Learning these TS Inter 1st Year Maths 1A Formulas Chapter 5 Products of Vectors will help students to solve mathematical problems quickly.

## TS Inter 1st Year Maths 1A Products of Vectors Formulas

→ The dot or scalar product of two vectors which are non zero denoted by a̅. b̅ and defineci by a̅. b̅ = |a̅| |b̅| cos θ where θ is the angle between a̅ and b̅ which is geometrically equal to product of magnitude of one of the vectors and the projection of the other on the first vector.

→ Dot product is a scalar, if a̅ = 0 or b̅ = 0 then wre define a̅ . b̅ = 0; If we write (a, b) = 9 then a̅ . b̅ = |a̅| |b̅| cos θ if a ≠ 0. b ≠ 0. a̅ .b̅ ⇔ a̅ and b̅ are perpendicular.

- Projection of b̅ on a̅ = \(\frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}|}\)
- Orthogonal projection b̅ on a̅ = \(\frac{(\bar{a} \cdot \bar{b}) \bar{a}}{|\bar{a}|^2}\); a̅ ≠ 0

(or) \(\left(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}|^2}\right)\)a̅ and its magnitude = \(\frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}|}\) - Component vector of b̅ along a̅ (or) parallel to a̅ is \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^2}\right)\)a̅.
- Component vector of b̅ along a̅ (or) parallel to a̅ is \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{b}|^2}\right)\) b̅ and component vector of a̅ perpendicular to b̅ is b̅ = a̅ – \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^2}\)

→ If i̅, j̅, k̅ are orthogonal unit vectors then i̅. j̅ = j̅.k̅ = k̅.i̅ = 0 and i̅ . i̅ = j̅ . j̅ = k̅ . k̅ = 1

→ If a̅, b̅, c̅ are any three vectors then

- (a̅ + b̅)
^{2}= |a̅|^{2}+ |b̅|^{2}+ 2(a̅ . b̅) - (a̅ – b̅)
^{2}= |a̅|^{2}– |b̅|^{2}+ 2(a̅ . b̅) - (a̅ + b̅)
^{2}+ (a̅ – b̅)^{2}= 2(|a̅|^{2}+ |b̅|^{2})

→ If a̅ = a_{1}i̅ + a_{2}j̅ + a_{3}k̅ and b̅ = b_{1}i̅ + b_{2}j̅ + b_{3}k̅ then

- a̅.b̅ = a
_{1}b_{1}+ a_{2}b_{2}+ a_{3}b_{3}

a̅ is perpendicular to b̅ ⇔ a_{1}b_{1}+ a_{2}b_{2}+ a_{3}b_{3}= 0 - cos θ = \(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|}=\frac{a_1 b_1+a_2 b_2+a_3 b_3}{\sqrt{\Sigma a_1^2} \sqrt{\Sigma b_1^2}}\)
- a̅ is parallel to b̅ ⇔ \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}\)
- a̅.a̅ ≥ 0; |a̅.b̅| ≤ |a̅||b̅|

|a̅ + b̅| ≤ |a̅| + |b̅|; |a̅ – b̅| ≤ |a̅| + |b̅|

→ a̅ × b̅ = |a̅||b̅| sin θ n̂ is the vector product of two vectors a̅ and b̅ and n̂ is a unit vector perpendicular to the plane containing a̅ and b̅.

sin θ = \(\frac{|\bar{a} \times \bar{b}|}{|\bar{a}||\bar{b}|}\); n = \(\frac{\bar{a} \times \bar{b}}{|\bar{a} \times \bar{b}|}\)

Also a̅ × b̅ ≠ b̅ × a̅ and a̅ × b̅ = -(b̅ × a̅)

→ (i) a̅ × a̅ = 0̅ , a̅, b̅ are parallel ⇒ a̅ × b̅ = 0

- i̅ × i̅ = j̅ × j̅ = k̅ × k̅ = 0̅
- i̅ × j̅ = k̅; j̅ × k̅ = i̅ , k̅ × i̅ = j̅

(ii) If a̅ = a_{1}i̅ + a_{2}j̅ + a_{3}k̅, b̅ = b_{1}i̅ + b_{2}j̅ + b_{3}k̅ ⇒ a̅ × b̅ = \(\left|\begin{array}{ccc}

\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\

\mathrm{a}_1 & \mathrm{a}_2 & \mathrm{a}_3 \\

\mathrm{~b}_1 & \mathrm{~b}_2 & \mathrm{~b}_3

\end{array}\right|\)

→ (i) Vector area of parallelogram with adjacent sides a̅, b̅ = |a̅ x b̅|

(ii) Vector area of parallelogram with diagonals \(\overline{\mathrm{d}}_1, \overline{\mathrm{d}}_2=\frac{1}{2}\left|\overline{\mathrm{d}}_1 \times \overline{\mathrm{d}}_2\right|\)

(iii) Area of the quadrilateral with diagonals \(\overline{\mathrm{AC}}, \overline{\mathrm{BD}}=\frac{1}{2}|\overline{\mathrm{AC}} \times \overline{\mathrm{BD}}|\)

(iv) Area of ΔABC = \(\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|\)

→ The vector equation of a plane in the normal form is r̅.n̅ = p, where n̅ is a unit normal vector from the origin to the plane and p is the perpendicular distance from the origin to the plane.

→ The vector equation of a plane passing through the point A(a̅) and perpendicular to n̅ is (r̅ – a̅)n̅ = 0 or r̅.n̅ = a̅. n̅

→ The angle 0 between the planes r̅,n̅_{1} = p_{1} and r̅ . n̅_{2} = p_{2} is 0 = cos^{-1}\(\frac{\bar{n}_1 \cdot \bar{n}_2}{\left|\bar{n}_1\right|\left|\bar{n}_2\right|}\)

→ The scalar triple product (STP) of the vectors a,b, c is (a̅ × b̅) . c̅ or a̅ . (b̅ × c̅) and is denoted by [a̅ b̅ c̅].

→ The magnitude |[a̅ b̅ c̅]| gives the volume of the parallelopiped with a̅, b̅, c̅ as its coterminus edges.

→ If a̅ = a_{1}i̅ + a_{2} j̅ + a_{3}k̅, b̅ = b_{1} i̅ + b_{2} j̅ + b_{3}k̅, c̅ = c_{1} i̅ + c_{2} j̅ + c_{3}k̅ then

[a̅ b̅ c̅] = \(\left|\begin{array}{lll}

a_1 & a_2 & a_3 \\

b_1 & b_2 & b_3 \\

c_1 & c_2 & c_3

\end{array}\right|\)

- Volume of the tetrahedron with a, b,c as its coterminus edges is V = \(\frac{1}{6}\)| [a̅ b̅ c̅]|
- Volume of tetrahedron ABCD is V = \(\frac{1}{6}|[\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]|\)

→ Three vectors a, b,c are coplanar ⇔ [a̅ b̅ c̅] = 0 and a, b,c are non – coplanar ⇔ [a̅ b̅ c̅] ≠ 0

→ The shortest distance between the skew lines r̅ = a̅ + tb̅ and r̅ = c̅ + sd̅ where s, t are scalars is \(\)

→ Vector triple product of three vectors a̅, b̅, c̅ is a vector defined by

(a̅ × b̅) × c̅ = (a̅. c̅)b̅ – (c̅ . b̅)a̅ (or) a̅ × (b̅ × c̅) = (a̅. c̅)b̅ – (a̅ . b̅)c̅

→ The vector equation of a plane passing through the point A(a̅) and parallel to two non – collinear vectors b̅ and c̅ is [r̅ b̅ c̅] = [a̅ b̅ c̅]

→ The vector equation of a plane passing through A(a̅), B(b̅) and parallel to the vector is [r̅ b̅ c̅] + [r̅ c̅ a̅] = [a̅ b̅ c̅]

→ The vector equation of a plane passing through three non colilnear points A(a̅), B(b̅) and C(c̅) is [r̅ b̅ c̅] + [r̅ c̅ a̅] + [r̅ a̅ b̅] = [a̅ b̅ c̅]

→ The vector equation of the plane containing the line r̅ = a̅ + tb̅; t ∈ R and perpendicular to the plane r̅.c̅ = q is [r̅ b̅ c̅] = [a̅ b̅ c̅]

- If a̅, b̅ are two non zero and non parallel vectors then |a̅ × b̅|
^{2}= a^{2}b^{2}– (a̅ . b̅) = \(\left|\begin{array}{cc}

\bar{a} \cdot \bar{a} & \bar{a} \cdot \bar{b} \\

\bar{a} \cdot \bar{b} & \bar{b} \cdot \bar{b}

\end{array}\right|\) - For any vector a̅, (a̅ × i̅) + (a̅ × j̅) + (a̅ × k̅) = 2|a̅|.
- If a̅, b̅, c̅ are the position vectors of the points A, B, C then the perpendicular distance from C to the line AB is = \(\frac{|\overline{\mathrm{AC}} \times \overline{\mathrm{AB}}|}{|\overline{\mathrm{AB}}|}=\frac{|(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})|}{|\overline{\mathrm{b}}-\overline{\mathrm{a}}|}\).