ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.2

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ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.2

Very short answer type questions (1 to 14) :

Question 1.
(i) If \(\vec{a}\) is a unit vector and \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 15, then find \(|\vec{x}|\).
(ii) If \(\vec{p}\) is a unit vector and \((\vec{x}-\vec{p}) \cdot(\vec{x}+\vec{p})\)= 80, then find \(|\vec{x}|\).
Solution:
(i) Given \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 15

(ii) Given \(\vec{p}\) is a unit vector and \((\vec{x}-\vec{p}) \cdot(\vec{x}+\vec{p})\)= 80

Question 2.
(i) If \((\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})\) = 12 and \(|\vec{a}|=2|\vec{b}|\), then find \(|\vec{a}| \text { and }|\vec{b}|\).
(ii) If \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 4 and \(\vec{a} \cdot \vec{b}\) = 1, then find \((\vec{a}-\vec{b})^2\).
(iii) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 5 and \(\vec{a} \cdot \vec{b}\) = 8, then find \(|\vec{a}-\vec{b}|\).
(iv) If \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 5 and \(\vec{a} \cdot \vec{b}\) = – 8 then find \(|\vec{a}+\vec{b}|\).
Solution:
(i) Given \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 12

(ii) Given \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 4
and \(\vec{a} \cdot \vec{b}\) = 1

(iii) Given \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 5
and \(\vec{a} \cdot \vec{b}\) = 8

(iv) Given \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 5
and \(\vec{a} \cdot \vec{b}\) = – 8

Question 3.
Find the magnitude of each of the two vectors \(\vec{a} \text { and } \vec{b}\), having same magnitude such that the angle between them is 60° and their scalar product is \(\frac{9}{2}\).
Solution:
Given \(|\vec{a}|=|\vec{b}|\) ;
θ = 60° ;
\(\vec{a} \cdot \vec{b}=\frac{9}{2}\)

Question 3 (old).
If the angle between two vectors \(\vec{a} \text { and } \vec{b}\) of equal magnitude is 60° and their scalar product is \(\frac{1}{2}\), then find their magnitudes. (NCERT)
Solution:
Given \(|\vec{a}|=|\vec{b}|\) ……………..(1)
Let θ be the angle between a and b such that θ = 60°
and \(\vec{a} \cdot \vec{b}=\frac{1}{2}\) [using (1)]
We know that
⇒ \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)
⇒ \(\frac{1}{2}=|\vec{a}| \cdot|\vec{b}| \cdot \cos \frac{\pi}{3}\)
⇒ \(\frac{1}{2}=|\vec{a}|^2 \cdot \frac{1}{2}\)
⇒ \(|\vec{a}|=1=|\vec{b}|\)

Question 4.
(i) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = √3 and \(\vec{a} \cdot \vec{b}\) = 3, then find the angle between \(\vec{a} \text { and } \vec{b}\).
(ii) Find the angle between the vectors \(\vec{a} \text { and } \vec{b}\) such that \(|\vec{a}|=|\vec{b}|\) = 3 and \(\vec{a} \cdot \vec{b}\) = 1.
(iii) Find the angle between two vectors having the same length √2 and scalar product – 1.
(iv) If \(|\vec{a}|\) = √3, \(|\vec{b}|\) = 2 and the angle between \(\vec{a} \text { and } \vec{b}\) is 60°, find \(\vec{a} \cdot \vec{b}\) = 1.
Solution:
(i) Given \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 2
and \(\vec{a} \cdot \vec{b}\) = 3
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\).
∴ \(\vec{a} \cdot \vec{b}\) = \(|\vec{a}||\vec{b}|\) cos θ
⇒ 3 = 3 × 2 × cos θ
⇒ cos θ = \(\frac{1}{2}\) ; 0 ≤ θ ≤ π
∴ θ = \(\frac{\pi}{3}\)

(ii) Given \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 3 ;
and \(\vec{a} \cdot \vec{b}\) = 1
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\).
∴ cos θ = \(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{b}|}\)
= \(\frac{1}{3 \times 3}=\frac{1}{9}\)
⇒ θ = cos^-1 (\(\frac{1}{9}\))

(iii) Let θ be the angle between two vectors \(\vec{a} \text { and } \vec{b}\) such that
\(|\vec{a}|=|\vec{b}|\) = √2
and \(\vec{a} \cdot \vec{b}\) = – 1
Since cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
= \(\frac{-1}{\sqrt{2} \times \sqrt{2}}=-\frac{1}{2}\)
⇒ cos θ = – cos \(\frac{\pi}{3}\)
= cos (180° – 60°)
= cos 120°
∴ θ = 120°

(iv) Given \(|\vec{a}|\) = √3 , \(|\vec{b}|\) = 2 ;
and θ = 60°
where θ be the angle between \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{a} \cdot \vec{b}\) = \(|\vec{a}||\vec{b}|\) cos θ
= √3 × 2 × cos 60°
= 2√3 × \(\frac{1}{2}\)
= √3

Question 4 (old).
(i) If \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 2 and \(\vec{a} \cdot \vec{b}\) = 3, then find the angle between \(\vec{a} \text { and } \vec{b}\).
(iv) Find the angle between the vectors \(\vec{a} \text { and } \vec{b}\) with magnitudes 1 and 2 respectively, and \(\vec{a} \cdot \vec{b}\) = 1
Solution:
(ii) Given \(|\vec{a}|\) = 2, \(|\vec{b}|\) = √3
and \(\vec{a} \cdot \vec{b}\) = 3
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\).
Then \(\vec{a} \cdot \vec{b}\) = \(|\vec{a}||\vec{b}|\) cos θ
⇒ 3 = 2 × √3 × cos θ
⇒ cos θ = \(\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}\) ; 0 ≤ θ ≤ π
∴ θ = 30°.

(iv) Given \(|\vec{a}|\) = 1, \(|\vec{b}|\) = 2 ;
and \(\vec{a} \cdot \vec{b}\) = 1
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\).
∴ cos θ = \(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{b}|}\)
= \(\frac{1}{1 \cdot 2}=\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

Question 5.
Find the angle between the vectors \(2 \hat{i}-\hat{j}+\hat{k}\) and \(3 \hat{i}+4 \hat{j}-\hat{k}\). (NCERT Exemplar)
Solution:
Let \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{b}=3 \hat{i}+4 \hat{j}-\hat{k}\)

Question 5 (old).
(i) Find the angle between the vectors \(\vec{a}=\hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\).
Solution:
Given \(\vec{a}=\hat{i}+\hat{j}-\hat{k}\)
and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\)

Question 6.
Find
(i) \((\vec{b}-\vec{a}) \cdot(3 \vec{a}+\vec{b})\) where \(\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}\), \(\overrightarrow{\vec{b}}=2 \hat{i}+\hat{j}-3 \hat{k}\). \((\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b})\)
(ii) where \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\), \(\vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}\)
Solution:
(i) Given \(\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}\),
and \(\overrightarrow{\vec{b}}=2 \hat{i}+\hat{j}-3 \hat{k}\)

(ii) Given \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}\)

Question 7.
(i) Find the projection of \(\vec{a} \text { on } \vec{b}\) if \(\vec{a} \cdot \vec{b}\) = 8 and \(\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}\).
(ii) Find the projection of \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(2 \hat{i}-3 \hat{j}+6 \hat{k}\).
(iii) Find the projection of \(\hat{i}-\hat{j}\) on the vector \(\hat{i}+\hat{j}\).
(iv) Find λ when the projection of \(\hat{i}+\lambda \hat{j}+\hat{k}\) on \(\hat{i}+\hat{j}\) is √2 units.
Solution:
(i) Given \(\vec{a} \cdot \vec{b}\) = 8
and \(\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}\)
∴ \(|\vec{b}|=\sqrt{2^2+6^2+3^2}\)
= \(\sqrt{49}\) = 7
Thus required projection of \(\vec{a} \text { on } \vec{b}\)
= \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{8}{7}\)

(ii) Let \(\vec{a}=\hat{i}+3 \hat{j}+7 \hat{k}\)
and \(\vec{b}=2 \hat{i}-3 \hat{j}+6 \hat{k}\)
∴ \(\vec{a} \cdot \vec{b}=(\hat{i}+3 \hat{j}+7 \hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})\)
= 1 (2) + 3 (- 3) + 7 (6)
= 2 – 9 + 42 = 35
and \(|\vec{b}|=\sqrt{2^2+(-3)^2+6^2}\)
= \(\sqrt{49}\) = 7
∴ required projection of \(\vec{a} \text { on } \vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
= \(\frac{35}{7}\) = 5

(iii) Given \(\vec{a}=\hat{i}-\hat{j}\)
and \(\vec{b}=\hat{i}+\hat{j}\)
∴ \(\vec{a} \cdot \vec{b}\) = 1 (1) – (1) (1) = 0
∴ scalar projection of \(\vec{a} \text { on } \vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) = 0

(iv) Let \(\vec{a}=\hat{i}+\lambda \hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}+\hat{j}\)
∴ \(\vec{a} \cdot \vec{b}=(\hat{i}+\lambda \hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}+0 \hat{k})\)
= 1 (1) + λ (1) + 1 (0)
= 1 + λ
\(|\vec{b}|=\sqrt{1^2+1^2}=\sqrt{2}\)
We know that projection \(\vec{a} \text { on } \vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
∴ √2 = \(\frac{1+\lambda}{\sqrt{2}}\)
⇒ 1 + λ = 2
⇒ λ = 1.

Question 8.
(i) If \(\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}\) and \(\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}\), then show that the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) are perpendicular to each other. (NCERT)
(ii) Find λ if \(\vec{a}=3 \hat{i}-\hat{j}+4 \hat{k}\) and \(\vec{b}=-\lambda \hat{i}+3 \hat{j}+3 \hat{k}\) are perpendicular to each other.
(iii) For what value of λ are the vectors \(\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}\) perpendicular to each other ?
Solution:
(i) Given \(\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}\)
and \(\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}\)

(ii) Given \(\vec{a}=3 \hat{i}-\hat{j}+4 \hat{k}\)
and \(\vec{b}=-\lambda \hat{i}+3 \hat{j}+3 \hat{k}\)
Since \(\vec{a}\) is ⊥ to \(\vec{b}\)
∴ \(\vec{a} \cdot \vec{b}\) = 0
⇒ \((3 \hat{i}-\hat{j}+4 \hat{k}) \cdot(-\lambda \hat{i}+3 \hat{j}+3 \hat{k})\) = 0
⇒ 3 (- λ) – 1 (3) + 4 (3) = 0
⇒ – 3λ + 9 = 0
⇒ λ = 3.

(iii) Given \(\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}\)
Now \(\vec{a} \text { and } \vec{b}\) are orthogonal or perpendicualr.
∴ \(\vec{a} \cdot \vec{b}\) = 0
⇒ 2 (1) + λ (- 2) + 1 (3) = 0
⇒ 2λ = 5
⇒ λ = \(\frac{5}{2}\)

Question 9.
(i) If \(\vec{a} \text { and } \vec{b}\) are unit vectors, then what is the angle between \(\vec{a} \text { and } \vec{b}\) so that \(\sqrt{3} \vec{a}-\vec{b}\) may be a unit vector. (NCERT Exemplar)
(ii) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3 and \(\vec{a} \cdot \vec{b}\) = 4, then what is the value of \(|\vec{a}+2 \vec{b}|\) ?
(iii) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3 and \(\vec{a} \cdot \vec{b}\) = 4, find \(|2 \vec{a}-3 \vec{b}|\).
(iv) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3 and \(|2 \vec{a}-\vec{b}|\) = 4, then find \(|2 \vec{a}+\vec{b}|\).
Solution:
(i) Given \(\vec{a} \text { and } \vec{b}\) are unit vectors
∴ \(|\vec{a}|\) = \(|\vec{b}|\) = 1
since \(\sqrt{3} \vec{a}-\vec{b}\) is a unit vector

(ii) Given \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3
and \(\vec{a} \cdot \vec{b}\) = 4

= 2² + 4 (4) + 4 (9)
= 4 + 16 + 36 = 56
∴ \(|\vec{a}+2 \vec{b}|=\sqrt{56}=2 \sqrt{14}\)

(iii) Given \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3
and \(\vec{a} \cdot \vec{b}\) = 4

(iv) Given \(|\vec{a}|\) = 2,
\(|\vec{b}|\) = 3
and \(|2 \vec{a}-\vec{b}|\) = 4

Question 10.
If \(\vec{a}, \vec{b}\) are unit vectors and c = \(|\vec{a}+\vec{b}|\), d = \(|\vec{a}-\vec{b}|\), then what is the value of c² + d² ?
Solution:
Given c = \(|\vec{a}+\vec{b}|\)

Question 11.
If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors such that \(\vec{c}=\vec{a}+\vec{b}\) and \(\vec{a} \cdot \vec{b}\) = 0, then show that c² = a² + b².
Solution:
Given \(\vec{c}=\vec{a}+\vec{b}\)

Question 12.
(i) If \(\vec{a}, \vec{b} \text { and } \vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), then find the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\). (NCERT)
(ii) If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors \(|\vec{a}|\) = 5, \(|\vec{b}|\) = 12, \(|\vec{c}|\) = 13 and \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\) then find the value of \(\).
Solution:
(i) Given \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) ;
on squaring, we ahve

(ii) Given \(|\vec{a}|\) = 5 ;
\(|\vec{b}|\) = 12 ;
\(|\vec{c}|\) = 13
and \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) ;
on squaring we have
∴

Question 13.
(i) We are given two vectors \(\vec{a} \text { and } \vec{b}\) such that \((\vec{a})^2=(\vec{b})^2\). Is it necessary that \(\vec{a}=\vec{b}\) ? Justify your answer.
(ii) For two non-zero vectors \(\vec{a} \text { and } \vec{b}\), state when \(|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2\) holds.
(iii) If \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\), then \(\vec{a} \cdot \vec{b}\) = 0. Is the converse true ? Justify your answer by an example.
(iv) If \(\vec{a}, \vec{b}\) are non – zero vectors and \(\vec{a} \cdot \vec{b}\) ≥ 0, then what can you say about the angle θ between the vectors \(\vec{a} \text { and } \vec{b}\).
Solution:
(i) Let \(\vec{a}=\hat{i}+2 \hat{j}\)
and \(\vec{b}=2 \hat{i}-\hat{j}\)

Here \(\vec{a}^2=|\vec{a}|^2\)
= \(\left(\sqrt{1^2+2^2}\right)^2\)
= \((\sqrt{5})^2\) = 5

and \(\vec{b}^2=|\vec{b}|^2\)
= \(\left(\sqrt{2^2+(-1)^2}\right)^2\)
= \((\sqrt{5})^2\) = 5
i.e. \(\vec{a}^2=\vec{b}^2\) but \(\vec{a} \neq \vec{b}\).

(ii) \(|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2\) holds

(iii) \(\vec{a} \cdot \vec{b}=\overrightarrow{0} \cdot \vec{b}\) = 0
if \(\vec{a}\) = 0
When \(\vec{b}\) = 0 ;
\(\vec{a} \cdot \vec{b}=\vec{a} \cdot \overrightarrow{0}\) = 0
but converse need not be true.
e.g. \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}+\hat{j}-\hat{k}\)
but \(\vec{a} \cdot \vec{b}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})\)
= 2 (1) – 1 (1) + 1 (- 1)
= 2 – 2 = 0
Here \(\vec{a} \cdot \vec{b}\) = 0 but neither \(\vec{a}=\overrightarrow{0}\) nor \(\vec{b}=\overrightarrow{0}\).

(iv) Since \(\vec{a} \text { and } \vec{b}\) are non-zero vectors.
∴ \(|\vec{a}| \cdot \mid \vec{b}\) > 0
also \(\vec{a} \cdot \vec{b}\) ≥ 0
Thus cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}\) ≥ 0
∴ 0 ≤ θ ≤ \(\frac{\pi}{2}\)

Question 14.
If \(\vec{a} \cdot \vec{a}\) = 0 and \(\vec{a} \cdot \vec{b}\) = 0, then what can you say about the vector \(\vec{b}\) ?
Solution:
Given \(\vec{a} \cdot \vec{a}\) = 0
and \(\vec{a} \cdot \vec{b}\) = 0
⇒ \(|\vec{a}|^2\) = 0 i.e. \(\vec{a}=\overrightarrow{0}\)
and (\(\vec{a} \perp \vec{b}\) or \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\))
i.e. \(\vec{a}=\overrightarrow{0}\) and \(\vec{b}\) can be any vector.

Question 15.
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 \(\vec{b}=3 \hat{i}+\hat{j}-2 \hat{k}\).
Solution:
Given \(\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}\)
and \(\vec{b}=3 \hat{i}+\hat{j}-2 \hat{k}\)

Question 16.
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 \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\).
Solution:
Given \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}\),
\(\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)

Here \(\vec{b} \cdot \vec{a}=(2 \hat{i}-2 \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}-2 \hat{k})\)
= 2 (1) – 2 (2) + 1 (- 2)
= 2 – 4 – 2 = – 4

\(\vec{c} \cdot \vec{a}=(2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})\)
= 2 (2) – 1 (- 2) + 4 (1)
= 4 + 2+ 4 = 10

∴ required projection of \(\vec{b}+\vec{c} \text { on } \vec{a}\)
= \(\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}\)
= = \(\frac{(\vec{b} \cdot \vec{a})+(\vec{c} \cdot \vec{a})}{|\vec{a}|}\)
= \(\frac{-4+10}{\sqrt{2^2+(-2)^2+1^2}}\)
= \(\frac{6}{3}\)
= 2

Question 17.
(i) If \(\vec{a}=\hat{i}-\hat{j}+7 \hat{k}\) and \(\vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}\), then find λ such that \(\vec{a}-\vec{b}\) and \(\vec{a}+\vec{b}\) are perpendicular to each other.
(ii) For any (non-zero) vectors \(\vec{a}\), and \(\vec{b}\) prove that the vectors \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}\) and \(|\vec{a}| \vec{b}-|\vec{b}| \vec{a}\) are perpendicular to each other. (NCERT)
Solution:
(i) Given \(\vec{a}=\hat{i}-\hat{j}+7 \hat{k}\)
and \(\vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}\)
Since \(\vec{a}-\vec{b}\) and \(\vec{a}-\vec{b}\) are ⊥ to each other.
∴ \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 0
⇒ \(\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}\) = 0
⇒ \(|\vec{a}|^2-|\vec{b}|^2\) = 0
[∵\(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\)]
⇒ {1² + (- 1)² + 7²} – {5² + (- 1)² + λ²} = 0
⇒ 51 – (26 + λ²) = 0
⇒ 25 – λ² = 0
⇒ λ = ± 5

(ii) Now,

Question 18.
If \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}\), \(\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}\) are such that \(\vec{a}+\lambda \vec{b}\) is perpendiculai to \(\vec{c}\), then find the value of λ ?
Solution:
Given \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}\),
\(\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\)
and \(\vec{c}=3 \hat{i}+\hat{j}\)

Now \(\vec{a}+\lambda \vec{b}\)
= \((2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})\)
= \((2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}\)

Now \(\vec{a}+\lambda \vec{b}\) is ⊥ to \(\vec{c}\)
∴ \(\vec{a}+\lambda \vec{b}\) . \(\vec{c}\) = 0
i.e. \(\) = 0
⇒ 3 (2 – λ) + (2 + 2λ) 1 = 0
⇒ 8 – λ = 0
⇒ λ = 8

Question 19.
(i) If \(\vec{c}\) is perpendicular to \(\vec{a} \text { and } \vec{b}\) both, then show that it is perpendicular to \(\vec{a}+\vec{b}\) as well as \(\vec{a}-\vec{b}\).
(ii) Show that the vectors \(\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})\), \(\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})\) and \(\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\) are mutually perpendicular unit vccotrs. (NCERT)
(iii) If \(\vec{a} \text { and } \vec{b}\) are unit vectors and θ is the angle between them, then prove that cos \(\frac{\theta}{2}=\frac{1}{2}|\vec{a}+\vec{b}|\).
Solution:
(i) Since \(\vec{c}\) is ⊥ to \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{c} \cdot \vec{a}\) = \(\vec{c} \cdot \vec{b}\)
Now \(\vec{c} \cdot(\vec{a}+\vec{b})=\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}\)
= 0 + 0 = 0
and \(\vec{c} \cdot(\vec{a}-\vec{b})=\vec{c} \cdot \vec{a}-(\vec{c} \cdot \vec{b})\)
= 0 – 0 = 0
Thus \(\vec{c}\) is ⊥ to both \(\vec{a}+\vec{b} \text { and } \vec{a}-\vec{b}\).

(ii) Given \(\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})\),
\(\vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})\)
and \(\vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\)

(iii) Given \(\vec{a} \text { and } \vec{b}\) s.t. \(|\hat{a}|=|\hat{b}|\) = 1
and given θ be the angle between them
Now, \(|\hat{a}+\hat{b}|^2=(\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b})\)
= \(|\hat{a}|^2+|\hat{b}|^2+2 \hat{a} \cdot \hat{b}\)
[∵ \(\hat{a}^2=|\vec{a}|^2, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\)]
= 1 + 1 + 2 \(|\hat{a}|=|\hat{b}|\) cos θ
= 2 + 2 cos θ
= 2 (1 + cos θ)
= 2 × 2 cos² \(\frac{\theta}{2}\)
= 4 cos² \(\frac{\theta}{2}\)
⇒ \(|\hat{a}+\hat{b}|=2 \cos \frac{\theta}{2}\)
\(\cos \frac{\theta}{2}=\frac{1}{2}|\hat{a}+\hat{b}|\)

Question 20.
Find the angles which the vector \(\vec{a}=3 \hat{i}-6 \hat{j}+2 \hat{k}\) makes with the co-ordinate axes.
Solution:
Given \(\vec{a}=3 \hat{i}-6 \hat{j}+2 \hat{k}\)
Let α, β and γ are the angle made by the vector with the coordinate axes.

Question 21.
(i) Dot products of a vector with vectors \(\), \(\) and \(\) are respectively – 1, 6 and 5. Find the vector.
(ii) The dot product of a vector with the vectors \(\), \(\)and \(\) are 0, 5 and 8 respectively. Find the vector.
Solution:
(i) Let the required vector be
\(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\) …………………..(1)
Now \(\vec{a} \cdot(3 \hat{i}-5 \hat{k})\) = – 1
⇒ 3a₁ – 5a₃ = – 1 ………………………..(2)
also \(\vec{a} \cdot(2 \hat{i}+7 \hat{j})\) = 6
⇒ 2a₁ + 7a₂ = 6 ……………………….(3)
and \(\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 5
⇒ a₁ + a₂ + a₃ = 5 ………………(4)
eqn. (3) – 7 × eqn. (4) ; we have
– 5a₁ – 7a₃ = – 29
⇒ 5a₁ + 7a₃ = 29 ………………………(5)
On solving (2) and (5) we get
46a₁ = 145 – 7 = 138
⇒ a₁ = 3 ;
a₃ = 2
∴ from (4) ;
3 + 2 + a₂ = 5
⇒ a₂ = 0

(ii) Let the required vector be
\(\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}\) ……………..(1)
given, \(\vec{a} \cdot(\hat{i}+\hat{j}-3 \hat{k})\) = 0

⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}-3 \hat{k})\) = 0
⇒ x + y – 3z = 0 ………………..(2)
also, \(\vec{a} \cdot(\hat{i}+3 \hat{j}-2 \hat{k})\) = 5

⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+3 \hat{j}-2 \hat{k})\) = 5
⇒ x + 3y – 2z = 5 ……………….(3)
also, \(\vec{a} \cdot(2 \hat{i}+\hat{j}+4 \hat{k})\) = 8

⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+\hat{j}+4 \hat{k})\) = 8
⇒ 2x + y + 4z = 8 ………………(4)

eqn. (3) – eqn. (2) gives ;
2y + z = 5 ………………….(5)
eqn. (4) – 2 × eqn. (3) gives ;
– 5y + 8z = – 2 ……………………(6)
eqn. (6) – 8 × eqn. (5) gives ;
– 5y – 16y = – 2 – 40
⇒ – 21 y = – 42
⇒ y = 2
∴ from(5) ;
4 + z = 5
⇒ z = 1
∴ from (2) ;
x + 2 – 3 = 0
⇒ x = 1
∴ required vector be \(\vec{a}=\hat{i}+2 \hat{j}+\hat{k}\).

Question 22.
Show that the vectors \(\vec{a}=\hat{i}-3 \hat{j}-5 \hat{k}\), \(\vec{b}=2 \hat{i}-\hat{j}+\hat{k}\) and \(\vec{c}=\hat{i}+2 \hat{j}+6 \hat{k}\) form a right-angled triangle.
Solution:
Given \(\vec{a}=\hat{i}-3 \hat{j}-5 \hat{k}\),
\(\vec{b}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{c}=\hat{i}+2 \hat{j}+6 \hat{k}\)

Now \(\vec{a}+\vec{c}=(\hat{i}-3 \hat{j}-5 \hat{k})+(\hat{i}+2 \hat{j}+6 \hat{k})\)
= \(2 \hat{i}-\hat{j}+\hat{k}=\vec{b}\)

∴ \(\vec{a}, \vec{b} \text { and } \vec{c}\) are coplanar.
also no two of these vectors are parallel.
∴ gives vectors forms a triangle.

Also \(\vec{a} \cdot \vec{b}=(\hat{i}-3 \hat{j}-5 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})\)
= 1 (2) – 3 (- 1) – 5 (1) = 0
These dot product of two non-zero vectors is zero.
Therefore these vectors \(\vec{a} \text { and } \vec{b}\) are ⊥ to each other.
Hence the sides represented by these vectors are perpendicular.
Thus, the given vectors forms a right angled triangle.

Question 23.
If A, B, C have position vectors \(\hat{j}+\hat{k}\), \(3 \hat{i}+\hat{j}+5 \hat{k}\) and \(3 \hat{j}+3 \hat{k}\) respectively, then prove that ∆ABC is right angled at C.
Solution:
Let \(\vec{a}=\hat{j}+\hat{k}\),
\(\vec{b}=3 \hat{i}+\hat{j}+5 \hat{k}\)
and \(\vec{c}=3 \hat{j}+3 \hat{k}\)
are the position vectors of points A, B, C respectively.

Question 24.
If the vertices of a AABC are A(- 1, 3, 2), B (2, 3, 5) and C (3, 5, – 2), then show that it is right angled at A. Also find the other two angles.
Solution:
The position vectors of the vertices A, B and C of a triangle are \(-\hat{i}+3 \hat{j}+2 \hat{k}\), \(2 \hat{i}+3 \hat{j}+5 \hat{k}\) and \(3 \hat{i}+5 \hat{j}-2 \hat{k}\)
‘
Hence, the three sides of the triangle are represented by \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}} \text { and } \overrightarrow{\mathrm{CA}}\).

∴ \(\overrightarrow{\mathrm{AB}}\) = P.V 0f B – P.V 0f A
= \((2 \hat{i}+3 \hat{j}+5 \hat{k})-(-\hat{i}+3 \hat{j}+2 \hat{k})\)
= \(3 \hat{i}+3 \hat{k}\)

\(\overrightarrow{\mathrm{BC}}\) = P.V 0f C – P.V 0f B
= \((3 \hat{i}+5 \hat{j}-2 \hat{k})-(2 \hat{i}+3 \hat{j}+5 \hat{k})\)
= \(\hat{i}+2 \hat{j}-7 \hat{k}\)

\(\overrightarrow{\mathrm{CA}}\) = P.V 0f A – P.V 0f C
= \((-\hat{i}+3 \hat{j}+2 \hat{k})-(3 \hat{i}+5 \hat{j}-2 \hat{k})\)
= \(-4 \hat{i}-2 \hat{j}+4 \hat{k}\)

∴ \(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CA}}\) = \((3 \hat{i}+3 \hat{k}) \cdot(-4 \hat{i}-2 \hat{j}+4 \hat{k})\)
= 3 (- 4) + 0 (- 2) + 3 (4) = 0

Thus \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CA}}\) are ⊥to each other.
∴ ∠A = 90
Therefore, ∆ ABC is right-angled at A.

Question 25.
Prove that two proper vectors \(\vec{a} \text { and } \vec{b}\) are at right angles iff \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2\). (NCERT)
Solution:

Question 26.
If the coordinates of four points are A (2, 3, 4) , B (5, 4, – 1), C (3, 6, 2) and D (1, 2, 0), then show that \(\overrightarrow{\mathrm{AB}}\) is perpendiculat to \(\overrightarrow{\mathrm{CD}}\). (NCERT Exemplar)
Solution:
Given P.V. of A = \(2 \hat{i}+3 \hat{j}+4 \hat{k}\) ;
P.V. of B = \(5 \hat{i}+4 \hat{j}-\hat{k}\) ;
P.V. of C = \(3 \hat{i}+6 \hat{j}+2 \hat{k}\)
and P.V. of D = \(\hat{i}+2 \hat{j}+0 \hat{k}\)

∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((5 \hat{i}+4 \hat{j}-\hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k})\)
= \(3 \hat{i}+\hat{j}-5 \hat{k}\)

∴ \(\overrightarrow{\mathrm{CD}}\) = P.V. of D – P.V. of C
= \((\hat{i}+2 \hat{j}+0 \hat{k})-(3 \hat{i}+6 \hat{j}+2 \hat{k})\)
= \(-2 \hat{i}-4 \hat{j}-2 \hat{k}\)

Here \(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}\)
= \((3 \hat{i}+\hat{j}-5 \hat{k}) \cdot(-2 \hat{i}-4 \hat{j}-2 \hat{k})\)
= 3 (- 2) + 1 (- 4) – 5 (- 2)
= – 6 – 4 + 10 = 0

Question 27.
If \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}\) and \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\), then find the angle between \(2 \vec{a}+\vec{b}\) and \(\vec{a}+2 \vec{b}\).
Solution:
Given \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}\)
and \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\)

Question 28.
If the points A, B and C with position vectors \(2 \hat{i}+\hat{j}+\hat{k}\), \(\hat{i}-3 \hat{j}-5 \hat{k}\) and \(\alpha \hat{i}-3 \hat{j}+\hat{k}\) respectively are the vertices of a right-angled triangle at C, then find the value (s) of a.
Solution:
Since the points A, B and C with position vectors \(2 \hat{i}+\hat{j}+\hat{k}\), \(\hat{i}-3 \hat{j}-5 \hat{k}\) and \(\alpha \hat{i}-3 \hat{j}+\hat{k}\) are the vertices of right angled triangle at C.

Question 29.
Let \(\vec{a} \text { and } \vec{b}\) be unit vectors. If the vectors \(\vec{c}=\vec{a}+2 \vec{b}\) and \(\vec{d}=5 \vec{a}-4 \vec{b}\) are perpendicualr to each other, then find the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:
Since \(\vec{a} \text { and } \vec{b}\) are unit vectors

Question 30.
Express the vector \(\vec{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}\) as the sum of two vectors such that one is parallel to the vector \(\vec{b}=3 \hat{i}+\hat{k}\) and the other is perpendicular to \(\).
Solution:
We want to find the vectors \(\vec{c} \text { and } \vec{d}\) such that
\(\vec{a}=\vec{c}+\vec{d}\) …………………………..(1)
given \(\vec{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}\)
Since \(\vec{c}\) is parallel to \(\vec{a}\)
∴ \(\vec{c}=\lambda \vec{b}\) for some scalar λ
⇒ \(\vec{c}=\lambda(3 \hat{i}+\hat{k})\)
∴ From (1) ; we have
\(\vec{d}=\vec{a}-\vec{c}\)
= \((5 \hat{i}-2 \hat{j}+5 \hat{k})-(3 \lambda \hat{i}+\lambda \hat{k})\)
⇒ \(\vec{d}=(5-3 \lambda) \hat{i}-2 \hat{j}+(5-\lambda) \hat{k}\)
also it is given that \(\vec{d} \perp \vec{b}\)
⇒ \(\vec{d} \cdot \vec{b}\) = 0
\([(5-3 \lambda) \hat{i}-2 \hat{j}+(5-\lambda) k] \cdot(3 \hat{i}+\hat{k})\) = 0
3 (5 – 3λ) – 2 (0) + (5 – λ) . 1 = 0
15 – 9λ + 5 – λ = 0
⇒ 10λ = 20
⇒ λ = 2.
Thus \(\vec{c}=2(3 \hat{i}+\hat{k})=5 \hat{i}+2 \hat{k}\)
and \(\vec{d}=[(5-6) \hat{i}-2 \hat{j}+(5-2) \hat{k}]\)
= \(-\hat{i}-2 \hat{j}+3 \hat{k}\)

Question 31.
If \(\vec{\alpha}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{\beta}=2 \hat{i}+\hat{j}-4 \hat{k}\) then express \(\vec{\beta}\) in the form \(\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2\). where \(\overrightarrow{\beta_1}\) is parallel to \(\vec{\alpha} \text { and } \overrightarrow{\beta_2}\) is perpendicualr to \(\vec{\alpha}\).
Solution:
Given \(\vec{\alpha}=3 \hat{i}+4 \hat{j}+5 \hat{k}\)
and \(\vec{\beta}=2 \hat{i}+\hat{j}-4 \hat{k}\)
Given \(\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2\) ……………….(1)
given \(\overrightarrow{\beta_1}\) is parallel to \(\vec{\alpha}\)
∴ \(\overrightarrow{\beta_1}=\lambda \vec{\alpha}\) for some scalar λ.

Question 32.
If the vectors \(\vec{a}=\hat{i}-\hat{j}+2 \hat{k}\), \(\vec{b}=2 \hat{i}+4 \hat{j}+\hat{k}\) and \(\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}\) are mutually orthogonal, then find the values of λ and μ.
Solution:
Given \(\vec{a}=\hat{i}-\hat{j}+2 \hat{k}\)
\(\vec{b}=2 \hat{i}+4 \hat{j}+\hat{k}\)
and \(\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}\)
Since given vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\) are mutually orthogonal.
∴ \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}\) = 0
Now \(\vec{a} \cdot \vec{c}\) = 0
⇒ \((\hat{i}-\hat{j}+2 \hat{k}) \cdot(\lambda \hat{i}+\hat{j}+\mu \hat{k})\) = 0
⇒ λ – 1 + 2µ = 0
⇒ λ + 2µ = 1 ………………………………(1)
and \(\vec{b} \cdot \vec{c}\) = 0
⇒ \((2 \hat{i}+4 \hat{j}+\hat{k}) \cdot(\lambda \hat{i}+\hat{j}+\mu \hat{k})\) = 0
⇒ 2λ + 4 + µ = 0
⇒ 2λ + µ = – 4 ………………..(2)
On solving (1) and (2) ; we have
λ = – 3 ; µ = 2

Question 33.
Find the values of λ and µ if the vectors \(\lambda \hat{i}-3 \hat{j}-6 \hat{k}\) and \(3 \hat{i}-\mu \hat{j}-2 \hat{k}\) are mutually perpendicular vectors of equal magnitude.
Solution:
Let \(\vec{a}=\lambda \hat{i}-3 \hat{j}-6 \hat{k}\)
and \(\vec{b}=3 \hat{i}-\mu \hat{j}-2 \hat{k}\)
Since \(\vec{a} \text { and } \vec{b}\) are mutually ⊥ to each other.
∴ \(\vec{a} \cdot \vec{b}\) = 0
⇒ \((\lambda \hat{i}-3 \hat{j}-6 \hat{k}) \cdot(3 \hat{i}-\mu \hat{j}-2 \hat{k})\) = 0
⇒ λ (3) – 3 (- µ) – 6 (- 2) = 0
⇒ 3λ + 3µ + 12 = 0
⇒ λ + µ + 4 = 0 ………………..(1)
also \(|\vec{a}|=|\vec{b}|\)
⇒ \(\sqrt{\lambda^2+(-3)^2+(-6)^2}=\sqrt{3^2+(-\mu)^2+(-2)^2}\)
⇒ \(\sqrt{\lambda^2+9+36}=\sqrt{9+\mu^2+4}\)
⇒ \(\sqrt{\lambda^2+45}=\sqrt{13+\mu^2}\)
On squaring ; we have
λ² – µ² = 13 – 45 = – 32
⇒ (λ – µ) (λ + µ) = – 32
⇒ (λ – µ) (- 4) = – 32 [using (1)]
⇒ λ – µ = 8 ……………………(2)
On adding (1) and (2) ; we have
2λ = 4
⇒ λ = 2
∴ from (1) ;
µ = – 6.

Question 34.
If \(\vec{a}=\hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\), then find a unit vector which is perpendicular to \(\vec{a}\) and is coplanar with \(\vec{a} \text { and } \vec{b}\).
Solution:
Let \(\vec{r}\) be any vector in the plane of \(\vec{a} \text { and } \vec{b}\).
Then \(\vec{r}=\lambda \vec{a}+\mu \vec{b}\),
where λ, µ are scalars.

Question 35.
Find a vector of magnitude 5 units which is coplanar with vectors \(3 \hat{i}-\hat{j}-\hat{k}\) and \(\hat{i}+\hat{j}-2 \hat{k}\) and is perpendicular to the vector \(2 \hat{i}+2 \hat{j}+\hat{k}\).
Solution:
Let the required vector be \(\vec{d}=a \hat{i}+b \hat{j}+c \hat{k}\)
Since \(|\vec{d}|\) = 5 units
⇒ a² + b² + c² = 25
Let given vectors are \(\vec{\alpha}=3 \hat{i}-\hat{j}-\hat{k}\)
and \(\vec{\beta}=\hat{i}+\hat{j}-2 \hat{k}\)
Since \(\vec{d}\) is coplanar with \(\vec{\alpha} \text { and } \vec{\beta}\).

⇒ 2a + 2b + c = 0 ………………(5)
Using eqn. (2), (3) and (4) in eqn. (5) ; we have
2 (3λ + μ) + 2 (- λ + μ) + (- λ – 2μ) = 0
⇒ 3λ + 2μ = 0
⇒ μ = – \(\frac{3 \lambda}{2}\) ………………(6)
∴ eqn. (2) ; (3) and (4) becomes ;

Question 36.
(i) If \(\vec{a}\),\(\vec{b}\) and \(\vec{c}\) are vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), then show taht the angle θ between \(\vec{b} \text { and } \vec{c}\) is given that cos θ = \(=\frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}\).
(ii) If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) and \(|\vec{a}|=5,|\vec{b}|=6 \quad \text { and } \quad|\vec{c}|=9\), then find the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:
(i)

(ii) Given \(|\vec{a}|=5\) ;
\(|\vec{b}|=6\) ;
and \(|\vec{c}|=9\)
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)

Question 38.
In any ∆ABC, prove by vector method that cos B = \(\frac{c^2+a^2-b^2}{2 c a}\). (ISC 2010).
Solution:
Let \(\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CA}}, \overrightarrow{\mathrm{AB}}\) represents the vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\) respectively
Now \(\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}+\overrightarrow{\mathrm{AB}}=\overrightarrow{0}\)

Question 39.
Prove by vector method that a diameter of a circle will subtend a right angle at a point on its circumference. (ISC 2003)
Solution:
Let O be the centre of circle with AB as diameter
and P be any point on the circumference of circle.
Take O as origin
let \(\overrightarrow{\mathrm{OP}}=\vec{b}\) ;
\(\overrightarrow{\mathrm{OB}}=\vec{a}\) ;
∴ \(\overrightarrow{\mathrm{OA}}=-\vec{a}\)
Now we want to prove that
∠APB = 90°

∴ ∠APB = 90°
Hence, the diameter of a circle will subtend a right angle at a point on its circumference.

Question 40.
Prove by vector method that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
Solution:
Let ABCD is a rectangle.