# Ncert Solutions For Class 12 Maths Ex 10.3

## Ncert Solutions For Class 12 Maths Chapter 10 Ex 10.3

Q.1 : Find the angle between two vectors $$\vec{ m }$$ and $$\vec{ n }$$ with magnitude $$\sqrt{ 3 }$$and 2 , respectively having $$\vec{ m }.\vec{ n } = \sqrt{ 6 }$$

Solution 1:

It is given that,

$$\left | \vec{ m } \right |= \sqrt{ 3 }$$ ,

$$\left | \vec{ n } \right |$$ = 2

And $$\vec{ m }.\vec{ n } = \sqrt{ 6 }$$

$$\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta$$

Now, we know that

Therefore,

$$\boldsymbol{\Rightarrow }$$ $$\sqrt{ 6 } = \sqrt{ 3 } \times 2 \times cos\theta$$

$$\boldsymbol{\Rightarrow }$$ $$\cos \theta = \frac{ \sqrt{ 6 }}{ \sqrt{ 3 } \times 2 }$$

$$\boldsymbol{\Rightarrow }$$ $$\cos \theta = \frac{ 1 }{ \sqrt{ 2 }}$$

$$\boldsymbol{\Rightarrow }$$ $$\theta = \frac{ \pi }{ 4 }$$

Therefore, the angle between the given vectors $$\vec{ m }$$ and $$\vec{ n }$$ is $$\frac{ \pi }{ 4 }$$

Q 2 : Find the angle between the vectors $$\hat{ a } – 2 \hat{ b } + 3 \hat{ c } and 3 \hat{ a } – 2 \hat{ b } + \hat{ c }$$

Solution 2:

The given vectors are:

$$\vec{ m } = \hat{ a } – 2 \hat{ b } + 3 \hat{ c } and \vec{ n } = 3 \hat{ a } – 2 \hat{ b } + \hat{ c }$$ $$\left |\vec{ m } \right | = \sqrt{ 1 ^{ 2 } + \left ( – 2 \right )^{ 2 } + 3 ^{ 2 }}$$ $$\left |\vec{ m } \right | = \sqrt{ 1 + 4 + 9 }$$ $$\left |\vec{ m } \right | = \sqrt{ 14 }$$ $$\left |\vec{ n } \right | = \sqrt{ 3 ^{ 2 } + \left ( – 2 \right )^{ 2 } + 1 ^{ 2 }}$$ $$\left |\vec{ n } \right | = \sqrt{ 9 + 4 + 1 }$$ $$\left |\vec{ n } \right | = \sqrt{ 14 }$$ $$Now, \vec{ m }.\vec{ n } = \left ( \hat{ a } – 2 \hat{ b } + 3 \hat{ c } \right )\left ( 3 \hat{ a } – 2 \hat{ b } + \hat{ c } \right )$$ $$Now, \vec{ m }.\vec{ n } = 1.3 + \left ( – 2 \right )\left ( – 2 \right ) + 3.1$$ $$Now, \vec{ m }.\vec{ n } = 3 + 4 + 3$$ $$Now, \vec{ m }.\vec{ n } = 10$$

Also, we know that

$$\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta$$

Therefore,

$$10 = \sqrt{ 14 } \sqrt{ 14 }\cos \theta$$ $$\cos \theta = \frac{ 10 }{ 14 }$$ $$\theta = \cos^{-1} \frac{ 5 }{ 7 }$$

Q 3. Find the projection of the vector $$\hat{ a } – \hat{ b }$$ on the vector $$\hat{ a } + \hat{ b }$$.

Solution 3:

Let, $$\hat{ i } = \hat{ a } – \hat{ b }$$

And $$\hat{ j } = \hat{ a } + \hat{ b }$$

Now, projection of vector $$\vec{ i }$$ on $$\vec{ j }$$ is given by,

$$\frac{ 1 }{ \left | \vec{ j} \right |} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 1 }{ \sqrt{ 1 + 1 }} \left \{ 1.1 + \left ( – 1 \right ) \left ( 1 \right ) \right \} = \frac{ 1 }{ \sqrt{ 2 }}\left ( 1 – 1 \right ) = 0$$

Hence the projection of vector $$\vec{ i }$$ on $$\vec{ j }$$ is 0

Q 4. Find the projection of the vector $$\hat{ a } + 3 \hat{ b } + 7 \hat{ c }$$ on the vector $$7\hat{ a } – \hat{ b } + 8 \hat{ c }$$

Solution 4:

Let $$\hat{ i } = \hat{ a } + 3 \hat{ b } + 7 \hat{ c }$$ and $$\hat{ j } = 7\hat{ a } – \hat{ b } + 8 \hat{ c }$$

Now, projection of vector $$\vec{ i }$$ on $$\vec{ j }$$ is given by,

$$\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 1 }{ \sqrt{ 7^{ 2 } + \left ( – 1 \right )^{ 2 } + 8 ^{ 2 }}} \left \{ 1 \left ( 7 \right ) + 3 \left ( – 1 \right ) + 7 \left ( 8 \right )\right \}$$ $$\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 7 – 3 + 56 }{ \sqrt{ 49 + 1 + 64 }}$$ $$\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 60 }{ \sqrt{ 114 }}$$

Q 5: Show that each of the given three vectors is a unit vector :

$$\frac{ 1 }{ 7 } \left ( 2 \hat{ a } + 3 \hat{ b } + 6 \hat{ c }\right ) , \frac{ 1 }{ 7 } \left ( 3 \hat{ a } – 6 \hat{ b } + 2 \hat{ c }\right ) , \frac{ 1 }{ 7 } \left ( 6 \hat{ a } + 2 \hat{ b } – 3 \hat{ c }\right )$$

Also, show that they are mutually perpendicular to each other.

Solution 5:

$$Let \vec{ i } = \frac{ 1 }{ 7 } \left ( 2 \hat{ a } + 3 \hat{ b } + 6 \hat{ c }\right ) = \frac{ 2 }{ 7 } \hat{ a } + \frac{ 3 }{ 7 } \hat{ b } + \frac{ 6 }{ 7 } \hat{ c } \\ Let \vec{ j } = \frac{ 1 }{ 7 } \left ( 3 \hat{ a } – 6 \hat{ b } + 2 \hat{ c }\right ) = \frac{ 3 }{ 7 } \hat{ a } – \frac{ 6 }{ 7 } \hat{ b } + \frac{ 2 }{ 7 } \hat{ c } \\ Let \vec{ k } = \frac{ 1 }{ 7 } \left ( 6 \hat{ a } + 2 \hat{ b } – 3 \hat{ c }\right ) = \frac{ 6 }{ 7 } \hat{ a } + \frac{ 2 }{ 7 } \hat{ b } – \frac{ 3 }{ 7 } \hat{ c }$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{ i } \right | = \sqrt{ \left (\frac{ 2 }{ 7 }\right )^{ 2 } + \left (\frac{ 3 }{ 7 }\right )^{ 2 } + \left (\frac{ 6 }{ 7 }\right )^{ 2 }}$$

$$\left | \vec{ i } \right | = \sqrt{ \frac{ 4 }{ 49 } + \frac{ 9 }{ 49 } + \frac{ 36 }{ 49 }}$$ $$\left | \vec{ i } \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{ j } \right | = \sqrt{ \left (\frac{ 3 }{ 7 }\right )^{ 2 } + \left (\frac{ – 6 }{ 7 }\right )^{ 2 } + \left (\frac{ 2 }{ 7 }\right )^{ 2 }}$$

$$\left | \vec{ j } \right | = \sqrt{ \frac{ 9 }{ 49 } + \frac{ 36 }{ 49 } + \frac{ 4 }{ 49 }}$$ $$\left | \vec{ j } \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{ k } \right | = \sqrt{ \left (\frac{ 6 }{ 7 }\right )^{ 2 } + \left (\frac{ 2 }{ 7 }\right )^{ 2 } + \left (\frac{ – 3 }{ 7 }\right )^{ 2 }}$$

$$\left | \vec{ k } \right | = \sqrt{ \frac{ 36 }{ 49 } + \frac{ 4 }{ 49 } + \frac{ 9 }{ 49 }}$$ $$\left | \vec{ k } \right | = 1$$

Thus, each of the given three vectors is a unit vector.

$$\boldsymbol{\Rightarrow }$$ $$\vec{ i }.\vec{ j } = \frac{ 2 }{ 7 }\times \frac{ 3 }{ 7 } + \frac{ 3 }{ 7 } \times \left ( \frac{ -6 }{ 7 } \right ) + \frac{ 6 }{ 7 } \times \frac{2}{7}$$

$$\vec{ i }.\vec{ j } = \frac{6}{49} – \frac{18}{49} + \frac{12}{49}$$ $$\vec{ i }.\vec{ j } = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{ j }.\vec{ k } = \frac{3}{ 7 }\times \frac{ 6 }{ 7 } + \frac{ -6 }{ 7 } \times \left ( \frac{ 2 }{ 7 } \right ) + \frac{ 2 }{ 7 } \times \frac{ -3}{7}$$

$$\vec{ j }.\vec{ k } = \frac{18}{49} – \frac{12}{49} – \frac{6}{49}$$ $$\vec{ j }.\vec{ k } = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{ k }.\vec{ i } = \frac{6}{ 7 }\times \frac{ 2 }{ 7 } + \frac{ 2 }{ 7 } \times \left ( \frac{ 3 }{ 7 } \right ) + \frac{ -3 }{ 7 } \times \frac{ 6}{7}$$

$$\vec{ k }.\vec{ i } = \frac{12}{49} – \frac{6}{49} – \frac{18}{49}$$ $$\vec{ k }.\vec{ i } = 0$$

Hence, the given three vectors are mutually perpendicular to each other.

Q 6: Find:

$$\left | \vec{m} \right | and \left | \vec{n} \right |$$, if $$\left ( \vec{m} + \vec{n} \right ).\left ( \vec{m} + \vec{n} \right ) = 8$$ and $$\left | \vec{m} \right | = 8\left | \vec{n} \right |$$

Solution 6:

$$\left ( \vec{m} + \vec{n} \right ).\left ( \vec{m} – \vec{n} \right ) = 8$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{ m}.\vec{ m} – \vec{ m}\vec{n} + \vec{ n}\vec{ m} – \vec{n}.\vec{ n} = 8$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{ m} \right |^{ 2} – \left | \vec{ n} \right |^{ 2} = 8$$

$$\boldsymbol{\Rightarrow }$$ $$8 \left | \vec{ n} \right |^{2} – \left | n \right |^{ 2} = 8$$ . . . . . . . . . $$\left [ \left | \vec{ m } = 8\left | \vec{ n} \right | \right | \right ]$$

$$\boldsymbol{\Rightarrow }$$ $$64 \left | \vec{n} \right |^{2} – \left | \vec{n} \right |^{2} = 8$$

$$\boldsymbol{\Rightarrow }$$ $$63 \left | \vec{ n } \right |^{2} = 8$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{n} \right |^{2} = \frac{ 8}{ 63}$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{n} \right |^{2} = \sqrt{\frac{ 8 }{ 63 }}$$ [ magnitude of a vector is non-negative]

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{n} \right |^{2} = \frac{ 2 \sqrt{2}}{ 3 \sqrt{7}}$$

$$\left | \vec{m} \right | = 8 \left | \vec{n} \right |$$ $$\left | \vec{m} \right | = \frac{ 8 \times 2\sqrt{2}}{ 3\sqrt{7}}$$ $$\left | \vec{m} \right | = \frac{ 16 \sqrt{2}}{ 3 \sqrt{7}}$$

Q 7: Find the product of the following : $$\left ( 3 \vec{m} – 5 \vec{n}\right ).\left ( 2\vec{m} + 7\vec{n}\right )$$

Solution 7:

$$\left ( 3 \vec{m} – 5 \vec{n}\right ).\left ( 2\vec{m} + 7\vec{n}\right )$$ $$= 3 \vec{m}.2 \vec{m} + 3 \vec{m}.7 \vec{n} – 5 \vec{n}.2 \vec{m} – 5 \vec{n}.7 \vec{n}$$ $$= 6 \vec{m}. \vec{m} + 21 \vec{m}. \vec{n} – 10 \vec{n}. \vec{m} – 35 \vec{n}. \vec{n}$$ $$= 6 \left | \vec{m} \right |^{2} + 11 \vec{m}.\vec{n} – 35 \left | \vec{n} \right |^{2}$$

Q 8: Find the magnitude of two vectors $$\vec{m} and \vec{n}$$, having the same magnitude and such that the angle between them is 60° and their scalar product is $$\frac{ 1 }{ 2 }$$

Solution 8:

Let θ be the angle between the vectors $$\vec{m} and \vec{n}$$.

As given in the question:

$$\left | \vec{m} \right | = \left | \vec{n} \right |, \vec{m}.\vec{n} = \frac{1}{2} and \theta = 60 ^{\circ}$$. . . . . . . . . . . . . . . . . . . . . . . ( 1 )

We know that:

$$\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta$$

Therefore,

$$\frac{1}{2} = \left | \vec{m} \right |\left | \vec{m} \right | cos 60 ^{\circ}$$ . . . . . . . . . . . . . . . . . . . [using ( 1 )]

$$\boldsymbol{\Rightarrow }$$ $$\frac{1}{2} = \left | \vec{m} \right |^{2} \times \frac{1}{2}$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{m} \right |^{2} = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{m} \right |^{2} = \left | \vec{n} \right |^{2} = 1$$

Q 9: Find:

$$\left | \vec{y} \right |$$ , if for a unit vector $$\vec{b}$$ , $$\left (\vec{y } – \vec{b} \right ).\left (\vec{y } + \vec{b} \right ) = 12$$

Solution 9:

$$\left (\vec{y } – \vec{b} \right ).\left (\vec{y } + \vec{b} \right ) = 12$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{y}.\vec{y} + \vec{y}.\vec{b} – \vec{b}.\vec{y} – \vec{b}.\vec{b} = 12$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{y} \right |^{ 2 } – \left |\vec{b} \right |^{ 2 } = 12$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{y} \right |^{ 2 } – 1 = 12 \left [ \left |\vec{b} \right | = 1 as \vec{b} is a unit vector \right ]$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{y} \right |^{ 2 } = 13$$

Therefore,

$$\left | \vec{ y } \right | = \sqrt{ 13 }$$

Q 10: If $$\vec{i} = 2\hat{a} + 2\hat{b} + 3\hat{c}$$ ,

$$\vec{j} = -\hat{a} + 2\hat{b} + \hat{c}$$

and $$\vec{k} = 3\hat{a} + \hat{b}$$ are such that $$\vec{i} + \lambda \vec{j}$$ is perpendicular to $$\vec{k}$$ then find the value of λ

Solution 10:

The given vectors are $$\vec{i} = 2\hat{a} + 2\hat{b} + 3\hat{c}$$ , $$\vec{j} = -\hat{a} + 2\hat{b} + \hat{c}$$ and $$\vec{k} = 3\hat{a} + \hat{b}$$

Now,

$$\vec{i} + \lambda \vec{j} = \left ( 2 \hat{a} + 2 \hat{b} + 3 \hat{c} \right ) + \lambda \left ( – \hat{a} + 2 \hat{b} + \hat{c} \right ) = ( 2 – \lambda ) \hat{a} + ( 2 + 2 \lambda )\hat{b} + \left ( 3 + \lambda \right )\hat{c}$$

If $$\left (\vec{i} + \lambda \vec{j} \right )$$ is perpendicular to $$\vec{k}$$ ,then

$$\left (\vec{i} + \lambda \vec{j} \right ).\vec{k} = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\left [\left ( 2 – \lambda \right )\hat{a} + \left ( 2 + 2\lambda \right )\hat{b} + \left ( 3 + \lambda \right )\hat{c} \right ]\left ( 3\hat{a} + \hat{b} \right ) = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\left [\left ( 2 – \lambda \right )3 + \left ( 2 + 2\lambda \right )1 + \left ( 3 + \lambda \right )0 \right ] = 0$$

$$\boldsymbol{\Rightarrow }$$ $$6 – 3\lambda + 2 + 2\lambda = 0$$

$$\boldsymbol{\Rightarrow }$$ $$– \lambda + 8 = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\lambda = 8$$

Hence, the required value of λ is 8.

Q 11: Show that:

$$\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a}$$ is perpendicular to $$\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a}$$ , for any two non zero vectors $$\vec{a} and \vec{b}$$

Solution 11:

$$\boldsymbol{\Rightarrow }$$ $$\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )$$

$$\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )$$ = $$\left |\vec{a} \right |^{2} \vec{b}.\vec{b} – \left |\vec{a} \right |\left |\vec{b} \right |\vec{b}.\vec{a} + \left |\vec{b} \right |\left |\vec{a} \right |\vec{a}.\vec{b} – \left |\vec{b} \right |^{2} \vec{a}.\vec{a}$$

$$= \left | \vec{a} \right |^{ 2 } \left |\vec{b} \right |^{ 2 }$$

$$\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )$$ = 0

Hence, $$\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a}$$ is perpendicular to $$\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a}$$

Q 12: If $$\vec{a}.\vec{a} = 0 and \vec{a}.\vec{b} = 0$$, then what can be concluded about the vector $$\vec{b}$$ ?

Solution:

It is given that $$\vec{a}.\vec{a} = 0 and \vec{a}.\vec{b} = 0$$

Now,

that $$\vec{a}.\vec{a} = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{a} \right |^{ 2 } = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{a} \right |= 0$$

Therefore, $$\vec{a}$$ is a zero vector

Hence, vector $$\vec{b}$$ satisfying $$\vec{a}.\vec{b} = 0$$ can be any vector

Q 13:

If $$\vec{a} , \vec{b} , \vec{c}$$ are unit vectors such that $$\vec{a} + \vec{b} + \vec{c}$$ = 0, find the value of $$\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}$$

Solution 13:

$$\left |\vec{a} + \vec{b }+ \vec{c} \right |^{ 2 } = \left (\vec{a} + \vec{b} + \vec{c} \right )\left (\vec{a} + \vec{b} + \vec{c} \right )$$

$$\left |\vec{a} + \vec{b }+ \vec{c} \right |^{ 2 }$$ = $$\left |\vec{a} \right | ^{2}+ \left |\vec{b} \right |^{ 2 } +\left | \vec{c} \right |^{ 2 } + 2 \left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right )$$

$$\boldsymbol{\Rightarrow }$$ $$0 = 1 + 1 + 1 + 2 \left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right )$$

$$\boldsymbol{\Rightarrow }$$ $$\left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right ) = \frac{ – 3 }{ 2 }$$

Q 14: If either vector $$\vec{a} = \vec{0} or \vec{b} = \vec{0 }$$ , then $$\vec{a}.\vec{b} = 0$$. But the converse need not be true. Justify your answer with an example.

Solution 14:

Consider $$\vec{a} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k} and \vec{b} = 3 \hat{i} + 3 \hat{j } – 6 \hat{k}$$

Then,

$$\vec{a}.\vec{b} = 2.3 + 4.3 + 3 \left ( – 6 \right ) = 6 + 12 – 18 = 0$$

We now observe that :

$$\left |\vec{a} \right | = \sqrt{ 2 ^{2} + 4 ^{ 2 } + 3 ^{2}} = \sqrt{ 29 }$$

Therefore,

$$\vec{a} \neq \vec{0}$$ $$\left |\vec{ b } \right | = \sqrt{ 3 ^{2} + 3 ^{ 2 } + ( -6 ) ^{2}} = \sqrt{ 54 }$$

Therefore,

$$\vec{b} \neq \vec{0}$$

Hence, the converse of the given statement need not be true

Q 15: If the vertices A, B, C of a triangle ABC are ( 1 , 2 , 3 ) , ( – 1 , 0 , 0 ) , ( 0 , 1 , 2 ) , respectively, then find ABC. [ ABC is the angle between the vectors $$\vec{BA} and \vec{BC}$$ ]

Solution 15:

The vertices of ∆ABC are given as A ( 1 , 2 , 3 ) , B ( – 1 , 0 , 0 ) , and C ( 0 , 1 , 2 ). Also, it is given that ∠ ABC is the angle between the vectors $$\vec{BA} and \vec{BC}$$

$$\vec{BA} = \left \{ 1 – \left ( – 1 \right ) \right \}\hat{ i} + \left ( 2 – 0 \right ) \hat{j} + \left ( 3 – 0 \right ) \hat{k} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$$ $$\vec{BC} = \left \{ 0 – \left ( – 1 \right ) \right \}\hat{ i} + \left ( 1 – 0 \right ) \hat{j} + \left ( 2 – 0 \right ) \hat{k} = \hat{i} + \hat{j} + 2 \hat{k}$$

Therefore,

$$\vec{BA }. \vec{BC} = \left ( 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \right ).\left ( \hat{i} + \hat{j} + 2 \hat{k} \right ) = 2 \times 1 + 2 \times 1 + 3 \times 2 = 2 + 2 + 6 = 10$$ $$\left |\vec{BA } \right | = \sqrt{ 2 ^{ 2 } + 2 ^{2} + 3^{ 2 }} = \sqrt{ 4 + 4 + 9 } = \sqrt{ 17 }$$ $$\left |\vec{BC } \right | = \sqrt{ 1 + 1 + 2 ^{ 2 }} = \sqrt{ 6 }$$

Now, it is known that :

$$\vec{BA} . \vec{BC} = \left |\vec{BA} \right |\left |\vec{BC} \right | cos \left ( \angle ABC \right )$$

Therefore,

$$10 = \sqrt{ 17 } \times \sqrt{ 6 } cos \left ( \angle ABC \right )$$

$$\boldsymbol{\Rightarrow }$$ $$cos \left ( \angle ABC \right ) = \frac{ 10 }{ \sqrt{ 17 } \times \sqrt{6}}$$

$$\boldsymbol{\Rightarrow }$$ $$\angle ABC = \cos^{-1}\left ( \frac{ 10 }{ \sqrt{ 102 }} \right )$$

Q 16: Show that the points A ( 1 , 2 , 7 ) , B ( 2 , 6 , 3 ) and C ( 3 , 10 , – 1 ) are collinear.

Solution 16:

The given points are A ( 1 , 2 , 7 ) , B ( 2 , 6 , 3 ) , and C ( 3 , 10 , – 1 ).

Therefore,

$$\vec{AB} = \left ( 2 – 1 \right ) \hat{i} + \left ( 6 – 2 \right ) \hat{j} + \left ( 3 – 7 \right ) \hat{k} = \hat{i} + 4 \hat{j} – 4 \hat{k}$$ $$\vec{BC} = \left ( 3 – 2 \right ) \hat{i} + \left ( 10 – 6 \right ) \hat{j} + \left ( – 1 – 3 \right ) \hat{k} = \hat{i} + 4 \hat{j} – 4 \hat{k}$$ $$\vec{AC} = \left ( 3 – 1 \right ) \hat{i} + \left ( 10 – 2 \right ) \hat{j} + \left ( – 1 – 7 \right ) \hat{k} = 2 \hat{i} + 8 \hat{j} – 8 \hat{k}$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{AB} \right | = \sqrt{ 1 ^{2} + 4 ^{2} + \left ( – 4 \right )^{ 2 }} = \sqrt{ 1 + 16 + 16 } = \sqrt{ 33 }$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{BC} \right | = \sqrt{ 1 ^{2} + 4 ^{2} + \left ( – 4 \right )^{ 2 }} = \sqrt{ 1 + 16 + 16 } = \sqrt{ 33 }$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{AC} \right | = \sqrt{ 2 ^{2} + 8 ^{2} + 8 ^{2} } = \sqrt{ 4 + 64 + 64 } = \sqrt{ 132 }$$ = $$2 \sqrt{33}$$

Therefore,

$$\left |\vec{AC} \right | = \left |\vec{AB} \right | + \left |\vec{BC} \right |$$

Hence, the given points A, B, and C are collinear.

Q 17: Show that the vectors $$2 \hat{i} + \hat{j} + \hat{k} , \hat{i} – 3 \hat{j } – 5 \hat{k} and 3 \hat{i} – 4 \hat{j} – 4 \hat{k}$$ form the vertices of a right angled triangle.

Solution 17:

Let vectors $$2 \hat{i} + \hat{j} + \hat{k} , \hat{i} – 3 \hat{j } – 5 \hat{k} and 3 \hat{i} – 4 \hat{j} – 4 \hat{k}$$ be position vectors of points A, B, and C respectively.

i.e , $$\vec{OA} = 2 \hat{i} – \hat{j} + \hat{k} , OB = \hat{i} – 3 \hat{j} – 5 \hat{k} and OC = 3 \hat{i} – 4 \hat{j } – 4 \hat{k}$$

Therefore,

$$\vec{AB} = \left ( 1 – 2 \right ) \hat{i} + \left ( – 3 + 1 \right ) \hat{j} + \left ( – 5 – 1 \right ) \hat{k} = – \hat{i} – 2 \hat{j} – 6 \hat{k}$$ $$\vec{BC} = \left ( 3 – 1 \right ) \hat{i} + \left ( – 4 + 3 \right ) \hat{j} + \left ( – 4 + 5 \right ) \hat{k} = – 2\hat{i} – \hat{j} – \hat{k}$$ $$\vec{AC} = \left ( 2 – 3 \right ) \hat{i} + \left ( – 1 + 4 \right ) \hat{j} + \left ( 1 + 4 \right ) \hat{k} = – \hat{i} + 3 \hat{j} + 5 \hat{k}$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{AB} \right | = \sqrt{ \left ( – 1 \right )^{ 2 } + \left ( – 2 \right )^{ 2 } + \left ( – 6 \right ) ^{ 2 }} = \sqrt{ 1 + 4 + 36 } = \sqrt{ 41 }$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{BC} \right | = \sqrt{ \left ( 2 \right )^{ 2 } + \left ( – 1 \right )^{ 2 } + \left ( 1 \right ) ^{ 2 }} = \sqrt{ 1 + 4 + 1 } = \sqrt{ 6 }$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{AC} \right | = \sqrt{ \left ( – 1 \right )^{ 2 } + \left ( 3 \right )^{ 2 } + \left ( 5 \right ) ^{ 2 }} = \sqrt{ 1 + 9 + 25 } = \sqrt{ 35 }$$

Therefore,

$$\left |\vec{BC} \right |^{ 2 } + \left |\vec{AC} \right | ^{2} = 6 + 35 = 41 = \left |\vec{AB} \right | ^{ 2 }$$

Hence, ∆ ABC is a right – angled triangle.

Q 18:

If $$\vec{a}$$ is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ $$\vec{a}$$ is unit vector if

(A) λ = 1

(B) λ = – 1

(C) $$a = \left |\lambda \right |$$

(D) $$a = \frac{ 1 }{ \left | \lambda \right |}$$

Solution 18:

Vector $$\lambda \vec{a}$$ is a unit vector if $$\left |\lambda \vec{a} \right | = 1$$

Now,

$$\left |\lambda \vec{a} \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\lambda \right | \left |\vec{a} \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{a} \right | = \frac{ 1 }{ \left |\lambda \right | }$$ . . . . . . . . . . . . [ λ ≠ 0 ]

$$\boldsymbol{\Rightarrow }$$ $$a = \frac{ 1 }{ \left |\lambda \right | }$$ . . . . . . . . . . . . . . . . . . [$$\left |\vec{a} \right | = a$$ ]

Therefore, vector $$\lambda \vec{a}$$ is a unit vector if $$a = \frac{ 1 }{ \left | \lambda \right |}$$