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

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