OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors (Continued) Ex 22(a)

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S Chand Class 12 ICSE Maths Solutions Chapter 22 Vectors (Continued) Ex 22(a)

Question 1.
If \(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\),
\(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and
\(\vec{c}\) = 5 \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{k}\), then find the value of
(i) \(\vec{a}\) . \(\vec{b}\)
(ii) \(\vec{b}\) . \(\vec{c}\)
(iii) \(\vec{c}\) \(\vec{a}\)
(iv) |\(\vec{a}\)|
(v) |\(\vec{b}\)|
(vi) |\(\vec{c}\)|
(vii) \(\vec{a}\) (\(\vec{b}\) + \(\vec{c}\)), \(\vec{a}\) \(\vec{b}\) + \(\vec{a}\) \(\vec{c}\) and show that they are equal.
(viii) (2 \(\vec{a}\) – \(\vec{b}\) + 2 \(\vec{c}\)) \(\vec{c}\)
(ix) (\(\vec{a}\) – 2 \(\vec{b}\)) (\(\vec{b}\) – 2 \(\vec{c}\))
(x) angle between \(\vec{a}\) and \(\vec{b}\), \(\vec{b}\) and \(\vec{c}\) and \(\vec{c}\) and \(\vec{a}\).
Answer:
Given \(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\);
\(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\);
\(\vec{c}\) = 5 \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{k}\)
(i) \(\vec{a}\) \(\vec{b}\)
= (\(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\)) (2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
= 1(2) + 2(1) – 1
(1) = 4 – 1 = 3
[∵ \(\hat{i}\) \(\hat{i}\) = 1 = \(\hat{j}\) \(\hat{j}\)
= \(\hat{k}\) \(\hat{k}\) and \(\hat{i}\) \(\hat{j}\)
= 0 = \(\hat{j}\) \(\hat{k}\) = \(\hat{k}\) \(\hat{i}\)

(ii) \(\vec{b}\) \(\vec{c}\)
= (2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) (5 \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{k}\))
= 2(5) + 1(-4) + 1(3) = 9

(iii) \(\vec{c}\) \(\vec{a}\)
= (5 \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{k}\)) (\(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\))
= 5(1) – 4(2) + 3(-1) = -6

(iv) |\(\vec{a}\)|
= \(\sqrt{1^2+2^2+(-1)^2}\)
= \(\sqrt{6}\)

(v) |\(\vec{b}\)|
= \(\sqrt{2^2+1^2+1^2}\)
= \(\sqrt{6}\)

(vi) |\(\vec{c}\)|
= \(\sqrt{5^2+(-4)^2+3^2}\)
= \(\sqrt{25+16+9}\)
= \(\sqrt{50}\)
= 5 \(\sqrt{2}\)

(vii) Now \(\vec{b}\) + \(\vec{c}\)
= (2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) + (5 \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{k}\))
= 7 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\)
∴ \(\vec{a}\)(\(\vec{b}\) + \(\vec{c}\))
= (\(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\))(7 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\))
= 1(7) + 2(-3) – 1
= 7 – 6 – 4 = -3
\(\vec{a}\) \(\vec{b}\) + \(\vec{a}\) \(\vec{c}\)
= 3 + (-6) = -3
From eqn. (1) and eqn. (2); \(\vec{a}\) (\(\vec{b}\) + \(\vec{c}\))
= \(\vec{a}\) \(\vec{b}\) + \(\vec{a}\) \(\vec{c}\)

(viii) 2 \(\vec{a}\) – \(\vec{b}\) + 2 \(\vec{c}\)
= 2(\(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\)) – (2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) + 2(5 \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{k}\))
= 10 \(\hat{i}\) – 5 \(\hat{j}\) + 3 \(\hat{k}\)
Thus (2 \(\vec{a}\) – \(\vec{b}\) + 2 \(\vec{c}\)) \(\vec{c}\)
= (10 \(\hat{i}\) – 5 \(\hat{j}\) + 3 \(\hat{k}\)) (5 \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{k}\))
= 10(5) – 5(-4) + 3(3)
= 50 + 20 + 9
= 79

(ix) \(\vec{a}\) – 2 \(\vec{b}\)
= (\(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\)) – 2(2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
= -3 \(\hat{i}\) – 3 \(\hat{k}\)
\(\vec{b}\) – 2 \(\vec{c}\)
= (2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) – 2(5 \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{k}\))
= -8 \(\hat{i}\) + 9 \(\hat{j}\) – 5 \(\hat{k}\)
Thus, (\(\vec{a}\) – 2 \(\vec{b}\))(\(\vec{b}\) – 2 \(\vec{c}\))
= (-3 \(\hat{i}\) – 3 \(\hat{k}\)) (-8 \(\hat{i}\) + 9 \(\hat{j}\) – 5 \(\hat{k}\))
= -3(-8) + 0(9) – 3(-5)
= 39

(x) Let θ₁ be the angle between \(\vec{a}\) and \(\vec{b}\)
∴ cosθ₁ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
= \(\frac{3}{\sqrt{6} \sqrt{6}}\)
[using (i), (iv) and (v)]
⇒ cos θ₁ = \(\frac{3}{6}\)
= \(\frac{1}{2}\)
⇒ θ₁ = \(\frac{\pi}{3}\)
Let θ₂ be the angle between \(\vec{b}\) and \(\vec{c}\)
∴ cos θ₂ = \(\frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|}\)
= \(\frac{9}{\sqrt{6} \times 5 \sqrt{2}}\)
[using (ii), (v) and (vi)]
⇒ cosθ₂ = \(\frac{9}{5 \times 2 \sqrt{3}}\)
⇒ θ₂ = cos^-1 (\(\frac{9}{10 \sqrt{3}}\))
Let θ₃ be the angle between \(\vec{c}\) and \(\vec{a}\)
∴ cos θ₃ = \(\frac{\vec{c} \cdot \vec{a}}{|\vec{c}||\vec{a}|}\)
= \(\frac{-6}{5 \sqrt{2} \times \sqrt{6}}\)
[using (iii), (iv) and (vi)]
⇒ cos θ₃ =
\(\frac{-6}{10 \sqrt{3}}\) = \(\frac{-3}{5 \sqrt{3}}\)
= \(\frac{-\sqrt{3}}{5}\)
⇒ θ₃ = cos^-1 (\(\frac{-\sqrt{3}}{5}\))

Question 2.
If \(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\),
\(\vec{b}\) = 3 \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\) and
\(\vec{c}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) show that \(\vec{a}\)
(\(\vec{b}\) + \(\vec{c}\)) = \(\vec{a}\) \(\vec{b}\) + \(\vec{a}\) \(\vec{c}\)
Answer:
Given \(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\);
\(\vec{b}\) = 3 \(\hat{i}\) –  \(\hat{j}\) + 4 \(\hat{k}\) and \(\vec{c}\) = \(\hat{i}\) –\(\hat{j}\) + \(\hat{k}\)
∴ \(\vec{b}\) + \(\vec{c}\) = (3 \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\)) + (\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
= 4 \(\hat{i}\) – 2 \(\hat{j}\) + 5 \(\hat{k}\)
L.H.S. = \(\vec{a}\) (\(\vec{b}\) + \(\vec{c}\))
= (\(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\))(4 \(\hat{i}\) – 2 \(\hat{j}\) + 5 \(\hat{k}\))
= 1(4) + 2(-2) + 1(5) = 5
Now
\(\vec{a}\) \(\vec{b}\) = (\(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)) (3 \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\))
= 1(3) + 2(-1) + 1(4) = 5
\(\vec{a}\) \(\vec{c}\) = (\(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)) .(\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
= 1(1) + 2(-1) + 1(1) = 0
R.H.S. = \(\vec{a}\) .\(\vec{b}\) + \(\vec{a}\).\(\vec{c}\)
= 5 + 0 = 5
∴ L.H.S. = R.H.S.
Thus \(\vec{a}\) (\(\vec{b}\) + \(\vec{c}\)) = \(\vec{a}\). \(\vec{b}\) + \(\vec{a}\).\(\vec{c}\)

Question 3.
Compute the angle between \(\vec{a}\) and \(\vec{b}\), where
(i) \(\vec{a}\) = 3 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\) and \(\vec{b}\)
= 2 \(\hat{i}\) – 2 \(\hat{j}\) + 4 \(\hat{k}\)
(ii) \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\),
\(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{k}\)
(iii) \(\vec{a}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\) and
\(\vec{b}\) = \(\hat{i}\) – 2 \(\hat{j}\) – 3 \(\hat{k}\).
Answer:
(i) Let θ be the angle between \(\vec{a}\) and \(\vec{b}\)

Question 4.
Prove that the following pair of vectors are perpendicular to each other :
(i) \(\vec{a}\) = 2 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\) and \(\vec{b}\)
= \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
(ii) \(\overrightarrow{\mathbf{A}}\) = a \(\hat{i}\) + b \(\hat{j}\) + c \(\hat{k}\) and
\(\overrightarrow{\mathbf{B}}\) = (b – c) \(\hat{i}\) + (c – a) \(\hat{j}\) + (a – b) \(\hat{k}\)
Answer:
(i) Given \(\vec{a}\) = 2 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\) and
\(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
∴ \(\vec{a}\). \(\vec{b}\) = 2(1) + 1(2) – 4(1) = 4 – 4 = 0
Thus \(\vec{a}\) ⊥ \(\vec{b}\)

(ii) \(\overrightarrow{\mathrm{A}}\) = a \(\hat{i}\) + b \(\hat{j}\) + c \(\hat{k}\) and
\(\overrightarrow{\mathrm{B}}\) = (b – c) \(\hat{i}\) + (c – a) \(\hat{j}\) + (a – b) \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{A}}\) .\(\overrightarrow{\mathrm{B}}\)
= (a \(\hat{i}\) + b \(\hat{j}\) + c \(\hat{k}\)).[(b – c) \(\hat{i}\) + (c – a) \(\hat{j}\) + (a – b) \(\hat{k}\)]
= a(b – c) + b(c – a) + c(a – b)
= a b – a c + b c – a b + c a – b c = 0
Thus \(\vec{A}\) is ⊥ to \(\vec{B}\)

Question 5.
(i) If |\(\vec{a}\)| and |\(\vec{b}\)| are two vectors such that |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3 and \(\vec{a}\) .\(\vec{b}\) = 3, find the angle between \(\vec{a}\) and \(\vec{b}\).
(ii) If \(\vec{a}\) and \(\vec{b}\) are two vectors such that |\(\vec{a}\)| = 1,
|\(\vec{b}\)| = 2 and \(\vec{a}\).\(\vec{b}\) = 0, find the angle between the two vectors.
Answer:
(i) Given |\(\vec{a}\)| = 2,
|\(\vec{b}\)| = 3 and \(\vec{a}\).\(\vec{b}\) = 3
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\)
∴ cos θ
= \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
= \(\frac{3}{2 \times 3}\)
= \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\) or 60°

(ii) Given |\(\vec{a}\)| = 1; |\(\vec{b}\)| = 2 and \(\vec{a}\).\(\vec{b}\) = 0
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\)
∴ cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\) = 0
⇒ θ = \(\frac{\pi}{2}\)

Question 6.
If A is (1, 3, 2), B is (-1, 1, 1) and C is (-3, 2, 3), prove that \(\overrightarrow{A B}\).\(\overrightarrow{B C}\) = 0. what can be said about ∆ ABC?
Answer:
Given P.V. of A = \(\hat{i}\) + 3 \(\hat{j}\) + 2\(\hat{k}\);
P.V. of B = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and
P.V. of C = -3 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of \(\mathrm{B}\)
P.V of A = (\(-\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) – (\(\hat{i}\) + 3 \(\hat{j}\) + 2 \(\hat{k}\))
= -2 \(\hat{i}\) – 2 \(\hat{j}\) – \(\hat{k}\)
\(\overrightarrow{\mathrm{BC}}\) = P.V. of C – P.V. of B
= (-3 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) – (\(-\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
= -2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\).\(\overrightarrow{\mathrm{BC}}\)
= (-2 \(\hat{i}\) – 2 \(\hat{j}\) – \(\hat{k}\)) (-2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\))
= (-2)(-2) – 2(1) – 1(2) = 0
Thus the ∆ABC is right angled ∆ and right angled at B.

Question 7.
(i) Given \(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\),
\(\vec{b}\) = 3 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\),
show that (\(\vec{a}\) + \(\vec{b}\)) is perpendicular to (\(\vec{a}\) – \(\vec{b}\)).
(ii) Find the value of x so that the pairs of vectors 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and 4 \(\hat{i}\) – 2 \(\hat{j}\) + x \(\hat{k}\) may be perpendicular to each other.
(iii) If \(\vec{a}\) = (5 \(\hat{i}\) – \(\hat{j}\) + 7 \(\hat{k}\)) and
\(\vec{b}\) = (\(\hat{i}\) – \(\hat{j}\) – λ\(\hat{k}\)), find the value of λ for which (\(\vec{a}\) + \(\vec{b}\)) and (\(\vec{a}\) – \(\vec{b}\)) are orthogonal.
(iv) Find the angle between the vectors (\(\vec{a}\) + \(\vec{b}\)) and (\(\vec{a}\) – \(\vec{b}\))
if \(\vec{a}\) = (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)) and \(\vec{b}\) = 3 \(\hat{i}\) + \(\hat{j}\) – 2 \(\hat{k}\).
Answer:
(i) Given \(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) and
\(\vec{b}\) = 3 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\vec{a}\) + \(\vec{b}\) = (\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) + (3 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= 4 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
\(\vec{a}\) – \(\vec{b}\) = (\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) – (3 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= -2 \(\hat{i}\) + 3 \(\hat{j}\) – 5 \(\hat{k}\)
Now (\(\vec{a}\) + \(\vec{b}\)) (\(\vec{a}\) – \(\vec{b}\))
= (4 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) (-2 \(\hat{i}\) + 3 \(\hat{j}\) – 5 \(\hat{k}\))
= 4(-2) + 1(3) – 1(-5) = 0
Thus \(\vec{a}\) + \(\vec{b}\) is ⊥ to \(\vec{a}\) – \(\vec{b}\).

(ii) Let \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and \(\vec{b}\)
= 4 \(\hat{i}\) – 2 \(\hat{j}\) + x \(\hat{k}\)
Now \(\vec{a}\) ⊥ \(\vec{b}\)
∴ \(\vec{a}\).\(\vec{b}\) = 0
⇒ (2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)).(4 \(\hat{i}\) – 2 \(\hat{j}\) + x \(\hat{k}\)) = 0
⇒ 2(4) + 3(-2) + 4(x) = 0
⇒ 2 + 4 x = 0
⇒ x = \(-\frac{1}{2}\)

(iii) Given \(\vec{a}\) = 5 \(\hat{i}\) – \(\hat{j}\) + 7 \(\hat{k}\) and \(\vec{b}\)
= \(\hat{i}\) – \(\hat{j}\) – λ \(\hat{k}\)
∴ \(\vec{a}\) + \(\vec{b}\) = (5 \(\hat{i}\) – \(\hat{j}\) + 7 \(\hat{k}\)) + (\(\hat{i}\) – \(\hat{j}\) – λ \(\hat{k}\))
= 6 \(\hat{i}\) – 2 \(\hat{j}\) + (7 – λ) \(\hat{k}\)
\(\vec{a}\) – \(\vec{b}\) = (5 \(\hat{i}\) – \(\hat{j}\) + 7 \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) – λ \(\hat{k}\))
= 4 \(\hat{i}\) + 0 \(\hat{j}\) + (7 + λ) \(\hat{k}\)
Since \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\) are orthogonal to each other.
∴(\(\vec{a}\) + \(\vec{b}\)).(\(\vec{a}\) – \(\vec{b}\)) = 0
⇒ [6 \(\hat{i}\) – 2 \(\hat{j}\) + (7 – λ) \(\hat{k}\)].[4 \(\hat{i}\) + 0 \(\hat{j}\) + (7 + λ) \(\hat{k}\)] = 0
⇒ 6(4) – 2(0) + (7 – λ)(7 + λ) = 0
⇒ 24 + 49 – λ² = 0
⇒ λ² = 73
⇒ λ = ± \(\sqrt{73}\)

(iv) Given \(\vec{a}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\) and \(\vec{b}\)
= 3 \(\hat{i}\) + \(\hat{j}\) – 2 \(\hat{k}\)
∴ \(\vec{a}\) + \(\vec{b}\)
= 5 \(\hat{i}\) + 0 \(\hat{j}\) + \(\hat{k}\);
\(\vec{a}\) – \(\vec{b}\)
= \(-\hat{i}\) – 2 \(\hat{j}\) + 5 \(\hat{k}\)
|\(\vec{a}\) + \(\vec{b}\)| = \(\sqrt{5^2+1^2}\) = \(\sqrt{26}\) ;
|\(\vec{a}\) – \(\vec{b}\)| = \(\sqrt{(-1)^2+(-2)^2+5^2}\)
= \(\sqrt{30}\)
and (\(\vec{a}\) + \(\vec{b}\)).(\(\vec{a}\) – \(\vec{b}\))
= 5(-1) + 0(-2) + 1(5) = 0
Lét θ be the angle between \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\)
∴ cosθ
= \(\frac{(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})}{|\vec{a}+\vec{b}||\vec{a}-\vec{b}|}\) = 0
⇒ θ = \(\frac{\pi}{2}\)

Question 8.
Find the angles of the triangle whose vertices are A(0, -1, -2), B(3, 1, 4) and C(5, 7, 1).
Answer:
Given P.V. of A = 0 \(\hat{i}\) – \(\hat{j}\) – 2 \(\hat{k}\);
P.V. of B = 3 \(\hat{i}\) + \(\hat{j}\) + 4 \(\hat{k}\)
and
P.V. of C = 5 \(\hat{i}\) + 7 \(\hat{j}\) + \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= (3 \(\hat{i}\) + \(\hat{j}\) + 4 \(\hat{k}\)) – (0 \(\hat{i}\) – \(\hat{j}\) – 2 \(\hat{k}\))
= 3 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\)
\(\overrightarrow{\mathrm{BC}}\) = P.V. of C – P.V. of B
= (5 \(\hat{i}\) + 7 \(\hat{j}\) + \(\hat{k}\)) – (3 \(\hat{i}\) + \(\hat{j}\) + 4 \(\hat{k}\))
= 2 \(\hat{i}\) + 6 \(\hat{j}\) – 3 \(\hat{k}\)
\(\overrightarrow{\mathrm{CA}}\) = P.V. of A – P.V. of C
= (0 \(\hat{i}\) – \(\hat{j}\) – 2 \(\hat{k}\)) – (5 \(\hat{i}\) + 7 \(\hat{j}\) + \(\hat{k}\))
= -5 \(\hat{i}\) – 8 \(\hat{j}\) – 3 \(\hat{k}\)

Clearly π – A be the angle between \(\overrightarrow{\mathrm{CA}}\) and \(\overrightarrow{\mathrm{AB}}\)

Question 9.
A is the point (1, 3, -2), B is (3, -4, 1), C is (-1, 0, 2). Find the lengths of the sides of the triangle A B C and the cosines of the angles A, B, C.
Answer:
Given
P.V. of A = \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\);
P.V. of B = 3 \(\hat{i}\) – 4 \(\hat{j}\) + \(\hat{k}\)
and
P.V. of C = \(-\hat{i}\) + 0 \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= (3 \(\hat{i}\) – 4 \(\hat{j}\) + \(\hat{k}\)) – (\(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\))
= 2 \(\hat{i}\) – 7 \(\hat{j}\) + 3 \(\hat{k}\)
\(\overrightarrow{\mathrm{BC}}\) = P.V. of C – P.V. of B
= (\(-\hat{i}\) + 0 \(\hat{j}\) + 2 \(\hat{k}\)) – (3 \(\hat{i}\) – 4 \(\hat{j}\) + \(\hat{k}\))
= -4 \(\hat{i}\) + 4 \(\hat{j}\) + \(\hat{k}\)
\(\overrightarrow{\mathrm{CA}}\) = P.V. of A – P.V. of C
= (\(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\)) – (\(-\hat{i}\) + 0 \(\hat{j}\) + 2 \(\hat{k}\))
= 2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)
|\(\overrightarrow{\mathrm{AB}}\)| = \(\sqrt{2^2+(-7)^2+3^2}\)
= \(\sqrt{62}\)
|\(\overrightarrow{\mathrm{CA}}\)| = \(\sqrt{2^2+3^2+(-4)^2}\)
= \(\sqrt{29}\)

Since π – A is the angle between \(\overrightarrow{\mathrm{CA}}\) and \(\overrightarrow{\mathrm{AB}}\)

Question 10.
Find the projection of:
(i) \(\vec{a}\) in the direction of \(\vec{b}\) if \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\),
\(\vec{b}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\).
(ii) \(\vec{b}\) in the direction of \(\vec{a}\) if \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\), \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\).
Answer:
(i) Given \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) and \(\vec{b}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\)
∴ Projection of \(\vec{a}\) on \(\vec{b}\)
= \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
= \(\frac{(2 \hat{i}+3 \hat{j}) \cdot(3 \hat{i}+2 \hat{j})}{\sqrt{3^2+2^2}}\)
= \(\frac{2(3)+3(2)}{\sqrt{13}}\)
= \(\frac{12}{\sqrt{13}}\)
(ii) Given \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\) and
\(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
∴ \(\vec{b}\).\(\vec{a}\)
= (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\))
= 1(2) + 2(3) + 3(-2) = 2
and |\(\vec{a}\)| = \(\sqrt{2^2+3^2+(-2)^2}\)
= \(\sqrt{17}\)
Thus, projection of \(\vec{b}\) on \(\vec{a}\)
= \(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}\)
= \(\frac{2}{\sqrt{17}} \times \frac{\sqrt{17}}{\sqrt{17}}\)
= \(\frac{2(\sqrt{17})}{17}\)

Question 11.
Find the projection of \(\vec{b}\) + \(\vec{c}\) on \(\vec{a}\) where \(\vec{a}\) = 2 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\),
\(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\) and
\(\vec{c}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\).
Answer:
Given \(\vec{a}\) = 2 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\);
\(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\) and
\(\vec{c}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\)
∴ \(\vec{b}\) + \(\vec{c}\)
= 3 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
Thus (\(\vec{b}\) + \(\vec{c}\)) \(\vec{a}\)
= (3 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\))(2 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\))
= 3(2) + 1(-2) + 2(1) = 6
and
|\(\vec{a}\)| = \(\sqrt{2^2+(-2)^2+1^2}\) = 3
Thus, projection of \(\vec{b}\) + \(\vec{c}\) on \(\vec{a}\)
= \(\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}\)
= \(\frac{6}{3}\) = 2

Question 12.
Using vectors, find the projection of the line joining the points (1, -1, 1) and (-2, 3, 4) on the line joining the points (4, -3, 5) and (7, 8, 7).
Answer:
Vector along the line joining A(1, -1, 1) and B(-2, 3, 4) be
\(\overrightarrow{A B}\) = P.V. of B – P.V. of A
= (-2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
= -3 \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\)
Vector along the line joining C(4, -3, 5)and D(7, 8, 7)be
\(\overrightarrow{C D}\) = P.V. of D – P . V. of C
∴ \(\overrightarrow{\mathrm{CD}}\)
= (7 \(\hat{i}\) + 8 \(\hat{j}\) + 7 \(\hat{k}\)) – (4 \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\))
= 3 \(\hat{i}\) + 11 \(\hat{j}\) + 2 \(\hat{k}\)
∴ |\(\overrightarrow{\mathrm{CD}}\)|
= \(\sqrt{3^2+11^2+2^2}\) = \(\sqrt{134}\)
\(\overrightarrow{\mathrm{AB}}\).\(\overrightarrow{\mathrm{CD}}\)
= (-3 \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\)) (3 \(\hat{i}\) + 11 \(\hat{j}\) + 2 \(\hat{k}\))
= -3(3) + 4(11) + 3(2)
= 41
Thus required projection of \(\overrightarrow{\mathrm{AB}}\) on \(\overrightarrow{\mathrm{CD}}\)
= \(\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}}{|\overrightarrow{\mathrm{CD}}|}\)
= \(\frac{41}{\sqrt{134}}\)

Question 13.
Find the vectors projection of the vector 7 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\) on
(i) 2 \(\hat{i}\) + 6 \(\hat{j}\) + 3 \(\hat{k}\)
(ii) 7 \(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\).
Answer:
(i) Let \(\vec{a}\) = 7 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\)
and \(\vec{b}\) = 2 \(\hat{i}\) + 6 \(\hat{j}\) + 3 \(\hat{k}\)
∴ Vector projection of \(\vec{a}\) on \(\vec{b}\)
= \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) \(\hat{b}\)
\(\vec{a}\).\(\vec{b}\)
= (7 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\)).(2 \(\hat{i}\) + 6 \(\hat{j}\) + 3 \(\hat{k}\))
= 7(2) + 1(6) – 4(3) = 14 + 6 – 12 = 8
|\(\vec{b}\)| = \(\sqrt{2^2+6^2+3^2}\)
= 7 and
\(\hat{b}\) = \(\frac{\vec{b}}{|\vec{b}|}\)
= \(\frac{2 \hat{i}+6 \hat{j}+3 \hat{k}}{7}\)
∴ from (1); Required vector projection \(\vec{a}\) on \(\vec{b}\)
= \(\frac{8}{7}\) × \(\frac{2 \hat{i}+6 \hat{j}+3 \hat{k}}{7}\)
= \(\frac{16}{49}\) \(\hat{i}\) + \(\frac{48}{49}\) \(\hat{j}\) + \(\frac{24}{49}\) \(\hat{k}\)

(ii) Given \(\vec{a}\) = 7 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\);
\(\vec{b}\) = 7 \(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\)
∴ \(\vec{a}\).\(\vec{b}\) = (7 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\))(7 \(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\))
= 7(7) + 1(1) – 4(-3) = 49 + 1 + 12 = 62
|\(\vec{b}\)| = \(\sqrt{7^2+1^2+(-3)^2}\) = \(\sqrt{59}\)
Thus, b = \(\frac{\vec{b}}{|\vec{b}|}\)
= \(\frac{7 \hat{i}+\hat{j}-3 \hat{k}}{\sqrt{59}}\)
Thus, vector projection of \(\vec{a}\) on \(\vec{b}\)
= \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) \(\hat{b}\)
= \(\frac{62}{\sqrt{59}} \frac{(7 \hat{i}+\hat{j}-3 \hat{k})}{\sqrt{59}}\)
= \(\frac{62}{59}\)(7 \(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\))

Question 14.
Let \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\) be vectors such that \(\vec{u}\) + \(\vec{v}\) + \(\vec{w}\) = \(\overrightarrow{0}\).
If |\(\vec{u}\)| = 3,|\(\vec{v}\)| = 4 and |\(\vec{w}\)| = 5, then find \(\vec{u}\).\(\vec{v}\) + \(\vec{v}\).\(\vec{w}\) + \(\vec{w}\).\(\vec{u}\).
Answer:
Given \(\vec{u}\) + \(\vec{v}\) + \(\vec{w}\)
= \(\overrightarrow{0}\)
⇒ (\(\vec{u}\) + \(\vec{v}\) + \(\vec{w}\))² = 0
⇒ (\(\vec{u}\) + \(\vec{v}\) + \(\vec{w}\)) .(\(\vec{u}\) + \(\vec{v}\) + \(\vec{w}\)) = 0
⇒ \(\vec{u}\)² + \(\vec{v}\)² + \(\vec{w}\)² + 2(\(\vec{u}\). \(\vec{v}\) + \(\vec{v}\).\(\vec{w}\) + \(\vec{w}\).\(\vec{u}\)) = 0
Given |\(\vec{u}\)| = 3;
|\(\vec{v}\)| = 4, |\(\vec{w}\)| = 5
∴ from (1); we have
|\(\vec{u}\)|² + |\(\vec{v}\)|² + |\(\vec{w}\)|² + 2(\(\vec{u}\). \(\vec{v}\) + \(\vec{v}\).\(\vec{w}\) + \(\vec{w}\).\(\vec{u}\)) = 0
⇒ 9 + 16 + 25 + 2(\(\vec{u}\).\(\vec{v}\) + \(\vec{v}\).\(\vec{w}\) + \(\vec{w}\).\(\vec{u}\)) = 0
⇒ \(\vec{u}\).\(\vec{v}\) + \(\vec{v}\).\(\vec{w}\) + \(\vec{w}\).\(\vec{u}\) = -25

Question 15.
Find the angles which the vector \(\vec{a}\) = \(\hat{i}\) – \(\hat{j}\) + \(\sqrt{2}\) \(\hat{k}\) makes with the coordinate axes.
Answer:
Let α, β and γ are the angles made by the vectors \(\vec{a}\) with x-axis, y-axis and z-axis.

Question 16.
Dot product of a vector with \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\), \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\) and 2 \(\hat{i}\) + \(\hat{j}\) + 4 \(\hat{k}\) are 0,5 and 8 respectively. Find the vector.
Answer:
Let the required vector be \(\vec{a}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) it is given that \(\vec{a}\) .(\(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\)) = 0
⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) .(\(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\)) = 0
⇒ x + y – 3 z = 0
Also,
\(\vec{a}\) (\(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\)) = 5
⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)).(\(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\)) = 5
⇒ x + 3 y – 2 z = 5
Also,
\(\vec{a}\) (2 \(\hat{i}\) + \(\hat{j}\) + 4 \(\hat{k}\)) = 8
⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (2 \(\hat{i}\) + \(\hat{j}\) + 4 \(\hat{k}\)) = 8
⇒ 2 x + y + 4 z = 8
eqn. (2) – eqn. (1) gives ;
2 y + z = 5
eqn. (3) -2 × eqn. (2); we have
-5 y + 8 z = -2
eqn. (5) – 8 eqn. (4); we have
∴ from (4);
-21 y = -2 – 40 = -42
⇒ y = 2
∴ from (1) ;
4 + z = 5
⇒ z = 1
x + 2 – 3 = 0
⇒ x = 1
Thus, the required vector be \(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)

Question 17.
If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference is \(\sqrt{3}\).
Answer:
Given |\(\vec{a}\)| = 1;
|\(\vec{b}\)| = 1 and
|\(\vec{a}\) + \(\vec{b}\)| = 1
We know that, |\(\vec{a}\) + \(\vec{b}\)|² + |\(\vec{a}\) – \(\vec{b}\)|²
= 2(|\(\vec{a}\)|² + |\(\vec{b}\)|² )
⇒ 1 + |\(\vec{a}\) – \(\vec{b}\)|² = 2(1 + 1)
⇒ |\(\vec{a}\) – \(\vec{b}\)|² = 4 – 1 = 3
⇒ |\(\vec{a}\) – \(\vec{b}\)| = \(\sqrt{3}\)
[using eqn. (1)]

Question 18.
If a and b are unit vectors inclined at the angle \theta, then prove that
(i) cos \(\frac{\theta}{2}\) = \(\frac{1}{2}\)|\(\hat{a}\) + \(\hat{b}\)|,
(ii) tan \(\frac{\theta}{2}\) = \(\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}\)
Answer:
(i) Given |\(\hat{a}\)| = 1 = |\(\hat{b}\)|
We know that, |\(\hat{a}\) + \(\hat{b}\)|² = (\(\hat{a}\) + \(\hat{b}\)) (\(\hat{a}\) + \(\hat{b}\))
= \(\hat{a}\)² + \(\hat{b}\)² + 2 \(\hat{a}\).\(\hat{b}\)
= |\(\vec{a}\)|² + |\(\hat{b}\)|² + 2|\(\hat{a}\)||\(\hat{b}\)|
cosθ where θ be the angle between \(\hat{a}\) and \(\hat{b}\).
⇒ |\(\hat{a}\) + \(\hat{b}\)|² = 1 + 1 + 2
cosθ = 2(1 + cosθ)
⇒ |\(\hat{a}\) + \(\hat{b}\)|²
= 2 × 2 cos² \(\frac{\theta}{2}\)
⇒ cos \(\frac{\theta}{2}\)
= \(\frac{1}{2}\)|\(\hat{a}\) + \(\hat{b}\)|

(ii) Also, |\(\hat{a}\) – \(\hat{b}\)|²
= (\(\hat{a}\) – \(\hat{b}\)).(\(\hat{a}\) – \(\hat{b}\))
= \(\hat{a}\)² + \(\hat{b}\)² – 2 \(\hat{a}\).\(\hat{b}\)
= |\(\vec{a}\)|² + |\(\hat{b}\)|² – 2|\(\hat{a}\)||\(\hat{b}\)|
cosθ = 1 + 1 – 2 × 1 × 1
cosθ = 2(1 – cosθ)
= 4 sin² \(\frac{\theta}{2}\)
⇒ sin \(\frac{\theta}{2}\)
= \(\frac{1}{2}\)|\(\hat{a}\) – \(\hat{b}\)|
On dividing eqn. (2) by eqn. (1); we have
tan \(\frac{\theta}{2}\) = \(\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}\)

Question 19.
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three mutually perpendicular unit vectors, then prove that
|\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)| = \(\sqrt{3}\)
Answer:
Since \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually ⊥ vectors.
∴ \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{c}\)
= \(\vec{c}\).\(\vec{a}\) = 0 and |\(\vec{a}\)|
=|\(\vec{b}\)| = |\(\vec{c}\)| = 1
Now |\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)|²
= (\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)) (\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\))
= \(\vec{a}\)² + \(\vec{b}\)² + \(\vec{c}\)² + 2(\(\vec{a}\).\(\vec{b}\) + \(\vec{b}\). \(\vec{c}\) + \(\vec{c}\).\(\vec{a}\))
= |\(\vec{a}\)|² + |\(\vec{b}\)|² + |\(\vec{c}\)|² + 2(\(\vec{a}\).\(\vec{b}\) + \(\vec{b}\).\(\vec{c}\) + \(\vec{c}\).\(\vec{a}\))
= 1 + 1 + 1 + 2 × 0 = 3
⇒ |\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)| = \(\sqrt{3}\)

Question 20.
If |\(\vec{a}\) + \(\vec{b}\)| = 60, |\(\vec{a}\) – \(\vec{b}\)| = 40 and |\(\vec{b}\)| = 46, find |\(\vec{a}\)|.
Answer:
Given |\(\vec{a}\) + \(\vec{b}\)| = 60;
|\(\vec{a}\) – \(\vec{b}\)| = 40;
|\(\vec{b}\)| = 46
We know that |\(\vec{a}\) + \(\vec{b}\)|² + |\(\vec{a}\) – \(\vec{b}\)|²
= 2(|\(\vec{a}\)|² + |\(\vec{b}\)|²|)
⇒ (60)² + (40)²
= 2(|\(\vec{a}\)|² + (46)²)
⇒ 3600 + 1600 = 2(|\(\vec{a}\)|² + 2116)
⇒ 2600 = |\(\vec{a}\)|² + 2116
⇒ |\(\vec{a}\)|² = 484
⇒ |\(\vec{a}\)|= 22

Question 21.
If a vector \(\vec{a}\) is perpendicular to two non-collinear vectors \(\vec{b}\) and \(\vec{c}\), then \(\vec{a}\) is perpendicular to every vector in the plane of \(\vec{b}\) and \(\vec{c}\).
Answer:
Given \(\vec{a}\) is ⊥ to two non-collinear vectors \(\vec{b}\) and \(\vec{c}\).
∴ \(\vec{a}\).\(\vec{b}\) = 0 = \(\vec{a}\).\(\vec{c}\)
Now \(\vec{z}\) = x \(\vec{b}\) + y \(\vec{c}\) be the vector in the plane of \(\vec{b}\) and \(\vec{c}\), where x and y are scalars.
Now \(\vec{a}\)(x \(\vec{b}\) + y \(\vec{c}\))
= x(\(\vec{a}\).\(\vec{b}\)) + y(\(\vec{a}\).\(\vec{c}\)) = x × 0 + y × 0 = 0
Thus \(\vec{a}\) is perpendicular to x \(\vec{b}\) + y \(\vec{c}\)
Hence \(\vec{a}\) is ⊥ to the plane of \(\vec{b}\) and \(\vec{c}\).

Question 22.
If \(\vec{A}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\),
\(\vec{B}\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\) and \(\vec{C}\) = 3 \(\hat{i}\) + \(\hat{j}\), find t such that \(\vec{A}\) + t \(\vec{B}\) is perpendicular to \(\overrightarrow{\mathbf{C}}\).
Answer:
Given \(\overrightarrow{\mathrm{A}}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\);
\(\overrightarrow{\mathrm{B}}\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\) and
\(\overrightarrow{\mathrm{C}}\) = 3 \(\hat{i}\) + \(\hat{j}\)
\(\overrightarrow{\mathrm{A}}\) + t \(\overrightarrow{\mathrm{B}}\)
= (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) + t(\(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\))
= (1 – t) \(\hat{i}\) + (2 + 2 t) \(\hat{j}\) + (3 + t) \(\hat{k}\)
Now \(\overrightarrow{\mathrm{A}}\) + t \(\overrightarrow{\mathrm{B}}\) is perpendicular to \(\overrightarrow{\mathrm{C}}\).
∴ (\(\overrightarrow{\mathrm{A}}\) + t \(\overrightarrow{\mathrm{B}}\)).\(\overrightarrow{\mathrm{C}}\) = 0
⇒ [(1 – t) \(\hat{i}\) + (2 + 2 t) \(\hat{j}\) + (3 + t) \(\hat{k}\)].[3 \(\hat{i}\) + \(\hat{j}\)] = 0
⇒ 3(1 – t) + (2 + 2 t).1 + (3 + t) 0 = 0
⇒ 3 – 3 t + 2 + 2 t = 0
⇒ t = 5

Question 23.
If A, B, C, D are any four points, show that \(\overrightarrow{A B}\).\(\overrightarrow{C D}\) + \(\overrightarrow{B C}\).\(\overrightarrow{A D}\) + \(\overrightarrow{C A}\).\(\overrightarrow{B D}\) = 0
Answer:

Aliter : Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\) are the position vectors of points A, B, C and D with O as origin of reference.
Then
\(\overrightarrow{\mathrm{AB}}\) = \(\vec{b}\) – \(\vec{a}\);
\(\overrightarrow{\mathrm{CD}}\) = \(\vec{d}\) – \(\vec{c}\);
\(\overrightarrow{\mathrm{BC}}\) = \(\vec{c}\) – \(\vec{b}\);
\(\overrightarrow{\mathrm{AD}}\) = \(\vec{d}\) – \(\vec{a}\);
\(\overrightarrow{\mathrm{CA}}\) = \(\vec{a}\) – \(\vec{c}\);
\(\overrightarrow{\mathrm{BD}}\) = \(\vec{d}\) – \(\vec{b}\)
L.H.S. = \(\overrightarrow{\mathrm{AB}}\).\(\overrightarrow{\mathrm{CD}}\) + \(\overrightarrow{\mathrm{BC}}\) \(\overrightarrow{\mathrm{AD}}\) + \(\overrightarrow{\mathrm{CA}}\).\(\overrightarrow{\mathrm{BD}}\)
= (\(\vec{b}\) – \(\vec{a}\))(\(\vec{d}\) – \(\vec{c}\)) + (\(\vec{c}\)– \(\vec{b}\)) (\(\vec{d}\) – \(\vec{a}\)) + (\(\vec{a}\) – \(\vec{c}\)).(\(\vec{d}\) – \(\vec{b}\))
= \(\vec{b}\).\(\vec{d}\) – \(\vec{b}\).\(\vec{c}\) – \(\vec{a}\).\(\vec{d}\) + \(\vec{a}\) \(\vec{c}\) + \(\vec{c}\).\(\vec{d}\)
– \(\vec{c}\).\(\vec{a}\) – \(\vec{b}\).\(\vec{d}\) + \(\vec{b}\).\(\vec{a}\) + \(\vec{a}\).[/latex] \(\vec{d}\) – \(\vec{a}\).\(\vec{b}\)
– \(\vec{c}\).\(\vec{d}\) + \(\vec{c}\).\(\vec{b}\) = 0

Question 24.
If \(\vec{a}\) and \(\vec{b}\) are vectors with a, b as their respective lengths, sho
(\(\frac{\vec{a}}{a^2}\) – \(\frac{\vec{b}}{b^2}\)) = \((\frac{\vec{a}-\vec{b}}{a b})^2\) .
Answer: