathrm{OA}}\) = ā, \(\overline{\mathrm{OB}}\) = b̄, \(\overline{\mathrm{OC}}\) = c̄
Since D is the midpoint of BC then the position vector of D is \(\overline{\mathrm{OD}}\) = \(\frac{\overline{\mathrm{OB}}+\overline{\mathrm{OC}}}{2}\) = \(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}\). The vector equation of the median through the vertex ‘A’ i.e., The vector equation of a line passing through A(ā) and D\(\left(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}\right)\) is r̄ = (1 – t) ā + tb̄, t ∈ R
r̄ = (1 – t)ā + t\(\left(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}\right)\), t ∈ R
⇒ r̄ = (1 – t)ā + (b̄ + c̄), t ∈ R

Question 12.
Find the vector equation of the line joining the points 2ī + j̄ + 3k̄ and – 4ī + 3j̄ – k̄. [Mar. 18, 16 (TS); Mar. 16 (AP); 11, 04, 95; May 09, 08, 95]
Answer:
Let ā = 2ī + j̄ + 3k̄, b̄ = – 4ī + 3j̄ – k̄
The vector equation of the line passing through the points 2ī + j̄ + 3k̄ and – 4ī + 3j̄ – k̄
r̄ = (1 – t)ā + tb̄, t ∈ R
r̄ = (1 – t)(2ī + j̄ + 3k̄) + t(- 4ī + 3j̄ – k̄)
= 2ī + j̄ + 3k̄ – 2tī – tj̄ – 3tk̄ – 4tī + 3tj̄ – tk̄
= 2ī + j̄ + 3k̄ – 6tī + 2tj̄ – 4tk̄
∴ r̄ = (2 – 6t)ī + (1 + 2t)j̄ + (3 – 4t)k̄, t ∈ R

Question 13.
Find the vector equation of the plane passing through the points ī – 2j̄ + 5k̄, – 5j̄ – k̄ and – 3ī + 5j̄. [Mar. 19, 17 (AP), May 15 (AP); May 14, 13, 11, 93; Mar. 06]
Answer:
Let ā = ī – 2j̄ + 5k̄, b̄ = – 5j̄ – k̄, c = – 3ī + 5j̄
Vector equation of the plane passing through the points ā, b̄, c̄ is
r̄ = (1 – s – t) ā + sb̄ + tc̄ where s, t ∈ R
= (1 – s – t) (ī – 2j̄ + 5k̄) + s (- 5j̄ – k̄) + t (- 3ī + 5j̄), s, t ∈ R
= (ī – 2j̄ + 5k̄) + s(- 5j̄ – k̄ – ī + 2j̄ – 5k̄) + t(- 3ī + 5j̄ – ī + 2j̄ – 5k̄)
r̄ = ī – 2j̄ + 5k̄ + s(- ī – 3j̄ – 6k̄) + t(- 4ī + 7j̄ – 5k̄)

Find the vector equation of the plane passing through the points (0, 0, 0), (0, 5, 0) and (2, 0. 1). [Mar. 18 (AP)]
Answer:
r̄ = (5t) j̄ + 5(2ī + k̄); t, s ∈ R.

Question 14.
If α β γ are the angles made by the vector 3ī – 6j̄ + 2k̄ with the positive directions of the co-ordinate axes, then find cos α, cos β, cos γ.
Answer:
Unit vectors along the coordinate axes are respectively ī, j̄, k̄.
Let p̄ = 3ī – 6j̄ + 2k̄
Given (p̄, ī) = α, (p̄, j̄) = β, and (p̄, k̄) = γ

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Spelling: Missing Letters Exercise Questions and Answers.

TS Inter 1st Year English Grammar Spelling: Missing Letters

Q.No. 15 (8 × 1/2- 4 Marks)

English spelling poses problems to learners. This is because of a) no one letter-one sound relationship b) silent letters and c) words borrowed from other languages and so on.

Though there are some rules and generalizations that MAY HELP improve one’s spelling, it is always good to develop a keen eye to the spelling of the words one comes across and to practise writing words as many times as possible, that too correctly.

• In the Intermediate Public Examination, usually the spelling of the words used in the selected lessons is tested.
• The words that have vowel clusters (as in rec£jve, entertain etc.) or the words that have doubled consonants (as in possible, success etc.) are generally set in the question paper.
• Careful observation of the spelling of commonly used but typically spelt words (like height, tuition etc.) will go a long way in helping the student improve spelling to a great extent.

Examples of some commonly misspelled words :

1. For adjectives that end in a consonant +y, ‘-y’ changes to ‘-ily’.
easy-easily
happy-happily
angry-angrily
Exceptions
shy-shyly
sly-slyly
coy-coyly

2. For adjectives that end in a consonant +le the ‘-le’ changes to ‘-ly’ after the consonant,
probable-probably
sensible-sensibly
idle-idly
noble-nobly
For adjectives that end in – ic, we add -ally.
tragic-tragically
automatic-automatically
Exception
public-publicly
The doubling of final consonants.
When a one syllable word ends with one vowel and consonant we double the final consonant before adding a suffix that begins with a vowel.
bat-batting
dub-dubbing
sad-sadder
dig-digger
travel-travelled
confer-conferred

3. Spelling of regular verbs.
The past tense and past participle are the same in regular verbs.
i) We simply add – ed to the base form of the verb.
clear-cleared
walk-walked
visit-visited

ii) If the word ends in -e, we add -d.
like-liked
smile-smiled
hope-hoped

iii) For verbs ending in y preceded by two consonants, we change y to i before adding the suffix – ed.
hurry-hurried
reply-replied
try-tried

iv) For the verbs that end in a consonant, the final consonant is doubled when a suffix with an initial vowel letter is added.
rob-robbed
mop-mopped
drop-dropped

4. For ‘ie’ and ‘ei’
The general rule is that ‘i’ comes before ‘e’.
chief
grief
relieve
But after ‘c’ ‘e’ comes before ‘i’.
receive
conceit
deceive
Note the exceptions
neighbour
weight
eight
weigh
seize
height

Here is a list of commonly misspelled words.

Exercise – A

Fill in the blanks with either “ei” or “ie”

1. n _ _ ther
2. br _ _ f
3. sh _ _ ld
4. cr _ _ d
5. tr _ _ d
6. fr _ _ nd
7. th _ _ f
8. gr _ _ f
9. l _ _ sure
10. c _ _ ling
11. s _ _ ze
Answer:
1. n e i ther
2. br i e f
3. sh o u ld
4. cr e e d / cr i e d
5. tr i e d
6. fr i e nd
7. th i e f
8. gr i e f
9. l e i sure
10. c e i ling
11. s e i ze

Exercise – B

Correct the following misspelled words.

1. foregn
2. lugage
3. liesure
4. knowlege
5. twelth
6. tomorow
7. gurantee
8. momonto
9. ilegal
10. restaurent
Answer:
1. foreign
2. luggage
3. leisure
4. knowledge
5. twelfth
6. tomorrow
7. guarantee
8. memento
9. illegal
10. restaurant

Exercise – C

1. Please ……………………….. (wait / weight) for me.
2. I went to market to ……………………….. (buy / bye) a toothbrush.
3. He may not ……………………….. (loose / lose) the match.
4. When Chief Justice came to our state, people from different walks of life went to ……………………….. (meat / meet) him.
5. Ahmad cut the paper into many a ……………………….. (piece / peace).
6. The letters on the board cannot be ……………………….. (scene / seen) clearly.
7. – My friend has a ……………………….. (stationery / stationary) shop.
8. It was a pretty ……………………….. (sight / cite).
Answer:
1. wait
2. buy
3. lose
4. meet
5. piece
6. seen
7. stationery
8. sight

Supply the missing letters in the following words.

Exercise – 1

i) sch _ _ l
ii) enc _ _ raging
iii) app _ _ rance
iv) exce _ _ ent
v) sp _ _ k
vi) a _ _ ention
vii) p _ _ pie
viii) kno _ _ edge
ix) di _ _ ipline
x) a _ _ ord
Answers
i) sch o o l
ii) enc o u raging
iii) app e q rance
iv) exe l l ent
v) sp e a k
vi) a t t ention
vii) p e o ple
viii) kno w l edge
ix) di s c ipline
x) a f f ord

Exercise – 2

i) tea _ _ er
ii) gl _ _ my
iii) le _ _ on
iv) re _ _ect
v) f _ _ thful
vi) infl _ _ nce
vii) le _ _ ers
viii) pl _ _ sant
ix) su _ _ est
x) si _ _ le
Answer:
i) tea c h er
ii) gl o o my
iii) le s s on
iv) re s p ect
v) f a i thful
vi) infl u e nce
vii) le a d ers
viii) pl e a sant
ix) su g g est
x) si m p le

Exercise – 3

i) childh _ _ d
ii) p _ _ ce
iii) frus _ _ ation
iv) ha _ _ en
v) gra _ _ ar
vi) col _ _ r
vii) ang _ _ sh
viii) li _ _ten
ix) obed _ _ nt
x) mu _ _ le
Answer:
i) child o o d
ii) p i e ce
iii) frs t r ation
iv) ha p p en
v) gra m m ar
vi) col o u r
vii) ang u i sh
viii) li g h ten
ix) obed i e nt
x) mu s c le

Exercise – 4

i) g _ _de
ii) ma _ _ er
iii) mar _ _ es
iv) fi _ _ ing
v) ye _ _ow
vi) h _ _ lithly
vii) sq _ _ re
viii) lau _ _ed
ix) su _ _ect
x) hi _ _ly
Answer:
i) g u i de
ii) ma n n er/ ma t t er/ ma d d er/ma p p er
iii) mar b l es / mar c h es
iv) fi b b ing
v) ye l l ow
vi) h e a lthy
vii) sq u a re
viii) lau g h ed
ix) su s p ect
x) hi g h ly

Exercise – 5

i) hi _ _ top
ii) ba _ _ an
iiii) r _ _ tine
iv) conc _ _ ve
v) m _ _ ntain
vi) mi _ _ ion
vii) in _ _ edible
viii) mons _ _ n
ix) ca _ _ y
x) reco _ _ ition
Answer:
i) hi l l top
ii) ba n y an
iii) r o u tine
iv) con e i ve
v) m o u ntain
vi) mi s s ion / mi l l ion
vii) in c r edible
viii) mons o o n
ix) ca r r y
x) reco g n ition

Exercise – 6

i) sa _ _ ling
ii) rup _ _ s
iii) hu _ _ and
iv) res _ _ rces
v) s _ _ rce
vi) su _ _ icient
vii) ma _ _ ive
viii) vi _ _ age
ix) init _ _ tive
x) a _ _ roval
Answer:
i) sa p p ling
ii) rup e e s
iii) hu s b and
iv) res o u rces
v) s o u rce
vi) su f f icient
viii) ma s s ive
ix) init i a tive
x) a p p roval

Exercise – 7

i) pers _ _ de
ii) flu _ _ er
iii) p _ _ nce
iv) ex _ _ ode
v) a _ _ empt
vi) br _ _th
vii) hu _ _ red
viii) pa _ _ ive
ix) co _ _ apse
x) thr _ _ ten
Answer:
i) per u a de
ii) flu t t er
iii) p o u nce
iv) ex p l ode
v) a t t empt
vi) br e a th
vii) hu n d red
viii) pa s s ive
ix) co l l apse
x) thr e a ten

 

Exercise – 8

i) thr _ _ gh
ii) sli _ _ tly
iii) gr _ _ nd
iv) wo _ _ y
v) sp _ _ d
vi) ang _ _ sh
vii) prev _ _ us
viii) mi _ _ t
ix) rec _ _ ve
x) p _ _ ce
Answer:
i) thr o u gh
ii) sli g h tly
iii) gr o u nd
iv) wo r r y
v) sp e e d
vi) ang u i sh
vii) prev i o us
viii) mi g h t
ix) rec e i ve
x) p i e ce / p e a ce

Exercise – 9

i) cro _ _ ed
ii) em _ _ atic
iii) tr _ _ ble
iv) consc _ _ us
v) pl _ _ sant
vi) dr _ _ dful
vii) chi _ _ ey
viii) pu _ _ le
ix) incr _ _ se
x) de _ _ ive / de _ _ ive
Answer:
i) cro p p ed / cro w n ed
ii) em p h atic
iii) tr o u ble
iv) cons i o us
v) pl e a sant
vi) dr e a dful
vii) chi m n ey
viii) pu z z le
ix) incr e a se
x) de p r ive / de c e ive

Exercise – 10

i) app _ _ r
ii) wo _ _ le
iii) tre _ _le
iv) con _ _ ary
v) cr _ _ m
vi) com _ _ ain
vii) f _ _ lt
viii) req _ _ st
ix) enc _ _ nter
x) acq _ _ int
Answer:
i) app e a r
ii) wo b b le
iii) tre m b le
iv) con t r ary
v) cr e a m
vi) com p l ain
vii) f a u lt
viii) req u e st
ix) enc o u nter
x) acq u a int

Telangana TSBIE TS Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory

Very Short Answer Type Questions

Question 1.
State the law of equipartition of energy. [TS Mar. ’18, ’16]
Answer:
Law of equipartition of energy :
The total energy of a gas is equally distributed in all possible energy modes, with each mode having an average energy equal to \(\frac{1}{2}\)K_BT. This is known as law of “equipartition of energy”.

Question 2.
Define mean free path. [AP Mar. ’19, ’18, ’17, ’15, May ’18; TS May ’18, Mar. ’17, ’15]
Answer:
Mean free path :
The average distance by can travel an atom or a molecule without colliding is called “mean free path”.
l = \(\frac{1}{\sqrt{2} \pi n d^2}\)
Where n is the number of molecules per unit volume and d is the diameter of the molecule.
(or)
Mean tee path of gas molecules is the aver-age distance covered by a molecule between two successive collisions.

Question 3.
How does kinetic theory justify Avogadro’s hypothesis and show that Avogadro Number in different gases is same?
Answer:
Avogadro’s law or Avogadro’s hypothesis:
It states that “equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules

Consider two gases distinguished by subscripts 1 and 2.

Let
m₁ and m₂ = masses of molecules of 1 and 2 respectively
N₁ and N₂ = no. of molecules of 1 and 2 respectively
P₁ and P₂ = pressures of 1 and 2 respectively
V₁ and V₂ = volumes of 1 and 2 respectively
T₁ and T₂ = absolute temperatures of 1 and 2 respectively
\(\overline{\mathrm{V}_1^2}\) and \(\overline{\mathrm{V}_2^2}\) = mean square velocities of molecules of 1 and 2 respectively.
According to kinetic theory of gases,
P₁V₁ = \(\frac{1}{3}\) m₁N₁ \(\overline{\mathrm{V}_1^2}\) and P₂V₂ = \(\frac{1}{3}\) m₂N₂ \(\overline{\mathrm{V}_2^2}\)
For equal volumes at the same pressure,
P₁V₁ = P₂ V₂
∴ \(\frac{1}{3}\) m₁N₁ \(\overline{\mathrm{V}_1^2}\) = \(\frac{1}{3}\) m₂N₂ \(\overline{\mathrm{V}_2^2}\) ……………. (1)

If the gases are at the same temperature, the average kinetic energies of their molecules are equal.
∴ equation (1) becomes : N₁ = N₂

Hence, equal volumes of all gases under similar conditions of temperature and pressure have the same number of molecules.

Question 4.
What are the units and dimensions of a specific gas constant? [AP Mar. ’14]
Answer:
For specific gas constant unit: Joule/Kelvin
Dimensional formula: r = \(\frac{PV}{T}\) =ML² T^-2 K^-1.

Question 5.
When does a real gas behave like an ideal gas? [AP Mar. ’19, ’14; TS Mar. ’19, ’16, May ’18, ’16; June ’15]
Answer:
No real gas is perfect or ideal. At extremely low pressures and high temperatures, some real gases (like H₂, O₂, N₂, He, etc.) obey the gas laws to a fair degree of accuracy and hence, behave as nearly ideal gas.

Question 6.
State Boyle’s Law and Charles Law. [AP Mar. 18, June 15; TS Mar. 15, May. 16]
Answer:
Boyle’s Law :
At constant temperature, the volume (V) of a given mass of a gas is inversely proportional to its pressure (P).
∴ V ∝ \(\frac{1}{P}\) or PV = constant = K.

Charles Law :
At constant pressure, the volume (V) of a given mass of a gas is directly proportional to its absolute temperature (T).
∴ V ∝ T or \(\frac{V}{T}\) = K (constant)

Question 7.
State Dalton’s law of partial pressures. [TS Mar. ’18, ’17; AP Mar. ’16, ’14]
Answer:
Dalton’s law of partial pressures :
For a mixture of non interacting ideal gases at same temperature and volume total pressure in the vessel is the sum of partial pressures of individual gases.
i.e., P = P₁ + P₂ + ………… where P is total pressure
P₁, P₂ ……….. etc, are individual pressures of each gas.

Question 8.
Define absorptive power of a body. What is the absorptive power of a perfect black body? [AP May ’14]
Answer:
Absorptive power of a body is defined as the ratio of energy absorbed by the body within the wave length range of λ and λ + dλ to the total energy flux following on the body.
Absorptive power,

Question 9.
Pressure of an ideal gas in container is independent of shape of the container – explain. [AP June ’15]
Answer:
Pressure exerted by a gas is due to continuous bombardment of gaseous molecules with the walls of the container. During each collision, certain momentum is transferred to the walls of the container and this transfer is independent of its shapes because area A and time interval ∆t do not appear in the final formula. Hence, pressure of an ideal gas in a container is independent of the shape of the container.

Question 10.
Explain the concept of degrees of freedom for molecules of a gas.
Answer:
The number of degrees of freedom of a dynamical system is defined as the total number of co-ordinates or independent quantities required to describe completely the position and configuration of the system.

Question 11.
What is the expression between the pressure and kinetic energy of a gas molecule?
Answer:
The pressure exerted by an ideal gas is numerically equal to \(\frac{2}{3}\) rd of the mean kinetic energy of translation per unit volume of gas.
P = \(\frac{2}{3}\)E

Question 12.
The absolute temperature of a gas is increased 3 times. What will be the increase in rms velocity of the gas molecule? [TS Mar. ’19, June ’15]
Answer:
The relation between r.m.s. velocity and absolute temperature of a gas is C ∝ √T.
Therefore, the r.m.s. velocity becomes √3 C.
Hence, increase in r.m.s. velocity
√3 C – C = 0.732 C = 73.2 %

Short Answer Questions

Question 1.
Explain the kinetic interpretation of Temperature.
Answer:
According to kinetic theory of gases, the pressure P exerted by one mole of an ideal gas is given by

Thus, the square root of the absolute temperature of an ideal gas is directly proportional to root mean square velocity of its molecules.
Also, from eqn. (1)

But, \(\frac{1}{2}\)mc² is average translational K.E per molecule of a gas.

Therefore, average kinetic energy of translation per molecule of a gas is directly proportional to the absolute temperature of the gas.

Question 2.
How can specific heat capacity of monoatomic, diatomic, and polyatomic gases be explained on the basis of Law of equipartition of Energy? [AP May ’17. ’16; Mar. ’13]
Answer:
From law of equipartition of energy, energy per each degree of freedom is \(\frac{1}{2}\) K_BT Per atom or molecule.

1) Monoatomic gas has three degrees of freedom.

But molar specific heat at constant volume C_v = \(\frac{dU}{dT}\)

2) A diatomic gas has 3 translational and two rotational degrees of freedom.
∴ Kinetic energy per molecule U₁ = 5.\(\frac{1}{2}\)K_B.T
For one gram mole total energy U = \(\frac{5}{2}\)K_B.T.N_A
Molar-specific heat at constant volume

3) A polyatomic gas has three translational, three rotational, and at least one vibrational degrees of freedom.
∴ Kinetic energy per molecule
U₁ = 3.\(\frac{1}{2}\)K_B.T + 3.\(\frac{1}{2}\)K_B.T + f = (3 + f)K_BT
f = Number of vibrational degrees of freedom

Kinetic energy
= per gram mole of molecules
= U₁N_A = U = (3 + f)K_B.N_A.T = (3 + f)RT
Molar specific heat C_v = \(\frac{dU}{dT}\) = (3 + f) R
∴ C_P = (u + f)R
∴ For polyatomic gases C_v = (3 + f) R & C_P = (4 + f)R

Question 3.
Explain the concept of absolute zero of temperature on the basis of kinetic theory.
Answer:
According to kinetic theory of gases, the pressure ‘P* exerted by one mole of an ideal gas is,

Absolute zero:
When T = 0, from equation (1), C = 0.

Hence, absolute zero of temperature may be defined as that temperature at which the root mean square velocity (C) of the gas molecules reduces to zero.

It means, molecular motion ceases at absolute zero.

This definition holds in cases of an ideal gas or perfect gas.

Question 4.
Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.
Answer:
According to kinetic theory of gases, the pressure ‘P’ exerted by one mole of an ideal gas is
P = \(\frac{1}{3}\) ρC² where p is density of the gas.
∴ P = \(\frac{1}{3}\frac{M}{V}\)C² ⇒ PV = \(\frac{1}{3}\)MC²
But PV = RT for one mole of ideal gas

But, \(\frac{1}{2}\) mC² is averaSe translational kinetic energy per molecule of a gas.

Therefore, average kinetic energy of molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Question 5.
Two thermally insulated vessels 1 and 2 of volumes V₁ and V₂ are joined with a valve and filled with air at temperatures (T₁, T₂) and pressures (P₁, P₂) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessels at equilibrium.
Answer:
According to Standard gas equation

As the vessels are thermally insulated and no work is done, total energy remains conserved.
Therefore, \(\frac{3}{2}\) (P₁V₁ + P₂V₂) = \(\frac{3}{2}\) P(V₁ + V₂)
where P is resultant pressure

For mixture of two gases,
(µ₁ + µ₂)RT = P(V₁ + V₂)
where T is resultant temperature.

Substitute equation (1) and equation (2) in the above equation, we have

This is the final equilibrium temperature of the air in the vessels.

Question 6.
What is the ratio of r.m.s speed of Oxygen and Hydrogen molecules at the same temperature?
Answer:
The r.m.s. speed of oxygen molecules at temperature T is

The r.m.s. speed of Hydrogen molecules at same temperature (T) is
C_H = \(\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}}}}\) But molecular weight of hydrogen = 2
Ratio of r.m.s. speed of Oxygen and Hy-drogen molecules is
∴ C_H = \(\sqrt{\frac{3 \mathrm{RT}}{2}}\)
Ratio of r.m.s. speed of Oxygen and Hydrogen molecules is

∴ C_ghy : C_H = 1 : 4

Question 7.
Four molecules of a gas have speeds 1,2,3 and 4 km/s. Find the rms speed of the gas molecule.
Answer:
Given, C₁ = 1 km/s ; C₂ = 2 km/s
C₃ = 3 km/s ; C₄ = 4 km/s

The root mean square speed of molecules is

C_rms = 2.74 km/s

Question 8.
If a gas has ‘f degrees of freedom, find the ratio of C_p and C_v.
Answer:
Suppose a polyatomic gas molecule has ‘f degrees degrees of freedom (0 and ratio of Cp and of freedom.
∴ Internal energy of one gram mole of the gas is
U = f × \(\frac{1}{2}\)k_BT × N_A = \(\frac{f}{2}\)RT

This is the relation between number of degrees of freedom (f) and ratio of C_p and C_v & (γ)

Question 9.
Calculate the molecular K.E of 1 gram of Helium (Molecular weight 4) at 127°C. Given R = 8.31 J mol^-1C^-1.
Answer:
n = number of moles of Helium = \(\frac{1gm}{4gm mol^{-1}}\) = 0.25 mole
R = 8.314 J mol^-1 KT^-1,
T = 127°C = 127 + 273 = 400 K
∴ Kinetic energy = \(\frac{3}{2}\) nRT = \(\frac{3}{2}\) × 0.25 mol × 8.314 J mol^-1 K^-1 × 400K = 1247.1 J

Question 10.
When pressure increases by 2%, what is the percentage decrease in the volume of a gas, assuming Boyle’s law is obeyed?
Answer:
Let ‘P’ be the initial pressure and ‘V’ be the volume.
When pressure is increased by 2%, new pressure, P¹ = P + \(\frac{2}{100}\)P ⇒ P¹ = \(\frac{102}{100}\)P
According to Boyle’s law, PV = constant ⇒ PV = P’V’

% decrease in volume = \(\frac{2}{100}\) × 100 = 1.96%

∴ When pressure increases by 2%, the volume decreases by 1.96%.

Long Answer Questions

Question 1.
Derive an expression for the pressure of an ideal gas in a container from Kinetic Theory and hence give Kinetic Interpretation of Temperature.
Answer:
Consider a cube of side ‘l’ and an ideal gas is filled in it. Let the co-ordinates X, Y and Z will coincide with the sides of the cube. Let velocities of gas molecules along these directions are V_x, V_y and V_z. Consider a gas molecule moving along X – axis with a velocity V_x. Its motion is perpendicular to Y – Z plane. Let the gas molecule suffered elastic collision with Y – Z plane and bounced back. In this case the velocities V_y and V_z are not considered because collision is along X – direction only.

Change in momentum along X-direction is final momentum (-mV_x) – Initial momentum (mV_x) = -2mV_x ………… (1)

Let area of one side of the cube is A. Then during the time ∆t only the molecules at a distance of V_x ∆t will collide with the walls. Let number of gas molecules in the volume AV∆t are say ‘n’. In these molecules half of the molecules will move towards the wall and remaining will move away from the wall.
∴ Amount of momentum transferred to the wall = Q
= 2mV_x × number of molecules collided with wall
∴ Q = 2mV_x(\(\frac{1}{2}\)nAV_x∆t)

Pressure on Y – Z plane =

If behaviour of gas is isotropic that is equal in all directions then V_x = V_y = V_z
∴ Root mean square velocity of gas

∴ Pressure of gas P = \(\frac{1}{3}\)nm\(\overline{\mathrm{V^2}}\)
From Pascal’s law pressure is same throughout the container so pressure in any direction (say x, y or z) is same.

Kinetic interpretation of temperature :
From gas equation

For Ideal gas internal energy is purely kinetic energy.

But \(\frac{E}{N}=\frac{1}{2}\)mV² and \(\frac{E}{N}=\frac{3}{2}\) K_BT

∴ \(\frac{E}{N}\) ∝ T is the average kinetic energy of gas molecule is proportional to absolute temperature of gas. But it does not depend on volume of container.

Additional Problems

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP, Take the diameter of an oxy-gen molecule to be 3 Å.
Solution:
Here, diameter, d – 3Å,
r = \(\frac{d}{2}=\frac{3}{2}\)Å = \(\frac{3}{2}\) × 10^-8
Molecular volume, V = \(\frac{4}{3}\) πr³ . N, where N is
Avogadro’s number = \(\frac{4}{3}\times\frac{22}{7}\) (1.5 × 10^-8)³ × (6.023 × 10²³) = 8.52 cc.
Actual volume occupied by 1 mole of oxygen at STP, V’ = 22400 cc
∴ \(\frac{V}{V’}=\frac{8.52}{22400}\) = 3.8 × 10^-4 ≈ 4 × 10^-4

Question 2.
Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°Q. Show that it is 22.4 litres.
Solution:
For one mole of an ideal gas, PV = RT
∴ V = \(\frac{RT}{P}\)
Put R = 8.31 .1 mole^-1 K^-1, T = 273 K,
P = 1 atmosphere = 1.013 × 10⁵ Nm^-2

= 0.0224 × 10⁶ cc = 22400 cc = 22.4 litre.

Question 3.
An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?
Solution:
V₁ = 1.0 cm³ = 1.0 × 10^-6m³;
T₁ = 12°C = 12 + 273 = 285 K;
P₁ = 1 atm. + h₁ p g = 1.01 × 10⁵ + 40 × 10³ × 9.8 = 493000 Pa.
When the air bubble reaches at the surface of lake, then
V₂ = ? ; T₂ = 35°C = 35 + 273 = 308 K ;
P₂ = 1 atm. = 1.01 × 10⁵ Pa

Question 4.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a he-lium gas atom at – 20°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Solution:
Let C and C’ be the r.m.s. velocities of argon and a helium gas atoms at temperature TK and T’ K respectively.
Here, M = 39.9 ; M’ = 4.0 ; T = ?
T’ = -20 + 273 = 253 K

Question 5.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas.
[Hint: Use Graham’s law of diffusion: R₁/R₂ = (M₂/ M₁)^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]
Solution:
According to Graham’s law of diffusion, = \(\frac{r_1}{r_2}=\sqrt{\frac{\mathrm{M}_2}{\mathrm{M}_1}}\)
Where, r₁ = diffusion rate of hydrogen = 28.7 cm³ s^-1
r₂ = diffusion rate of unknown gas = 7.2 cm³ s^-1
M₁ = molecular mass of hydrogen = 2 u
M₂ = ?

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Silent Letters Exercise Questions and Answers.

TS Inter 1st Year English Grammar Silent Letters

Q.No. 16 (8 × 1/2 = 4 Marks)

Silent letters are a peculiar feature of English spelling. Letters are used in writing but the corresponding sounds are not uttered in pronunciation. This poses problems to learners both in spelling and pronunciation. A study says that 60% of the English words have silent letters !

Careful observation of both the spelling and pronunciation is the only way out to overcome this problem. There are, however, some generalizations about silent letters that will help learners understand this feature to some extent.

In the Intermediate Public Examination, normally, the words that appear in the prescribed selections with a silent consonant letter are set under this question.

Some Guidelines :

• ‘b’ before word final ‘t’ is silent – doubt debt.
• ‘b’ in final ‘mb’ clusters is also silent.
  comb
  womb
  tomb
• ‘g’ before word final ‘n’ is silent.
  sign
  resign
  foreign
  campaign
  reign
  sovereign
• ‘g’ in most ‘ough’, ‘augh’ combinations is silent.
  (But in some words with ‘augh’ it may sound as /f/ as in ‘laughter),
  though
  taught
  through
  benign
  thorough
• ‘h’ in word beginning position (not always) is silent.
  honest
  honour
• ‘k’ in word beginning followed by ‘n’ is silent.
  knowledge
  know
  knit
  knee
  knave
  knuckle
• ‘l’ before word final’m’ is silent.
  calm
  palm (but in realm and film – T is not silent)
• Word final ‘n’ preceded by ‘m’ is silent,
  condemn
  column
  autumn
• ‘p’ in word beginning position followed by ‘s’ is silent.
  ’psychology
  psalm
  pseudo
• ‘p’ in words like receipt is also silent.
• ‘r’ followed by a consonant letter is silent,
  world
  card
• ‘f is silent in clusters like ‘astle’, ‘isten’, ‘istle’ as in
  castle
  whistle
  listen
• ‘W’ is silent in words like
  Write
  wrap
  know
  fawn
  lawn

(These guiding principles are by no way exhaustive. They help the learner to a great extent in identifying the silent letters. About vowels – the problem is not this simple or unambiguous. So no generalisation is given here.)

TEXTUAL EXAMPLES

Silent letter – Example words

b – lamb, bomb, tomb, climb, dumb, subtle, plumber, womb, crumb, thumb
c – muscle, blackguard, yacht, indict, scene
d – Wednesday, handkerchief, handbag, handsome, adjourn, adjective, judge
g – gnaw, gnome, phlegm, foreign, resign, campaign, align, sovereign
h – honour, heir, ghost, night, rhyme, rhythm, when, where, hour
k – know, knee, knock, knot, kneel, knowledge
t – talk, folk, salmon, colonel, calf, calm, half, walk, baulk
m – mnemonic . . .
n – hymn, solemn, damn, autumn, column
p – recepit, psychic, pneumonia, psyche
q(u) – lacquer
r – girl, bird, card, teacher, leader, curd, world
s – isle, aisle, viscount, island
t – thistle, fasten, mortgage, soften, watch, tsunami
w – whole, sword, two, who, wrist, wrong
y – prayer, mayor
z – rendezvous
gh – sigh, high, daughter, naughty, caught, brought, bought, night, haughty

Exercise -1

I. Underline the silent letters in the following words.

1) knell
2) consign
3) thorough
4) yellow
5) often
6) delight
7) benign
8) almond
9) yolk
10) limb
11) pseudonym
12) cupboard
13) Indict
14) climb
15) half
16) bouquet
17) wreath
18) dumb
19) depot
20) answer
21) aisle
22) exhibition
23) night
24) condemn
25) pneumonia
26) design
27) christmas
28) reign
29) palm
30) debut

31) precis
32) whistle
33) poignant
34) knead
35) castle
36) subtle
37) feign
38) debt
39) knight
40) knack
41) debris
42) comb
43) succumb
44) fight
45) deign
46) chalk
47) folk
48) bustle
49) align
50) malign
51) honest
52) lodge
53) pawn
54) doubt
55) bridge
56) coup
57) rapport
58) ghost
59) through
60) listen
Answer:
1) knell
2) consign
3) thorough
4) yellow
5) often
6) delight
7) benign
8) almond
9) yolk
10) limb

11) pseudonym
12) cupboard
13) Indict
14) climb
15) half
16) bouquet
17) wreath
18) dumb
19) depot
20) answer
21) aisle
22) exhibition
23) night
24) condemn
25) pneumonia
26) design
27) christmas
28) reign
29) palm
30) debut
31) precis
32) whistle
33) poignant
34) knead
35) castle
36) subtle
37) feign
38) debt
39) knight
40) knack
41) debris
42) comb
43) succumb
44) fight
45) deign
46) chalk
47) folk
48) bustle
49) align

50) malign
51) honest
52) lodge
53) pawn
54) doubt
55) bridge
56) coup
57) rapport
58) ghost
59) through
60) listen

II. Identify the silent consonant letters in the following words.

Exercise – 1

i) bright
ii) scene
iii) hour
iv) neighbour
v) wrong
vi) knell
vii) wreath
viii) palm
ix) limb
x) design
Answer:
i) bright – gh
ii) scene – c
iii) hour – h – r
iv) neighbour – g, h, r
v) wrong – w
vi) knell – k
vii) wreath – w
viii) palm – l
ix) limb – b
x) design – g

Exercise – 2

i) chalk
ii) knock
iii) depot
iv) teacher
v) often
vi) thought
vii) honest
viii) almond
ix) know
x) talk
Answer:
i) chalk – 1
ii) knock – k
iii) depot – t
iv) teacher – r
v) often – t
vi) thought – gh
vii) honest – h
viii) almond – l
ix) know – k, w
x) talk – l

Exercise – 3

i) lodge
ii) castle
iii) feign
iv) laugh
v) debut
vi) malign
vii) talk
viii) psyche
ix) lighten
x) muscle
Answer:
i) lodge – d
ii) castle – l
iii) feign – g
iv) laugh – …
v) debut – t
vi) malign – g
vii) talk – l
viii) psyche – p
ix) lighten – gh
x) muscle – c

Exercise – 4

i) yolk
ii) would
iii) pneumonia
iv) consign
v) drawing
vi) what
vii) knead
viii) doubt
ix) island
x) aisle
Answer:
i) yolk – l
ii) would – l
iii) pneumonia – p
iv) consign – g
v) drawing – w
vi) what – h
vii) knead – k
viii) doubt – b
ix) island – s
x) aisle – s

Exercise – 5

i) thorough
ii) who
iii) benign
iv) receipt
v) rhythm
vi) diversity
vii) nursery
viii) column
ix) curd
x) kneel
Answer:
i) thorough – gh
ii) who – w
iii) benign – g
iv) receipt – p
v) rhythm – h
vi) diversity – r
vii) nursery – r
viii) column – n
ix) curd – r
x) kneel – k

Exercise – 6

i) bustle
ii) although
iii) parliament
iv) fight
v) knee
vi) brought
vii) bomb
viii) could
ix) hymn
x) which
Answers
i) bustle – t
ii) although – gh
iii) parliament – r, i
iv) fight – gh
v) knee – k
vi) brought – gh
vii) bomb – b
viii) could – l
ix) hymn – n
x) which – h

Exercise – 7

i) align
ii) ghost
iii) leader
iv) straight
v) calf
vi) plumber
vii) wrap
viii) thistle
ix) attempt
x) burden
Answer:
i) align – g
ii) ghost – h
iii) leader – r
iv) straight – gh
v) calf – l
vi) plumber – b, r
vii) wrap – w
viii) thistle – t
ix) attempt – t
x) burden – r

Exercise – 8

i) through
ii) sovereign
iii) slightly
iv) tsunami
v) watch
vi) tomb
vii) caught
viii) naughty
ix) half
x) leopard
Answer:
i) through – gh
ii) sovereign – g
iii) slightly – gh
iv) tsunami – t
v) watch – t
vi) tomb – b
vii) caught – gh
viii) naughty – gh
ix) half – l
x) leopard – o, r

Exercise – 9

i) wrist
ii) daughter
iii) receipt
iv) solemn
v) hatter
vi) mnemonic
vii) 4umb
viii) damn
ix) should
x) folk
Answer:
i) wrist – w
ii) daughter – gh
iii) receipt – p
iv) solemn – n
v) hatter – r
vi) mnemonic – m
vii) dumb – b
viii) damn – n
ix) should – l
x) folk – l

Exercise – 10

i) knock
ii) autumn
iii) cupboard
iv) tight
v) walk
vi) sword
vii) subtle
viii) psalm
ix) handsome
x) gnaw
Answer:
i) knock – k
ii) autumn – n
iii) cupboard – p
iv) tight – gh
v) walk – l
vi) sword – w, r
vii) subtle – b
viii) psalm – p, l
ix) handsome – d
x) gnaw – g, w

Telangana TSBIE TS Inter 1st Year Physics Study Material 13th Lesson Thermodynamics Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 13th Lesson Thermodynamics

Very Short Answer Type Questions

Question 1.
Define Thermal equilibrium. State the zeroth law of thermodynamics (or) How does it lead to Zeroth Law of Thermodynamics? [AP May ’13]
Answer:
Thermal Equilibrium:
Two systems are said to be in thermal equilibrium with each other if they are at same temperature.

Explanation :
If two different temperature of systems are kept in contact. Heat transfer from hot to cold body. If they are at same, then they are said to be in thermo equilibrium.

Generally at thermal equilibrium, the temperature of two systems is same. This concept leads to Zeroth law of thermodynamics.

Zeroth law of thermodynamics :
It states that if two systems say A & B are in thermal equilibrium with a third system ‘C’ separately then the two systems A and Bare also in thermal equilibrium with each other.

Question 2.
Define Calorie. What is the relation between calorie and mechanical equivalent of heat?
Answer:
Calorie:
The amount of heat required to rise the temperature of 1g of water by 1°C is called calorie.

Relation between calorie and Joule, 1 calorie = 4.2 J.

Question 3.
What thermodynamic variables can be defined by a) Zeroth Law b) First Law?
Answer:
Zeroth law refers temperature and first law refers internal energy.

Question 4.
Define specific heat capacity of the substance. On what factors does it depend?
Answer:
Specific heat capacity:
The quantity of heat required to rise the temperature of unit mass of the substance through 1 °C or IK is called the “specific heat capacity of the substance.”
S = \(\frac{dQ}{m.dT}\)
The specific heat capacity depends upon the factors like temperature and nature of the substance.

Question 5.
Define molar specific heat capacity. [AP May ’13]
Answer:
Molar specific heat capacity:
It is defined as the amount of heat required to rise the temperature of one mole of a gas through 1 °C or IK.

Question 6.
For a solid, what is the total energy of an oscillator?
Answer:
For a solid, the total energy of an oscillator can be expressed as the sum of its potential and kinetic energies.

Question 7.
Indicate the graph showing the variation of specific heat of water with temperature. What does it signify?
Answer:
The variation of specific heat of water with the temperature is as shown in the graph.

From the graph, we find that at T = 15°C, specific heat of water, S = 1 calg^-1 °C^-1

At T = 0°C, S = 1.008 Calg^-1 °C^-1, the highest and at T = 30°C, s = 0.9976 Calg^-1 °C^-1, the lowest and beyond 30°C, specific heat of water increases slightly with rise in temperature.

At T = 100°C, specific heat of water is 1.0057 calg^-1 °C^-1.

It signifies that the specific heat of water decreases with increase in temperature from 0° to 30°C and increases from 30c – 100°C,

Question 8.
Define state variables and equation of state.
Answer:
State variables:
The variables which deter mine the thermodynamic behaviour of a sys tern are called “state variables.”

If the system is a gas, then P, V, and T (for a given mass) are called state variables.

Equation of state:
The general relationship between pressure, volume, and temperature for a given mass of the system (eg., gas) is called “equation of the state.”

For n moles of an ideal gas, the equation of state is, PV = nRT.

Question 9.
Why a heat engine with 100% efficiency can never be realised in practise?
Answer:
The efficiency of heat engine, η = 1 – \(\frac{T_2}{T_1}\)

The efficiency will be 100%, or 1, if T₂ = OK or T₁ = ∞.

Since, both these conditions cannot be attained practically, a heat engine cannot have 100% efficiency.

Question 10.
In summer, when the valve of a bicycle tube is opened, the escaping air appears cold. Why?
Answer:
This happens due to adiabatic expansion of the air in the tube of the bicycle. Hence the air cools.

Question 11.
Why does the brake drum of an automobile get heated up while moving down at constant speed?
Answer:
When an automobile moving down with constant speed its potential energy decreases. This decrease in potential energy is converted in the form of heat energy. As a result, the break drum of an automobile get heated.

Question 12.
Can a room be cooled by leaving the door of an electric refrigerator open?
Answer:
No. When a refrigerator is working in a closed room with its door closed, it is rejecting heat from inside to the air in the room. So, temperature of room increases gradually.

When the door of refrigerator is kept open, heat rejected by the refrigerator to the room will be more than the heat taken by the refrigerator from the room. Therefore, temperature of room will increase at a slower rate compared to the first case.

Hence, a room cannot be cooled by leaving the door of an electric refrigerator open.

Question 13.
Which of the two will increase the pressure more, an adiabatic or an isothermal process, in reducing the volume to 50%?
Answer:
In an adiabatic process, no exchange of heat is allowed between the system and surroundings. Hence, the work done during reducing the volume to 50% results in the increase in the temperature of the system thereby further increase in the pressure (∵ PV = RT). In case of isothermal compression, the excess heat is exchanged with the surroundings, maintaining constant temperature. Hence the increase in pressure is only due to decrease in volume obeying Boyle’s law.

Hence adiabatic compression increases the pressure more than isothermal compression.

Question 14.
A thermos flask containing a liquid is shaken vigorously. What happens to its temperature?
Answer:
Temperature of the liquid increases, because work is done in shaking the liquid. W ∝ Q.

Question 15.
A sound wave is sent into a gas pipe. Does its internal energy change?
Answer:
Yes, the internal energy changes when a sound wave is sent into a gas pipe. Because the sum of all the energies contained in the system in equilibrium is called its internal energy.

Question 16.
How much will be the internal energy change in
i)isothermal process ii) adiabatic process
Answer:
i) In an isothermal process,
T = constant i.e., dT = 0 ∴ dU = 0
So, in a isothermal process, the internal energy does not change.

ii) In an adiabatic process,
Q = constant i.e., dQ = 0 ∴ dU = -dW ≠ 0

So, in an adiabatic process, the change in internal energy is equal to the amount of work done.

Question 17.
The coolant in a chemical or a nuclear plant should have high specific heat. Why?
Answer:
“Specific heat of a substance is the amount of heat required to raise the temperature of unit mass of the substance through 1 °C or IK”. The coolant in a chemical or nuclear plant should be able to absorb more amount of heat released from the plant. Hence, the coolant should have high specific heat.

Question 18.
Explain the following processes i) Isochoric process ii) Isobaric process
Answer:
i) Isochoric process:
The process that occurs at constant volume is called “Isochoric process” or “Isovolumic process”.
∴ dV = 0.

ii) Isobaric process :
The process that occurs at constant pressure is called “Isobaric process.”
∴ dP = 0

Short Answer Questions

Question 1.
State and explain first law of thermodynamics.
Answer:
First law of thermodynamics :
The heat energy (dQ) supplied to a system is equal to the sum of the increase in the internal energy (dU) of the system and external work done (dW) by it
i.e., dQ = dU + dW ……….. (1)

If dV is the increase in the volume under constant pressure (P) then dW = PdV.
∴ dQ = dU + PdV ………….. (2)

The importance of this law is that it defines, the thermodynamic quantity, internal energy which has a fixed value in a state.

Increase in internal energy dU = nC_vdT

Where n is the number of moles of the gas. This equation helps to calculate change of internal energy of the system when the temperature change by ∆T.

Limitations of 1st law of thermody namics:

1. It does not tell about the direction of heat flow. That is it does not specify the conditions under which a body can use the heat energy to produce the work.
2. It does not give any information about the efficiency with which heat can be converted into work.

Question 2.
Define two principal specific heats of a gas. Which is greater and why? [TS June ’15]
Answer:
i) Specific heat of a gas at constant vok ume (C_v) :
It is defined as the amount of heat energy required to raise the temperature of one gram of gas through 1°C of 1K, when volume of the gas is kept constant.

It is measured in cal.g^-1.K^-1 or J.g^-1. K^-1,

ii) Specific heat of a gas at constant pressure (C_p) :
It is defined as the amount of heat energy required to raise the temperature of one gram of gas through 1 °C or IK, when pressure of the gas is kept constant.
It is also measured in cal. g^-1.K^-1 or J.g^-1.K^-1.

Out of the two principal specific heats of a gas, C_p > C_v. This can be justified as follows:

a) When heat is given to a gas at constant volume, it is only used in increasing the internal energy of the gas, i.e., in raising the temperature of the gas, and no heat is spent in the expansion of the gas.

b) When heat is given to a gas at constant pressure, it is spent in two ways :

1. Part of the heat is increasing the internal energy of the gas and hence the temperature of the gas.
2. Remaining amount of heat is used in doing work i.e., in the expansion of the gas against the external pressure.

Therefore to raise the temperature of 1 mole of a gas through 1°C or 1K, more heat is required at constant pressure (C_p) than at constant volume (C_p).

Hence, C_p > C_v.

Question 3.
Derive a relation between the two specific heat capacities of gas on the basis of first law of thermodynamics. [TS June ’15]
Answer:
Relationship between the two specific heat capacities of gas:
To derive C_p – C_v = R:
Let one gram mole of given mass of gas is enclosed within a cylinder with a frictionless air tight-piston. Let P, V be the pressure and volume of the gas at a temperature T.

i) In specific heat at constant volume :
The volume of the given mass of gas must remain constant. Hence, the piston is fixed in position AB.

Let C_v be the amount of heat energy supplied. It is utilised only to raise the temperature of the gas by 1°C.
∴ dU = C_vdT

ii) In specific heat at constant pressure, the pressure must remain constant. Hence the piston is allowed to move freely. The amount of heat energy supplied is used not only to do external work but also to increase the temperature of the gas by 1°C.

In this case the piston moves forward and work is done against external pressure. Suppose by the time the piston moves from AB to CD, the temperature increases by 1 °C.

Let the work done in moving the piston through a distance ‘dl’ be dW = P dV.

The energy supplied has to increase the internal energy and to do external work.
∴ C_pdT = dQ = dU + dW

From first law of thermodynamics
dQ = dU + dW
∴ C_PdT = C_vdT + dW
(C_P – C_v) dT = dW = PdV
But work done, dW = F × S
= P × A × dl (where A is area of cross-section of the piston) = PdV
But PV = RT (for one mole of gas).

∴ dW = PdV = RdT OR (C_P – C_v) dT = RdT
But change of temperature = dT = 1°C (from definition of specific heat)
∴ C_p – C_v = R

So difference of molar specific heats of the gas is equals to universal gas constant R.

Question 4.
Obtain an expression for the work done by an ideal gas during isothermal change.
Answer:
Work done during isothermal process:
Consider n mole of a perfect gas contained in a cylinder. When the piston moves through a small distance dx, then small work dW will be done by it
∴ dW = P A dx = P dV,
where ‘A’ = the area of cross-section of the piston.

Therefore, when the system goes from initial state A (P₁, V₁) to the final state B (P₂, V₂), the amount of work done,
W = \(\int_{v_1}^{v_2} P d V\) ………….. (1)
But PV = nRT (or) P = \(\frac{nRT}{V}\)
Substitute P in equation (1),
W = \(\int_{v_1}^{v_2} \frac{n R T}{V} d V\)
During an isothermal process, temperature remains constant.

All the heat supplied to the gas is used only to do work since temperature remains constant, internal energy does not change.
∴ dQ = PdV

Question 5.
Obtain an expression for the work done by an ideal gas during adiabatic change and explain.
Answer:
Work done during Adiabatic process :
Consider n mole of perfect gas contained in a cylinder having insulating walls. When piston moves through a small distance dx, then small work (dW) will be done.

∴ dW = (Pa) dx = PdV,
where a is area of cross-section of the piston. Therefore, when the system goes from initial state A (P₁, V₁) to the final state B ( P₂, V₂) the amount of work done,
W = \(\int_{v_1}^{v_2} P d V\) ………….. (1)
For an adiabatic change,
PV^γ = K (a constant ) or P = KV^-γ

Substituting for P in equation (1), the work done in an adiabatic process

∴ Work done in adiabatic process,

∴ Work done in adiabatic change
W = \(\frac{nR}{(\gamma -1)}\)(T₁ – T₂)

Since heat is not supplied to the gas, it can do work only by expanding its internal energy. dQ = 0 = dU + PdV or PdV = – dU.
So the gas cools in adiabatic expansion.

Question 6.
Compare isothermal and an adiabatic process.
Answer:

Isothermal changes	Adiabatic changes
1. Temperature (T) remains constant, i.e., ∆T = 0	1. Heat content (Q) remains constant, i.e., ∆Q = 0.
2. System is thermally conducting to the surroundings.	2. System is thermally insulated from the surroundings.
3. The changes occur slowly.	3. The changes occur suddenly.
4. Internal energy (U) remains constant, i.e., ∆U = 0.	4. Internal energy changes, i.e., U ≠ constant ∴ ∆U ≠ 0.
5. Specific heat becomes infinite.	5. Specific heat becomes zero.
6. Equation of isothermal changes is PV = constant.	6. Equation of adiabatic changes is PVγ = constant
7. Slope of isothermal curve, \(\frac{dP}{dV}\) = -(P/V)	7. Slope of adiabatic curve, \(\frac{dP}{dV}\) = -γ(P/V)
8. Coefficient of Isothermal elasticity; Ei = P	8. Coefficient of adiabatic elasticity; Ea = γP

Question 7.
Explain the following processes
i) Cyclic process with example
ii) Non-cyclic process with example
Answer:
i) Cyclic process :
A process in which the system after passing through various stages such as change in pressure, volume and temperature etc. returns to its initial state is defined as “cyclic process”.

For a thermodynamic system the internal energy of the system depends on thermodynamic variables such as pressure, volume, temperature, etc. In cyclic process the system finally returns to the initial state and it is in thermal equilibrium with surroundings. So change in internal energy of the system dU = 0.

Hence, in a cyclic process work done is equal to energy absorbed in the cyclic process.

So for cyclic process dU = 0 and dQ = dW.

Example:
Generally, all heat engines (or) refrigerators are operated in cyclic process.

ii) Non-cyclic process:
A non-cyclic process consists of a series of changes involved do not return the system back to its initial state.

Example:
Suppose a gas with variables P₁, V₁, T₁ is taken through a series of different states subjecting to a number of changes including isothermal expansions and compressions. In the final state, if the system does not come back to P₁V₁T₁, then the gas is said to be undergo a non-cyclic process.

Work done in a non-cyclic process depends upon the path chosen or the series of changes involved.

Question 8.
Write a short note on Quasistatic process.
Answer:
Quasi-static process:
A Quasistatic process can be defined as an infinitesimally show process in which at each and every intermediate stage the system remains in thermal and mechanical (thermodynamic) equilibrium with the surroundings through out the entire process.

Explanation:
A non-equilibrium thermodynamic system can be treated as an idealized process in which at every stage the system is in equilibrium state.

In a thermodynamic system let the piston moves in a frictionless manner. Instead of sudden compression of piston imagine the piston moves very very slowly i.e., it appears almost static. Then the pressure inside the cylinder P + ∆P and temperature T + ∆T are almost equal to external pressure P and temperature T.

Since for extremely slow process the values of ∆P and ∆T are so small that we can treat P + ∆P = P and T + ∆T = T. Such type of process is called Quasi static process.

For Example, to take a gas from the state (P, T) to another state (P¹, T¹), via a quasistatic process, we change the external pressure / temperature by a very small amount and allow the system to equalise its pressure / temperature with the surroundings. Continue the process infinitely slowly till the final state (P¹, T¹) is attained.

A quasi-static process is a hypothetical construct. The process must be infinitely slow, should not involve large temperature differences or accelerated motion of the piston of the container.

Question 9.
Explain qualitatively the working of a heat engine.
Answer:
Heat engine :
A heat engine is a device used to convert heat energy into mechanical work.

Generally heat engines will work in a cyclic process. Heat engine consists of three important units.

1) Source :
Which is an object or system at high temperature. A heat engine will absorb heat energy Q₁ from source.

2) Working substance :
Every heat engine requires a working substance to do work Generally the working substance is like steam or fuel vapour and air mixture etc. A part of heat energy of working substance is converted into mechanica work.

3) Sink :
In every heat engine heat, energy content of working substance is not converted into work totally. So some energy (Q₂) is wasted or rejected by the engine. This rejected energy (Q₂) is delivered to some other body or system at low temperature. This body with low temperature is called “sink”.

Efficiency of heat engine,

where
Q₁ = heat energy supplied by source
Q₂ = heat energy delivered to sink
Block diagram of heat engine is as shown.

Long Answer Questions

Question 1.
Explain reversible and irreversible processes. Describe the working of Carnot engine. Obtain an expression for the efficiency. [AP Mar. ’18, ’17, ’16, ’14; May 18. 17, ’16; TS Mar. ’19, ’17, 15, May ’17]
Answer:
Reversible process :
In reversible process, a thermodynamic system can be retraced back in opposite direction to the changes that take place in the direct process or in forward process.

A reversible process is only an ideal concept.
Examples for reversible process :

1. Peltier effect and Seebeck effect.
2. Fusion of ice and vapourisation of water.

Irreversible process:
A thermodynamic process that cannot be taken back in opposite direction is called an “irreversible process.”

Examples:

1. Work done against friction.
2. Magnetization of materials.

Carnot’s Engine :
Carnot’s engine works on the principle of reversible process within the temperatures T₁ and T₂.

It consists of four continuous processes. The total process is known as Carnot Cycle.

Step 1 :
In Carnot cycle, the 1st step consists of isothermal expansion of gases. So temperature T is constant, P, V changes are

Step 2 :
In this stage gases will expand adiabatically. So energy to the system Q is constant.

Step 3 :
In this stage gases will be compressed isothermally. So P₁V change are

Step 4 :
In the fourth stage the gas suffers adiabatic compression and returns to original stage.

The total work done W = Q₁ – Q₂ i.e., the difference to heat energy absorbed from source and heat energy given to sink
Efficiency of Carnot engine

Question 2.
State second law of thermodynamics. How is heat engine different from a refrigerator? Explain. [TS Mar. ’18, ’16, May ’18, ’16; AP Mar. ’19, ’16, ’15, ’13; June ’15; May ’14]
Answer:
Second Law of Thermodynamics :
First law of thermodynamics is based on “Law of conservation of energy”, while second law of thermodynamics gives “information about the transformation of heat energy”.

So, there are two conventional statements of second law depending on common experience.

1) Kelvin-Plank statement:
It is impossible for an engine working in a cyclic process to extract heat from a hot body and to convert it completely into work.

2) Clausius Statement:
It is impossible for a self-acting machine, unaided by any external agency to transfer heat from a cold body to a hot reservoir. In other words, heat cannot by itself flow from a colder body to a hotter body.

Heat engine:
A heat engine is a device used to convert heat energy into mechanical work.

Generally heat engines will work in a cyclic process. Heat engine consists of three important units.

1) Source :
Which is an object or system at high temperature. A heat engine will absorb heat energy Q₁ from source.

2) Working substance :
Every heat engine requires a working substance to do work. Generally, the working substance is like steam or fuel vapour and air mixture etc. A part of heat energy of working substance is converted into mechanical work.

3) Sink :
In every heat engine heat energy content of working substance is not converted into work totally. So some energy (Q₂) is wasted or rejected by the engine. This rejected energy (Q₂) is delivered to some other body or system at low temperature. This body with low temperature is called sink.

Efficiency of heat engine,

where
Q₁ = heat energy supplied by source
Q₂ = heat energy delivered to sink
Block diagram ot heat engine is as shown.

Refrigerator :
A refrigerator works in the reverse process of heat engine.

It extracts heat energy Q₂ from sink i.e., from low temperature body with the help of external work and delivers heat energy Q₁ to high temperature body called source.
Work done W = Q₁ – Q₂.
In refrigerators external work is done on working substance.
A block diagram of refrigerator is as shown.

Difference between heat engine and refrigerator :
Refrigerator extracts heat energy from sink i.e., from low temperature body with the help of external work and delivers heat energy to high temperature body called source. A heat engine will absorb heat energy from source and reject heat energy to the sink. Heat engine will work in a reversible process but the refrigerator works in the reverse process of heat engine.

Question 3.
State second Saw of thermodynamics. Describe the working of Carnot engine. Obtain an expression for the efficiency. [AP Mar. ’16]
Answer:
Second Law of Thermodynamics :
First law of thermodynamics is based on Law of conservation of energy, while second law of thermodynamics gives information about the transformation of heat energy. So, there are two conventional statements of second law depending on common experience.

1) Kelvin-Plank statement:
It is impossible for an engine working in a cyclic process to extract heat from a hot body and to convert it completely into work.

2) Clausius Statement:
It is impossible for a self-acting machine, unaided by any external agency to transfer heat from a cold body to a hot reservoir. In other words, heat cannot by itself flow from a colder body to a hotter body.

Carnot’s Engine :
Carnot’s engine works on the principle of reversible process within the temperatures T₁ and T₂.

It consists of four continuous processes. The total process is known as Carnot Cycle.

Step 1 :
In Carnot cycle, the 1st step consists of isothermal expansion of gases. So temperature T is constant, P, V changes are

Work done in isothermal process

Step 2 :
In this stage gases will expand adiabatically. So energy to the system Q is constant.

Step 3 :
In this stage gases will be compressed isothermally. So PjV changes are

Step 4 :
In the fourth stage the gas suffers adiabatic compression and returns to original stage.

Total work done in Carnot Cycle
W = W_1,2 + W_{2, 3} + W_{3, 4} + W_{4, 1}

The total work done W = Q₁ – Q₂ i.e., the difference to heat energy absorbed from source and heat energy given to sink Efficiency of Carnot engine

Question 4.
What is the difference between heat engine and refrigerator. [TS May. ’16]
Answer:
Differences between heat engine and refrigerator:
Refrigerator extracts heat energy from sink i.e., from low temperature body with the help of external work and delivers heat energy to high temperature body called source.

A heat engine will absorb heat energy from source and reject heat energy to the sink. Heat engine will work in a reversible process but the refrigerator works in the reverse process of heat engine.

Problems

Question 1.
If a monoatomic ideal gas of volume 1 litre at N.T.P. is compressed (I) adiabatically to half of its volume, find the work done on the gas. Also find (ii) the work done if the compression is isothermal. (γ = 5/3)
Solution:
i) During an adiabatic process T₁V₁^γ-1 = T₂V₂^γ-1

ii) Work done during isothermal compression is
W = 2.3026 nRT log₁₀ \(\frac{V_2}{V_1}\)
n = number of moles = \(\frac{1}{22.4}\)
T = 273 K; R = 8.314 J mol^-1K^-1

Question 2.
Five moles of hydrogen when heated through 20 K expand by an amount of 8.3 × 10^-3m³ under a constant pressure of 10⁵ N/m². If C_v = 20 J/mole K, find C_p.
Solution:
We know that C_p – C_v = R.
Multiplying throughout by n ∆ T
nC_p ∆T – nC_v ∆T = nR ∆T
n ∆ T (C_p – C_v) = P ∆ V
5 × 20 (C_p – 20) = 10⁵ × 8.3 × 10^-3 (∵ nR∆T = P∆V)
C_p – 20 = 8.3
C_p = 28.3 J/mole K.
(∵ n = 5, ∆T = 20 K, P = 1 × 10⁵N/m² & C_v = 20 J/mole K and ∆V = 8.3 × 10³ m³)

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Matrices Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Matrices Important Questions Long Answer Type

Question 1.
Without expanding the determinant show that \(\left|\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} \\
\mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a}
\end{array}\right|\) = 2\(\left|\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{b} & \mathbf{c} & \mathbf{a} \\
\mathbf{c} & \mathbf{a} & \mathbf{b}
\end{array}\right|\) [Mar. 15 (AP); May 98, 96, 91]
Answer:

Question 2.
Show that \(\left|\begin{array}{ccc}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\) = (a – b) (b – c) (c – a) (ab + bc + ca). [Mar. 17(AP), 09: May 15 (AP); 02]
Answer:

= (a – b) (b – c) (c – a) [0 (c³ – c² (a + b + c) – (a + b) (0 – a – b – c) + (a² + ab + b²) (0 – 1)]
= (a – b) (b – c) (c – a) [0 + a² + ab + ac + ab + b² + bc – a² – ab – b²]
= (a – b) (b – c) (c – a) (ab + bc + ca) = RHS.

Question 3.
Show that \(\left|\begin{array}{ccc}
\mathbf{a}-\mathbf{b}-\mathbf{c} & \mathbf{2 a} & \mathbf{2 a} \\
\mathbf{2 b} & \mathbf{b}-\mathbf{c}-\mathbf{a} & \mathbf{2 b} \\
\mathbf{2 c} & \mathbf{2 c} & \mathbf{c}-\mathbf{a}-\mathbf{b}
\end{array}\right|\) = (a + b + c)³. [Mar. 11; May 11]
Answer:

Question 4.
Find the value of x if \(\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|\) = 0 [Mar 15 (TS); Mar. 06]
Answer:

⇒ (x – 2) (30 – 24) – (2x – 3) (10 – 6) + (3x – 4) (4 – 3) = 0
⇒ (x – 2)6 – (2x – 3)4 + (3x – 4)(1)
⇒ 6x – 12 – 8x + 12 + 3x – 4 = 0
⇒ x – 4 = 0
⇒ x = 4.

Question 5.
Show that \(\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|\) = 2 (a + b + c)³ [Mar. 18, 16 (AP); Mar. 16 (TS), 10 Mar.19 (TS), May 12, 10, 08, 03, 99]
Answer:
= 2 (a + b + c)² [1{(c + a + 2b) – 0} – a (0 – 0) + b (0 – 1)]
= 2(a + b + c)² [c + a + 2b – b]
= 2(a + b + c)² (a + b + c) = 2(a + b + c)³

Question 6.
Show that \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|^2=\left|\begin{array}{ccc}
2 b c-a^2 & c^2 & b^2 \\
c^2 & 2 a c-b^2 & a^2 \\
b^2 & a^2 & 2 a b-c^2
\end{array}\right|\) = (a³ + b³ + c³ – 3abc)². [Mar. 19 (AP) Mar. 18 (TS): May 14. 09; Mar. 12, 01]
Answer:

Question 7.
Show that \(\left|\begin{array}{ccc}
a^2+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|\) = (a – 1)³ [Mar. 13, 07]
Answer:

= (a – 1)² [(a + 1) (1 – 0) – 1 (2 – 0) + 0 (6 – 3)]
= (a – 1)² [a + 1 – 2] = (a – 1)² (a – 1) = (a – 1)³ = R.H.S.

Question 8.
Show that \(\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{a}^2 & \mathbf{b}^2 & \mathbf{c}^2 \\
\mathbf{a}^3 & \mathbf{b}^3 & \mathbf{c}^3
\end{array}\right|\) = abc (a – b) (b – c) (c – a). [May. 06]
Answer:

= abc(a – b)(b – c) [0(c² – bc – c²) – 0(c² – ac – bc) + 1(b + c – a – b)]
= abc (a – b) (b – c) (c – a) = R.H.S

Question 9.
If A = \(\left[\begin{array}{lll}
\mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\
\mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \\
\mathbf{a}_{\mathbf{3}} & \mathbf{b}_3 & \mathbf{c}_3
\end{array}\right]\) is a non-singular matrix, then show that A is invertiable and A^-1 = \(\frac{{Adj} \mathbf{A}}{{det} \mathbf{A}}\). [Mar. 17 (AP). May 15 (AP). 13, 10, 07, 06, 02, Mar. 07, 02, 99, 94, 82, 80]
Answer:

Question 10.
Solve the system of equations 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20 by using Cramer’s rule. [Mar. 12, 03; May 09]
Answer:
Given system of linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20
Let A = \(\left[\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and D = \(\left[\begin{array}{l}
18 \\
13 \\
20
\end{array}\right]\)
Then we can write the given equations in the form of matrix equation as AX = D.
Δ = det A = \(\left|\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right|\) = 3 (- 7 + 16) – 4 (14 – 40) + 5 (- 4 + 5)
= 3(9) – 4 (- 26) + 5 (1)
= 27 + 104 + 5 = 136 ≠ 0
Hence, we can solve the given equations by using Cramer’s rule.
Δ₁ = \(\left|\begin{array}{rrr}
18 & 4 & 5 \\
13 & -1 & 8 \\
20 & -2 & 7
\end{array}\right|\) = 18(- 7 + 16) – 4(91 – 160) + 5(- 26 + 20) = 18(9) – 4(- 69) + 5(- 6) = 162 + 276 – 30 = 408
Δ₂ = \(\left|\begin{array}{lll}
3 & 18 & 5 \\
2 & 13 & 8 \\
5 & 20 & 7
\end{array}\right|\) = 3(91 – 160) – 18(14 – 40) + 5(40 – 65) = 3(- 69) – 18(- 26) + 5(- 25) = – 207 + 468 – 125 = 136
Δ₃ = \(\left|\begin{array}{rrr}
3 & 4 & 18 \\
2 & -1 & 13 \\
5 & -2 & 20
\end{array}\right|\) = 3(- 20 + 26) – 4(40 – 65) + 18(- 4 + 5) = 3(6) – 4(- 25) + 18(1)
= 18 + 100 + 18 = 136
Hence, by Cramer’s rule,
x = \(\frac{\Delta_1}{\Delta}=\frac{408}{136}\) = 3, y = \(\frac{\Delta_2}{\Delta}=\frac{136}{136}\) = 1, z = \(\frac{\Delta_3}{\Delta}=\frac{136}{136}\) = 1
∴ The solution of the giveñ system of equations is x = 3, y = 1, z = 1.

Solve the system of equations 2x – y + 3z =9, x + y + z = 6, x – y + z = 2 by using Cramer’s rule. [Mar. 17 (TS), 16 (AP), 02; May 13]
Answer:
x = 1, y = 2, z = 3

Question 11.
Solve: 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by using the matrix inversion method. [Mar. ‘19 (TS): Mar. ‘15 (AP) ; Mar. ‘13. ‘08, ‘01, ‘00, 96]
Answer:
Given system of linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20
Let A = \(\left[\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
18 \\
13 \\
20
\end{array}\right]\)
This can be represented as AX = B and X = A1B is a solution.
∆ = det A = \(\left|\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right|\) = 3(- 7 + 16) – 4(14 – 40) + 5(- 4 + 5) = 3(9) – 4(- 26) + 5(1) = 27 + 104 + 5 = 136
Cofactor of 3 is A₁ = +(- 7 + 16) = 9
Cofactor of 5 is A₃ = +(32 + 5) = 37
Cofactor of -1 is B₂ = +(21 – 25) = – 4
Cofactor of 5 is C₁ = +(- 4 + 5) = 1
Cofactor of 7 is C₃ = +(- 3 – 8) = – 11
Cofactor of 2 is A₂ = – (28 + 10) = – 38.
Cofactor of 4 is B₁ = – (14 – 40) = 26
Cofactor of -2 is B₃ = – (24 – 10) = – 14
Cofactor of 8 is C₂ = – (- 6 – 20) = 26

Solve 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by using matrix Inversion method. [May 15 (AP); May. 12]
Answer:
x = 2, y = 2, z = 2

Solve x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3 by using matrix Inversion method. [May 03, 93]
Answer:
x = 7, y = – 10, z = 4

Question 12.
Solve the equations 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by Gauss-Jordan method. [May 15(TS): May 06, 01: Mar. 01]
Answer:
Given system of Linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20

In this case, system of equations have unique solution.
i.e., x = 3, y = 1, z = 1.
∴ The solution of the given system of equations is x = 3, y = 1, z = 1.

Question 13.
Solve the equation 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 by Gauss-Jordan method. [Mar. 18(AP): Mar. 11, 10; May 11]
Answer:
The given system of linear equations are 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2

In this case, system of equations has unique solution i.e., x = 1; y = 2; z = 3.
∴ The solution of given system of equations is x = 1; y = 2; z = 3.

Question 14.
Solve the following system of equations by Gauss – Jordan method : x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0. [May 10, 07; Mar. 09, 1, 99]
Answer:
Given system of linear equations are x + y + z = 9, 2x + 5y + 7z = 52; 2x + y – z = 0. The matrix form is AX = D

In this case, the system of equations has unique solution i.e., x = 1; y = 3; z 5.
∴ The solution of given system of equations is x = 1; y = 3; z = 5.

Solve the equations 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by Gauss-Jordan method. [Mar. 07]
Answer:
x = 2, y = 2, z = 2

Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by Gauss-Jordan method. [May 09; Mar. 03]
Answer:
x = 3, y = 1, z = 1

Question 15.
Examine whether the system of equations x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0 are consistent or Inconsistent and If consistent, find the complete solution. [May 15(TS); May 11]
Answer:
The given system of linear equations are x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0.

∴ Rank [A] = 3
Now, Rank [AD] = 3
Since, the 3 × 3 sub matrix is \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) whose det is 1 ≠ 0.
∴ Rank [A] = Rank [AD] = 3.
In this case, system of equations has unique solution. i.e., x = 1, y = 3, z = 5.
Hence, the given system is consistent.

Question 16.
Examine whether the system of equations x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9 are consistent or inconsistent and if consistent, find the complete solution. [Mar.11, 05]
Answer:
Given system of linear equations are x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9
The given system of equations can be written as AX = D

∴ Rank of AD = 3.
Rank [A] = Rank [AD] = 3.
In this case, system of equations has unique solution. i.e., x = 1, y = 2, z = 3.
Hence, given system is consistent.

Question 17.
Examine whether the system of equations x + y + z = 1, 2x + y + z = 2, x + 2y + 2z = 1 are consistent or Inconsistent and if consistent, find the complete solution. [Mar. 15 (TS); May 05]
Answer:
Given system of linear equations are x + y + z = 1; 2x + y + z = 2; x + 2y + 2z = 1
The system of equations can be written as AX = D

∴ Rank [A] = Rank [AD]
In this case, system of equations has infinitely many solutions. x = 1; y + z = 0
∴ The solution of the given system of equations is x = 1, y + z = 0.
Hence, the given system is consistent.

Question 18.
Examine whether the following system of equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + 4z = 1 are consistent or Inconsistent and if consistent, find the complete solution. [May. 02]
Answer:
The given system of linear equations are x + y + z = 6; x – 2y + 3z = 10; x + 2y – 4z = 1
The system of equations can be written as AX = D

∴ Rank [AD] = 3
∴ Rank [A] = Rank [AD] = 3
∴ In this case, system of equations has unique solution.
i.e., x = – 7; y = 22; z = – 9
Hence, the given system is consistent.

Question 19.
If A = \(\left[\begin{array}{ccc}
3 & 2 & -1 \\
2 & -2 & 0 \\
1 & 3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & -1 & 0 \\
2 & 1 & 3 \\
4 & -1 & 2
\end{array}\right]\) and X = A + B then find X.
Answer:

Question 20.
If \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\), then find the values of x, y, z and a.
Answer:
Given \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\)
From equality of matrices
x – 1 = 1 ⇒ x = 2
5 – y = 3 ⇒ y = 2
z – 1 = 4 ⇒ z = 5
a – 5 = 0 ⇒ a = 5
∴ x = 2, y = 2, z = 5, a = 5

Question 21.
If \(\left[\begin{array}{ccc}
x-1 & 2 & y-5 \\
z & 0 & 2 \\
1 & -1 & 1+a
\end{array}\right]=\left[\begin{array}{ccc}
1-x & 2 & -y \\
2 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\) then find the values of x, y, z and a.
Answer:
Given
\(\left[\begin{array}{ccc}
x-1 & 2 & y-5 \\
z & 0 & 2 \\
1 & -1 & 1+a
\end{array}\right]=\left[\begin{array}{ccc}
1-x & 2 & -y \\
2 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\)
From the equality of matrices,
x – 1 = 1 – x ⇒ 2x = 2 ⇒ x = 1
y – 5 = – y ⇒ 2y = 5 ⇒ y = 5/2
z = 2
1 + a = 1 ⇒ a = 0
∴ x = 1, y = 5/2, z = 2, a = 0

Question 22.
find the trace of A
if A = \(\left[\begin{array}{ccc}
1 & 2 & -1 / 2 \\
0 & -1 & 2 \\
-1 / 2 & 2 & 1
\end{array}\right]\).
Answer:
Given A = \(\left[\begin{array}{ccc}
1 & 2 & -1 / 2 \\
0 & -1 & 2 \\
-1 / 2 & 2 & 1
\end{array}\right]\)
The elements of the principal diagonal of ‘A’ are 1, – 1, 1
Hence, the trace of A = 1 + (- 1) + 1
= 1 – 1 + 1 = 1

Question 23.
If A = \(\left[\begin{array}{ccc}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) find B – A and 4A – 5B.
Answer:

Question 24.
If A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & -1 \\
-1 & 0 & 3
\end{array}\right]\) find A – B and 4B – 3A.
Answer:

Question 25.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\) do AB and BA exist? If they exist find them. Do A and B commute with respect to multiplication?
Answer:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\)
The order of matrix A is 2 × 3
The order of matrix B is 3 × 2
The no.of columns in A The no.of rows in B.
∴ AB is defined

The no.of columns in B = The no.of rows in A.
∴ BA is defined.

∴ AB ≠ BA
∴ A and B is not commute with respect to multiplication.

Question 26.
Find A² where A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\).
Answer:

Question 27.
If A = \(\left[\begin{array}{ll}
\mathrm{i} & 0 \\
0 & \mathrm{i}
\end{array}\right]\) find A².
Answer:

Question 28.
If A = \(\left[\begin{array}{ccc}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\) then find A³.
Answer:

Question 29.
If A = \(\left[\begin{array}{cc}
-1 & 2 \\
0 & 1
\end{array}\right]\) then find AA’. Do A and A’ commute with respect to multiplication of matrices?
Answer:

∴ A and A’ do not commute with respect to multiplication of matrices.

Question 30.
Find the determinant of the matrix \(\left[\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right]\)
det A = a(bc – f²) – h(ch – gf) + g(hf – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

Question 31.
Find the determinant of the matrix \(\left[\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right]\)
det A = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ – b³ + abc + abc – c³
= 3abc – a³ – b³ – c³

Question 32.
Find the adjoint and the inverse of the matrix A = \(\).
Answer:
Given A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -5
\end{array}\right]\)
Cofactor of 1 is A₁ = + (- 5) = – 5
Cofactor of 2 is B₁ = – (3) = – 3
Cofactor of 3 is A₂ = – (2) = – 2
Cofactor of – 5 is B₂ = +(1) = 1
∴ The cofactor matrix of A is

Now det A = ad – bc = 1(- 5) – 2(3)
= – 5 – 6 = – 11 ≠ 0
Hence A is invertiable.

Question 33.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) then show that A² – 4A – 5I = 0. [Mar. 16(AP)]
Answer:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\)

Question 34.
Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\)
Cofactor of 1, A₁ =+(1 – 0) = 1
Cofactor of 0, B₁ = – (2 – 0) = – 2
Cofactor of 2, C₁ = + (4 – 3) = 1
Cofactor of 2, A₂ = – (0 – 4) = 4
Cofactor of 1, B₂ = + (1 – 6) = – 5
Cofactor of 0, C₂ = – (2 – 0) = – 2
Cofactor of 3, A₃ = + (0 – 2) = – 2
Cofactor of 2, B₃ = – (0 – 4) = 4
Cofactor of 1, C₃ = + (1 – 0) = 1
∴ Cofactor matrix of A = B

Question 35.
Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\)
Cofactor of 2, A₁ = + (0 – 2) = – 2
Cofactor of 1, B₁ = – (1 – 2) = 1
Cofactor of 2, C₁ = + (2 – 0) = 2
Cofactor of 1, A₂ = – (1 – 4) = 3
Cofactor of 0, B₂ = + (2 – 4) = – 2
Cofactor of 1, C₂ = – (4 – 2) = —2
Cofactor of 2, A₃ = + (1 – 0) = 1
Cofactor of 2, B₃ = – (2 – 2) = 0
Cofactor of 1, C₃ = + (0 – 1) = – 1
∴ Cofactor matrix of A = B

Question 36.
Solve the system of equations
2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 by using Cramer’s rule.
Answer:

Question 37.
Solve 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by using matrix inversion method.
Answer:
The given system of linear equations are
2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0
Let A = \(\left[\begin{array}{crr}
2 & -1 & 3 \\
-1 & 2 & 1 \\
3 & 1 & -4
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
8 \\
4 \\
0
\end{array}\right]\)
Then we can write the given equations in the form of AX = D.
detA = 2(- 8 – 1) + 1(4 – 3) + 3(- 1 – 6)
= 2(- 9) + 1 (1) + 3(- 7)
= – 18 + 1 – 21 = – 38 ≠ 0
Hence, we can solve the given equations ¡n matrix inversion method.
Cofactor of 2 is A₁ = + (- 8 – 1) = – 9
Cofactor of – 1 is B₁ = – (4 – 3) = – 1
Cofactor of 3 is C₁ = + (- 1 – 6) = – 7
Cofactor of – 1 is A₂ = – (4 – 3) = – 1
Cofactor of 2 is B₂ = + (- 8 – 9) = – 17
Cofactor of 1 is C₂ = – (2 + 3) = – 5
Cot actor of 3 is A₃ = + (- 1 – 6) = – 7
Cofactor of 1 is B₃ = – (2 + 3) = – 5
Cofactor of – 4 is C₃ = + (4 – 1) = 3

Question 38.
Solve x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3 by using matrix inversion method.
Answer:
Given system of linear equations are
x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
1 \\
6 \\
3
\end{array}\right]\)
Then we can write the given equations in the form of AX = D.
det A = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right|\) = 1 (18 – 12) – 1 (18 – 3) + 1 (8 – 2) = 1(6) – 1(15) + 1(6)
= 6 – 15 + 6 = – 3 ≠ 0
Hence we can solve the given equations using matrix inversion method.
Cofactor of 1 is A₁ = + (18 – 12) = 6
Cofactor of 1 is B₁ = – (18 – 3) = – 15
Col actor of 1 is C₁ = + (8 – 2) = 6
Cofactor of 2 is A₂ = – (9 – 4) = – 5
Cofactor of 2 is B₂ = + (9 – 1) = 8
Cofactor of 3 is C₂ = – (4 – 1) = – 3
Cofactor of 1 is A₃ = + (3 – 2) = 1
Cofactor of 4 is B₃ = – (3 – 2) = – 1
Cofactor of 9 is C₃ = + (2 – 2) = O
∴ Cofactor matrix of A = B

∴ The solution of given system of equations is x = 7, y = – 10, z = 4.

Question 39.
Solve the equations 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by Gauss – Jordan method.
Answer:
Given system of linear equations are 2x – y + 3z = 8; – x + 2y + z = 4; 3x + y – 4z = 0.
Matrix equation form is AX = D

In this case, the system of equations has unique solution. i.e., x = y = z = 2
∴ The solution of given system of equations is x = 2; y = 2; z = 2.

Question 40.
Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by Gauss – Jordan method.
Answer:
Given system of linear equations are 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20.
Matrix equation form is AX = D

In this case, the system of equations has unique solution. i.e., x = 3, y = 1, z = 1.
∴ The solution of given system of equations is x = 3; y = 1; z = 1.

Some More Maths 1A Matrices Important Questions

Question 1.
If A = \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
6 & -1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -1 & 3
\end{array}\right]\) then find the matrix X such that A + B – X = 0. What is the order of the matrix X?
Answer:
Given A = \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
6 & -1 & 5
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -1 & 3
\end{array}\right]\)
A and B are matrices of same order 2 × 3.
If A + B – X is to be defined the order of X also must also be 2 × 3.
Given A + B – X = 0 ⇒ X = A + B

∴ Order of X is 2 × 3.

Question 2.
Construct a 3 × 2 matrix whose elements are defined by a_ij = \(\frac{1}{2}\) |i – 3j|.
Answer:
In general a 3 × 2 matrix is given by

Question 3.
If A = \(\left[\begin{array}{cc}
-1 & 3 \\
4 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{ll}
\mathbf{x}_1 & \mathbf{x}_2 \\
\mathbf{x}_3 & \mathbf{x}_4
\end{array}\right]\) and A + B = X, then find the values of x₁, x₂, x₃ and x₄.
Answer:

Question 4.
A certain book shop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs. 80, Rs. 60 and Rs. 40 each respectively. Using matrix algebra, find the total value of the books in the shop.
Answer:
Number of 3 types of books is expressed by the row matrix A

= [120 96 120]
Selling price of 3 types of books is expressed by the column matrix B

Total value of the books in the shop is given by AB
AB = [120 96 120] \(\left[\begin{array}{l}
80 \\
60 \\
40
\end{array}\right]\)
= [120 × 80 + 96 × 60 + 120 × 40]
= [9600 + 5760 + 4800]
= [20160]
∴ Total value of the books = Rs. 20160

Question 5.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\), find AB and BA, if it exists.
Answer:
Given A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\)
The order of matrix A is 2 × 2
The order of matrix 13 is 2 × 3
The no.of columns in A = The no.of rows in B
∴ AB is defined

The no.of columns in B ≠ The no.of rows in A
∴ BA is not defined.

Question 6.
Give examples of two square matrices A and B of the same order for which AB = 0. But BA ≠ 0.
Answer:

Question 7.
If A = \(\left[\begin{array}{rr}
7 & -2 \\
-1 & 2 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-2 & -1 \\
4 & 2 \\
-1 & 0
\end{array}\right]\) then find AB’ and BA’. [Mar. 18 (AP)]
Answer:

Question 8.
Find the minors of – 1 and 3 in the matrix \(\left[\begin{array}{ccc}
2 & -1 & 4 \\
0 & -2 & 5 \\
-3 & 1 & 3
\end{array}\right]\).
Answer:

Question 9.
Find the cofactors of the elements 2, – 5 in the matrix \(\left[\begin{array}{ccc}
-1 & 0 & 5 \\
1 & 2 & -2 \\
-4 & -5 & 3
\end{array}\right]\).
Answer:

Question 10.
Show that the determinant of skew – symmetric matrix of order three is always zero.
Answer:
Let A = \(\left[\begin{array}{ccc}
0 & -c & -b \\
c & 0 & -a \\
b & a & 0
\end{array}\right]\) is a skew – symmetric matrix of order ‘3’.
det A = 0(0 + a²) + c(0 + ab) – b(ac – 0)
= 0 + abc – abc = 0 + 0 = 0
∴ The determinant of skew symmetric matrix of order 3 is always zero.

Question 11.
Show that \(\left[\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & z & \mathrm{x}+\mathbf{y}
\end{array}\right]\) = 4xyz.
Answer:
LHS = \(\left[\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & z & \mathrm{x}+\mathbf{y}
\end{array}\right]\)
= (y + z) [(z + x) (x + y) – yz] – x[y(x + y) – yz] + x[yz – z(z + x)]
= (y + z) [zx + xy + zy + x² – yz] – x[xy + y² – yz] + x[yz – z² – zx]
= xyz + xy² + zy² + x²y – y²z + z² x + xyz + z²y + x²z – yz² – x²y – xy² + xyz + xyz – xz² – zx²
= 4xyz
= RHS.

Question 12.
If Δ₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \gamma & 1
\end{array}\right|\), Δ₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\) and Δ₁ = Δ₂, then show that cos²α + cos²β + cos²γ = 1.
Answer:
Δ₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \gamma & 1
\end{array}\right|\)
= 1(1 – cos²γ) – cos α(cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos² γ – cos² α + cos α cos β cos γ + cos α cos β cos γ – cos²β
= 1 – cos² γ – cos²α – cos²β + 2 cos α cos β cos γ

Δ₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\)
= 0(0 – cos²γ) – cos α (0 – cos γ cos β) + cos β (cos α cos γ – 0)
= cos α cos β cos γ + cos α cos β cos γ
= 2 cos α cos β cos γ

Given Δ₁ = Δ₂
1 – cos² α – cos²β – cos²γ + 2 cos α cos β cos γ = 2 cos α cos β cos γ
1 – cos²α – cos²β – cos²γ = 0
cos²α + cos² β + cos² γ = 1.

Question 13.
Show that \(\left|\begin{array}{ccc}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|\) = 0.
Answer:
LHS = \(\left|\begin{array}{ccc}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|\)
= 1(bc² – ab² – b²c + c²a) – a(c² – ab – b² + ac) + (a² – bc) (c – b)
= bc² – ab² – b²c + c²a – ac² + a²b + ab² – a²c + a²c – a²b – bc² . cb² = 0
= RHS.

Question 14.
Solve the following system of equations by using Cramer’s rule. [Mar. 15 (TS)]
x – y + 3z = 5, 4x + 2y – z = 0, – x + 3y + z = 5
Answer:
Given system of equations can be written as:

Question 15.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{b}-\mathbf{a} \\
\mathbf{c}-\mathbf{b} & \mathbf{c}+\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{b}-\mathbf{c} & \mathbf{a}-\mathbf{c} & \mathbf{a}+\mathbf{b}
\end{array}\right]\) then show that ABA^-1 is a diagonal matrxi.
Answer:
Given A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\),
B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{b}-\mathbf{a} \\
\mathbf{c}-\mathbf{b} & \mathbf{c}+\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{b}-\mathbf{c} & \mathbf{a}-\mathbf{c} & \mathbf{a}+\mathbf{b}
\end{array}\right]\)
Cot actor of 0 is A₁ = + (0 – 1) = – 1
Cofactor of ‘1’ is B₁ = – (0 – 1) = 1
Cofactor of 1 is C₁ = + (1 – 0) = 1
Cofactor of 1 is A₂ = – (0 – 1) = 1
Cofactor of 0 is B₂ = + (0 – 1) = – 1
Cofactor of 1 is C₂ = – (0 – 1) = 1
Cofactor of 1 is A₃ = + (1 – 0) = 1
Cofactor of 1 is B₃ = – (0 – 1) = 1
Cofactor of 0 is C₃ = +(0 – 1) = – 1
∴ Cofactor matrix of

det A = 0(0 – 1) – 1 (0 – 1) + 1 (1 – 0)
= 0 + 1 + 1 = 2 ≠ 0
∴ A is invertiable.

Question 16.
If A = \(\left[\begin{array}{rrr}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\) then show that A^-1 = A³
Answer:
Given A = \(\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\)

Cofactor of 3 is A₁ =+(- 3 + 4) = 1
Cofactor of – 3 is B₁ = – (2 – 0) = – 2
Cofactor of 4 is C₁ = (- 2 + 0) = – 2
Cofactor of 2 is A₂ = – (- 3 + 4) = – 1
Cofactor of – 3 is B₂ = + (3 – 0) = 3
Cofactor of 4 is C₂ = – (- 3 + 0) =3
Cofactor of 0 is A₃ = + (- 12 + 12) = 0
Cofactor of – 1 is B₃ = – (12 – 8) = – 4
Cofactor of 1 is C₃ = + (- 9 + 6) = – 3

Question 17.
For any square matrix A, show symmetric. [Mar. 15 (AP)]
Answer:
Let ‘A” be a square matrix
(AA’)’ = (A’)’ A’ = AA’
∴ (AA’)’ = AA’
⇒ AA’ is a symmetric matrix.

Question 18.
Find the rank of the matrix \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
det A = 1(6 – 0) – 4(4 – 0) – 1(2 – 0)
= 6 – 16 – 2 – 12 ≠ 0
∴ A is a non – singular.
Hence Rank (A) = 3.

Question 19.
Find the rank of the matrix \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\). [Mar. 19 (AP), Mar. 15 (TS)]
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\)
det A = 1 (6 – 4) – 2 (4 – 0) + 3 (2 – 0) = 2 – 8 + 6 = 0
Since det A = 0, Rank (A) ≠ 3.
Now, \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\) is a sub matrix of ‘A’ whose determinant is 3 – 4 = – 1 ≠ 0.
Hence Rank (A) = 2.

Question 20.
Solve the following system of homogeneous equations x – y + z = 0, x + 2y – z = 0, 2x + y + 3z = 0. [Mar.16 (TS)]
Answer:
The coefficient matrix is \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
1 & 2 & -1 \\
2 & 1 & 3
\end{array}\right]\)
Its determinant is 1(6 + 1) + 1(3 + 2) + 1(1 – 4) = 1(7) + 1(5) + 1(- 3) = 7 + 5 – 3 = 9
Hence the system has the trivial solution x = y = z = 0 only.

Question 21.
Solve the following system of equations by using Matrix inversion method.
2x – y + 3z = 9, x + y + z = 6, x – y + z = 2. [Mar. 16 (TS)]
Answer:

Question 22.
Solve x + y + z = 9, 2x + 5y + 7z = 52 and 2x + y – z = 0 by using matrix inversion method. [Mar. 17 (AP)]
Answer:

Question 23.
Solve the following system of equations by Cramer’s rule: 2x – y + k = 8, – x + 2y + z = 4, 3x + y – 4z = 0 [Mar. 18 (TS)]
Answer:
Given equations are
2x – y + 3z = 8,
– x + 2y + z = 4,
3x + y – 4z = 0