NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

Vectors Class 12 NCERT Solutions introduces students to the concept of vectors and their algebraic operations. It covers topics such as scalar and vector products, triple products, and their applications in geometry and physics.

The chapter is divided into four exercises and one miscellaneous exercise. Here is a brief summary of each exercise:

Exercise 10.1: This exercise has ten questions that ask students to find the direction cosines of a given vector and vice versa.

Exercise 10.2: This exercise has eleven questions that ask students to find the magnitude and direction of a given vector, as well as to perform vector operations such as addition, subtraction, and scalar multiplication.

Exercise 10.3: This exercise has seven questions that ask students to find the scalar and vector products of two given vectors.

Exercise 10.4: This exercise has six questions that ask students to find the scalar and vector triple products of given vectors.

Miscellaneous Exercise: This exercise has six questions that ask students to apply the concepts of vector algebra to solve problems in geometry and physics.

Overall, this chapter is essential for students who want to pursue higher education in Mathematics and Physics. It provides a solid foundation for students to understand the concepts of vectors and their applications in real-world problems.

Access NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra

Exercise 10.1

1. Represent graphically a displacement of $40\text{km},{30}^{\circ }$ east of north.

Ans:

Here, the vector OP is representing the displacement of $40\text{km},{30}^{\circ }$ in East of North direction.

2. Classify the following measures as scalars and vectors.

i. $10\text{kg}$

Ans: $10\text{kg}$is a scalar quantity because it has only magnitude not direction.

ii. 2 meters north-west

Ans: 2 meters north-west is a vector quantity because it has both magnitude as well as direction.

iii. ${40}^{\circ }$

Ans: ${40}^{\circ }$ is a scalar quantity because it has only magnitude, not direction.

iv. $40$ watt

Ans: 40 watts is a scalar quantity because it has only magnitude, not direction.

v. ${10}^{-19}$coulomb

Ans: ${10}^{-19}$Coulomb is a scalar quantity because it has only magnitude not direction.

vi. $20\text{m/}{\text{s}}^2$

Ans: $20\text{m/}{\text{s}}^2$ is a vector quantity because it has both magnitude as well as direction.

3.  Classify the following as scalar and vector quantities:-

i. Time period

Ans: Time period is a scalar quantity because it has only magnitude.

ii. distance

Ans: Distance is a scalar quantity because it has only magnitude.

iii. force   

Ans: Force is a vector quantity because it has both magnitude as well as direction.

iv. velocity

Ans: Velocity is a vector quantity because it has both magnitude as well as direction.

v. work done

Ans: Work done is a scalar quantity because it has only magnitude.

4.  In Figure, identify the following vectors.

i. Co-initial

Ans: Vectors $\overrightarrow{a}$ and $\overrightarrow{d}$ are co-initial.

ii. Equal   

Ans: Vectors $\overrightarrow{b}$ and $\overrightarrow{d}$ are equal.

iii. Collinear but not equal 

Ans: Vectors $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear but not equal.

5. Answer the following as true or false.

i.$\overrightarrow{a}$ and $-\overrightarrow{a}$ and are collinear.

Ans: True

ii. Two collinear vectors are always equal in magnitude.

Ans: False

iii. Two vectors having same magnitude are collinear.

Ans: False

iv. Two collinear vectors having the same magnitude are equal.

Ans: False

Exercise 10.2

1.  Compute the magnitude of the following vectors:-

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k};\overrightarrow{b}=2\hat{i}-7\hat{j}-3\hat{k};\overrightarrow{c}={\displaystyle \frac{1}{\sqrt{3}}}\hat{i}+{\displaystyle \frac{1}{\sqrt{3}}}\hat{j}-{\displaystyle \frac{1}{\sqrt{3}}}\hat{k}$

Ans:  $|\overrightarrow{a}|=\sqrt{{(1)}^2+{(1)}^2+{(1)}^2}=\sqrt{3}$

$|\overrightarrow{b}|=\sqrt{{(2)}^2+{(-7)}^2+{(-3)}^2}$

$=\sqrt{4+49+9}$

$=\sqrt{62}$

$|\overrightarrow{c}|=\sqrt{{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2-{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2}$

$=\sqrt{{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}}=1$

2. Write two different vectors having the same magnitude. 

Ans: $\overrightarrow{a}=(\hat{i}-2\hat{j}+3\hat{k})$  and  $\overrightarrow{b}=(2\hat{i}+\hat{j}-3\hat{k})$.

$|\overrightarrow{a}|=\sqrt{1^2+{(-2)}^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$

$|\overrightarrow{b}|=\sqrt{2^2+1^2+{(-3)}^2}=\sqrt{4+1+9}=\sqrt{14}$

But $\overrightarrow{a}\ne \overrightarrow{b}$

3. Write two different vectors having same direction. 

Ans:  $\overrightarrow{p}=(\hat{i}+\hat{j}+\hat{k})$  and  $\overrightarrow{q}=(2\hat{i}+2\hat{j}+2\hat{k})$.

The Direction Cosines of $\overrightarrow{p}$ are $a={\displaystyle \frac{1}{\sqrt{1^2+1^2+1^2}}}={\displaystyle \frac{1}{\sqrt{3}}},b={\displaystyle \frac{1}{\sqrt{1^2+1^2+1^2}}}={\displaystyle \frac{1}{\sqrt{3}}},c={\displaystyle \frac{1}{\sqrt{1^2+1^2+1^2}}}={\displaystyle \frac{1}{\sqrt{3}}}$

The Direction Cosines of $\overrightarrow{q}$ are$a={\displaystyle \frac{2}{\sqrt{2^2+2^2+2^2}}}={\displaystyle \frac{2}{2\sqrt{3}}}={\displaystyle \frac{1}{\sqrt{3}}},b={\displaystyle \frac{2}{\sqrt{2^2+2^2+2^2}}}={\displaystyle \frac{2}{2\sqrt{3}}}={\displaystyle \frac{1}{\sqrt{3}}},c={\displaystyle \frac{2}{\sqrt{2^2+2^2+2^2}}}={\displaystyle \frac{2}{2\sqrt{3}}}={\displaystyle \frac{1}{\sqrt{3}}}$

The Direction Cosines of $\overrightarrow{p}$ and $\overrightarrow{q}$ are equal but $\overrightarrow{p}\ne \overrightarrow{q}$.

4. Find the values of x and y so that the vectors $2\hat{i}+3\hat{j}$ and $x\hat{i}+y\hat{j}$ are equal. 

Ans:  $2\hat{i}+3\hat{j}=x\hat{i}+y\hat{j}\Rightarrow x=2,y=3$

5.  Find the scalar and vector components of the vector with initial point $(2,1)$ and terminal point $(-5,7)$.

Ans:  Let the points be $\text{P}(2,1)$ and $\text{Q}(-5,7)$ 

$\overrightarrow{PQ}=(-5-2)\hat{i}+(7-1)\hat{j}$

$\Rightarrow \overrightarrow{PQ}=-7\hat{i}+6\hat{j}$

So, scalar components of required vector  are $-7$ and 6 and the vector components are $-7\hat{i}$ and $6\hat{j}$.

6. Find the sum of the vectors $\overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k},\overrightarrow{b}=-2\hat{i}+4\hat{j}+5\hat{k}$ and $\overrightarrow{c}=\hat{i}-6\hat{j}-7\hat{k}$.

Ans:  $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}=0\hat{i}-4\hat{j}-1\hat{k}=-4\hat{j}-\hat{k}$

7.  Find the unit vector in the direction of the vector $\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}$. 

Ans: $|\overrightarrow{a}|=\sqrt{1^2+1^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$

$\therefore \hat{a}={\displaystyle \frac{\overrightarrow{a}}{|\overrightarrow{a}|}}={\displaystyle \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}}={\displaystyle \frac{1}{\sqrt{6}}}\hat{i}+{\displaystyle \frac{1}{\sqrt{6}}}\hat{j}+{\displaystyle \frac{2}{\sqrt{6}}}\hat{k}$

8. Find the unit vector in the direction of vector $\overrightarrow{PQ}$, where $\text{P}$ and $\text{Q}$ are the points $(1,2,3)$ and $(4,5,6)$, respectively.

Ans:  $\overrightarrow{PQ}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)k=3\hat{i}+3\hat{j}+3k$

$|\overrightarrow{PQ}|=\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=\sqrt{27}=3\sqrt{3}$

So, unit vector $={\displaystyle \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}}={\displaystyle \frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}}={\displaystyle \frac{1}{\sqrt{3}}}\hat{i}+{\displaystyle \frac{1}{\sqrt{3}}}\hat{j}+{\displaystyle \frac{1}{\sqrt{3}}}\hat{k}$

9. For given vectors, $\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{b}=-\hat{i}+\hat{j}-\hat{k}$, find the unit vector in the direction of the vector $\overrightarrow{a}+\overrightarrow{b}$. 

Ans: $\overrightarrow{a}+\overrightarrow{b}=(2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}=1\hat{i}+0\hat{j}+1\hat{k}=\hat{i}+\hat{k}$

$|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{1^2+1^2}=\sqrt{2}$

So, unit vector $={\displaystyle \frac{(\overrightarrow{a}+\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}|}}={\displaystyle \frac{\hat{i}+\hat{k}}{\sqrt{2}}}={\displaystyle \frac{1}{\sqrt{2}}}\hat{i}+{\displaystyle \frac{1}{\sqrt{2}}}\hat{k}$

10. Find a vector in the direction of vector $5\hat{i}-\hat{j}+2\hat{k}$ which has magnitude 8 units.

Ans: $\overrightarrow{a}=5\hat{i}-\hat{j}+2\hat{k}$

$|\overrightarrow{a}|=\sqrt{5^2+{(-1)}^2+2^2}=\sqrt{25+1+4}=\sqrt{30}$

$\therefore \hat{a}={\displaystyle \frac{\overrightarrow{a}}{|\overrightarrow{a}|}}={\displaystyle \frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}}$

So, a vector in direction of  $5\hat{i}-\hat{j}+2\hat{k}$ with magnitude 8 units is:$8\hat{a}=8\left({\displaystyle \frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}}\right)={\displaystyle \frac{40}{\sqrt{30}}}\hat{i}-{\displaystyle \frac{8}{\sqrt{30}}}\hat{j}+{\displaystyle \frac{16}{\sqrt{30}}}\hat{k}$

11. Show that the vectors $2\hat{i}-3\hat{j}+4\hat{k}$ and $-4\hat{i}+6\hat{j}-8\hat{k}$ are collinear.

Ans:  $\overrightarrow{a}=2\hat{i}-3\hat{j}+4\hat{k}$ and $\overrightarrow{b}=-4\hat{i}+6\hat{j}-8\hat{k}$

$\overrightarrow{b}=-4\hat{i}+6\hat{j}-8\hat{k}=-2(2\hat{i}-3\hat{j}+4\hat{k})=-2\overrightarrow{a}$

$\therefore \overrightarrow{b}=\lambda \overrightarrow{a},\lambda =-2$

So, the given vectors are collinear.

12. Find the direction cosines of the vector $\hat{i}+2\hat{j}+3\hat{k}$.

Ans: $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$.

$|\overrightarrow{a}|=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$

So, the Direction cosines of $\overrightarrow{a}$ are $\left({\displaystyle \frac{1}{\sqrt{14}}},{\displaystyle \frac{2}{\sqrt{14}}},{\displaystyle \frac{3}{\sqrt{14}}}\right)$

13.  Find the direction cosines of the vector joining the points $\text{A}(1,2,-3)$ and $\text{B}(-1,-2,1)$ directed from A to b.

Ans: $\overrightarrow{\text{AB}}=(-1-1)\hat{i}+(-2-2)\hat{j}+\{1-(-3)\}\hat{k}$

$\Rightarrow \overrightarrow{\text{AB}}=-2\hat{i}-4\hat{j}+4\hat{k}$

$|\overrightarrow{\text{AB}}|=\sqrt{{(-2)}^2+{(-4)}^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6$

So, the Direction cosines of $\overrightarrow{\text{AB}}$ are $\left(-{\displaystyle \frac{2}{6}},-{\displaystyle \frac{4}{6}},{\displaystyle \frac{4}{6}}\right)=\left(-{\displaystyle \frac{1}{3}},-{\displaystyle \frac{2}{3}},{\displaystyle \frac{2}{3}}\right)$

14.  Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes OX, OY and OZ .

Ans: $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

$|\overrightarrow{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

So, the Direction Cosines of $\overrightarrow{a}$ are $\left({\displaystyle \frac{1}{\sqrt{3}}},{\displaystyle \frac{1}{\sqrt{3}}},{\displaystyle \frac{1}{\sqrt{3}}}\right)$

$\cos \alpha ={\displaystyle \frac{1}{\sqrt{3}}},\cos \beta ={\displaystyle \frac{1}{\sqrt{3}}},\cos \gamma ={\displaystyle \frac{1}{\sqrt{3}}}$

So, the vector is equally inclined to OX, OY, and $\text{OZ}$.

15. Find the position vector of a point $\text{R}$ which divides the line joining two points P and Q whose position vectors are $\hat{i}+2\hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ratio 2 : 1 ,

i. Internally 

Ans: $\overrightarrow{\text{OP}}=\hat{i}+2\hat{j}-\hat{k}$ and $\overrightarrow{\text{OQ}}=-\hat{i}+\hat{j}+\hat{k}$

 The position vector of $\text{R}$ is $\overrightarrow{\text{OR}}={\displaystyle \frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2\hat{j}-\hat{k})}{2+1}}={\displaystyle \frac{(-2\hat{i}+2\hat{j}+2\hat{k})+(\hat{i}+2\hat{j}-\hat{k})}{3}}$

$={\displaystyle \frac{-\hat{i}+4\hat{j}+\hat{k}}{3}}=-{\displaystyle \frac{1}{3}}\hat{i}+{\displaystyle \frac{4}{3}}\hat{j}+{\displaystyle \frac{1}{3}}\hat{k}$

ii. Externally.

Ans: The position vector of $R$ is $\overrightarrow{\text{OR}}={\displaystyle \frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2\hat{j}-\hat{k})}{2-1}}=(-2\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-\hat{k})$

$=-3\hat{i}+3\hat{k}$

16.  Find the position vector of the mid point of the vector joining the points P(2,3,4)  and Q(4,1,-2).

Ans: The position vector of $R$ is

$\overrightarrow{\text{OR}}={\displaystyle \frac{(2\hat{i}+3\hat{j}+4\hat{k})+(4\hat{i}+\hat{j}-2\hat{k})}{2}}={\displaystyle \frac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2}}$

$={\displaystyle \frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}}=3\hat{i}+2\hat{j}+\hat{k}$

17. Show that the points $\text{A},\text{B}$ and $\text{C}$ with position vectors, $\overrightarrow{a}=3\hat{i}-4\hat{j}-4\hat{k},\overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{c}=\hat{i}-3\hat{j}-5\hat{k}$, respectively form the vertices of a right angled triangle. 

Ans: 

$\overrightarrow{\text{AB}}=\overrightarrow{b}-\overrightarrow{a}=(2-3)\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}=-\hat{i}+3\hat{j}+5\hat{k}$

$\overrightarrow{\text{BC}}=\overrightarrow{c}-\overrightarrow{b}=(1-2)\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}=-\hat{i}+2\hat{j}-6\hat{k}$

$\overrightarrow{\text{CA}}=\overrightarrow{a}-\overrightarrow{c}=(3-1)\hat{i}+(-4+3)\hat{j}+(-4+5)\hat{k}=2\hat{i}-\hat{j}+\hat{k}$

$|\overrightarrow{\text{AB}}{|}^2=(-1{)}^2+3^2+5^2=1+9+25=35$

$|\overrightarrow{\text{BC}}{|}^2=(-1{)}^2+(-2{)}^2+(-6{)}^2=1+4+36=41$

$|\overrightarrow{CA}{|}^2=2^2+(-1{)}^2+1^2=4+1+1=6$

$|\overrightarrow{\text{AB}}{|}^2+|\overrightarrow{\text{CA}}{|}^2=35+6=41=|\overrightarrow{\text{BC}}{|}^2$

So, ABC is a right angled triangle.

18. In triangle ABC, which of the following is not true?

a. $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\overrightarrow{0}$

b. $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\overrightarrow{0}$

c. $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\overrightarrow{0}$

d. $\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\overrightarrow{0}$

Ans:

$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$

$\Rightarrow \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=-\overrightarrow{\text{CA}}$

$\Rightarrow \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\overrightarrow{0}$

$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$

$\Rightarrow \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\overrightarrow{0}$

$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\overrightarrow{0}$

$\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\overrightarrow{0}$

If $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\overrightarrow{0}$

$\overrightarrow{\text{AC}}=\overrightarrow{\text{CA}}$

$\Rightarrow \overrightarrow{\text{AC}}=-\overrightarrow{\text{AC}}$

$\Rightarrow \overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}=\overrightarrow{0}$

$\Rightarrow 2\overrightarrow{\text{AC}}=\overrightarrow{0}$

$\Rightarrow \overrightarrow{\text{AC}}=\overrightarrow{0}$, which is not true.

So, the equation given in option $\text{C}$ is False.

Hence, the correct answer is $\text{C}$.

19. If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two collinear vectors, then which of the following are incorrect?

a. $\overrightarrow{b}=\lambda \overrightarrow{a}$, for some scalar $\lambda$

b. $\overrightarrow{a}=\pm \overrightarrow{b}$

c. the respective components of $\overrightarrow{a}$ and $\overrightarrow{b}$ are proportional

d. both the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ have same direction, but different magnitudes 

Ans: If $\overrightarrow{a}$ and $\overrightarrow{b}$ are collinear vectors, they are parallel. $\overrightarrow{b}=\lambda \overrightarrow{a}($ scalar $\lambda )$

If $\lambda =\pm 1$, then $\overrightarrow{a}=\pm \overrightarrow{b}$

If $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$,

$\overrightarrow{b}=\lambda \overrightarrow{a}\Rightarrow b_1\hat{i}+b_2\hat{j}+b_3\hat{k}=\lambda \left(a\hat{i}+a_2\hat{j}+a_3\hat{k}\right)$

$\Rightarrow b_1\hat{i}+b_2\hat{j}+b_3\hat{k}=\left(\lambda a_1\right)\hat{i}+\left(\lambda a_2\right)\hat{j}+\left(\lambda a_3\right)\hat{k}$

$\Rightarrow b_1=\lambda a_1,b_2=\lambda a_2,b_3=\lambda a_3$

$\Rightarrow {\displaystyle \frac{b_1}{a_1}}={\displaystyle \frac{b_2}{a_2}}={\displaystyle \frac{b_3}{a_3}}=\lambda$

Thus, respective components of $\overrightarrow{a}$ and $\overrightarrow{b}$ are proportional. 

But, $\overrightarrow{a}$ and $\overrightarrow{b}$ may have different directions. So, option D is incorrect. The correct answer is $\text{D}$.

Exercise 10.3

1. Find the angle between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ with magnitudes $\sqrt{3}$ and 2, respectively having $\overrightarrow{a}\cdot \overrightarrow{b}=\sqrt{6}$

Ans: $|\overrightarrow{a}|=\sqrt{3},|\overrightarrow{b}|=2,\overrightarrow{a}.\overrightarrow{b}=\sqrt{6}$

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

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

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

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

2.  Find the angle between the vectors $i-2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$. 

Ans: $|\overrightarrow{a}|=\sqrt{1^2+{(-2)}^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$

$|\overrightarrow{b}|=\sqrt{3^2+{(-2)}^2+1^2}=\sqrt{9+4+1}=\sqrt{14}$

$\overrightarrow{a}.\overrightarrow{b}=(i-2\hat{j}+3\hat{k})(3\hat{i}-2\hat{j}+\hat{k})$

$=1.3+(-2)(-2)+3.1$

$=3+4+3$

$=10$

$\therefore 10=\sqrt{14}\sqrt{14}\cos \theta$

$\Rightarrow \cos \theta ={\displaystyle \frac{10}{14}}$

$\Rightarrow \theta ={\cos }^{-1}\left({\displaystyle \frac{5}{7}}\right)$

3. Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $i+\hat{j}$. 

Ans: $\overrightarrow{a}=\hat{i}-\hat{j}$ and $\overrightarrow{b}=\hat{i}+\hat{j}$

Projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is ${\displaystyle \frac{1}{\mid \overrightarrow{b}|}}(\overrightarrow{a}\cdot \overrightarrow{b})={\displaystyle \frac{1}{\sqrt{1+1}}}\{1.1+(-1)(1)\}={\displaystyle \frac{1}{\sqrt{2}}}(1-1)=0$

4. Find the projection of the vector $\hat{i}+3\hat{j}+7\hat{k}$ on the vector $7\hat{i}-\hat{j}+8\hat{k}$. 

Ans: $\overrightarrow{a}=\hat{i}+3\hat{j}+7\hat{k}$ and $\overrightarrow{b}=7\hat{i}-\hat{j}+8\hat{k}$

Projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is ${\displaystyle \frac{1}{|\overrightarrow{b}|}}(\overrightarrow{a}\cdot \overrightarrow{b})={\displaystyle \frac{1}{\sqrt{7^2+{(-1)}^2+8^2}}}\{1(7)+3(-1)+7(8)\}={\displaystyle \frac{7-3+56}{\sqrt{49+1+64}}}={\displaystyle \frac{60}{\sqrt{114}}}$

5. Show that each of the given three vectors is a unit vector, which are mutually perpendicular to each other. ${\displaystyle \frac{1}{7}}(2\hat{i}+3\hat{j}+6\hat{k}),{\displaystyle \frac{1}{7}}(3\hat{i}-6\hat{j}+2\hat{k}),{\displaystyle \frac{1}{7}}(6\hat{i}+2\hat{j}-3\hat{k})$

Ans: $\overrightarrow{a}={\displaystyle \frac{1}{7}}(2\hat{i}+3\hat{j}+6\hat{k})={\displaystyle \frac{2}{7}}\hat{i}+{\displaystyle \frac{3}{7}}\hat{j}+{\displaystyle \frac{6}{7}}\hat{k}$,

$\overrightarrow{b}={\displaystyle \frac{1}{7}}(3\overline{i}-6\hat{j}+2\hat{k})={\displaystyle \frac{3}{7}}\hat{i}-{\displaystyle \frac{6}{7}}\overline{j}+{\displaystyle \frac{2}{7}}\hat{k}$

$\overrightarrow{c}={\displaystyle \frac{1}{7}}(6\hat{i}+2\hat{j}-3\hat{k})={\displaystyle \frac{6}{7}}\hat{i}+{\displaystyle \frac{2}{7}}\hat{j}-{\displaystyle \frac{3}{7}}\hat{k}$

$|\overrightarrow{a}|=\sqrt{{\left({\displaystyle \frac{2}{7}}\right)}^2+{\left({\displaystyle \frac{3}{7}}\right)}^2+{\left({\displaystyle \frac{6}{7}}\right)}^2}=\sqrt{{\displaystyle \frac{4}{49}}+{\displaystyle \frac{9}{49}}+{\displaystyle \frac{36}{49}}}=1$

$|\overrightarrow{b}|=\sqrt{{\left({\displaystyle \frac{3}{7}}\right)}^2+{\left(-{\displaystyle \frac{6}{7}}\right)}^2+{\left({\displaystyle \frac{2}{7}}\right)}^2}=\sqrt{{\displaystyle \frac{9}{49}}+{\displaystyle \frac{36}{49}}+{\displaystyle \frac{4}{49}}}=1$

$|\overrightarrow{c}|=\sqrt{{\left({\displaystyle \frac{6}{7}}\right)}^2+{\left({\displaystyle \frac{2}{7}}\right)}^2+{\left(-{\displaystyle \frac{3}{7}}\right)}^2}=\sqrt{{\displaystyle \frac{36}{49}}+{\displaystyle \frac{4}{49}}+{\displaystyle \frac{9}{49}}}=1$

So, each of the vectors is a unit vector.

$\overrightarrow{a}.\overrightarrow{b}={\displaystyle \frac{2}{7}}\times {\displaystyle \frac{3}{7}}+{\displaystyle \frac{3}{7}}\times \left({\displaystyle \frac{-6}{7}}\right)+{\displaystyle \frac{6}{7}}\times {\displaystyle \frac{2}{7}}={\displaystyle \frac{6}{49}}-{\displaystyle \frac{18}{49}}+{\displaystyle \frac{12}{49}}=0$

$\overrightarrow{b}\cdot \overrightarrow{c}={\displaystyle \frac{3}{7}}\times {\displaystyle \frac{6}{7}}+\left({\displaystyle \frac{-6}{7}}\right)\times {\displaystyle \frac{2}{7}}+{\displaystyle \frac{2}{7}}\times \left({\displaystyle \frac{-3}{7}}\right)={\displaystyle \frac{18}{49}}-{\displaystyle \frac{12}{49}}-{\displaystyle \frac{6}{49}}=0$

$\overrightarrow{c}.\overrightarrow{a}={\displaystyle \frac{6}{7}}\times {\displaystyle \frac{2}{7}}+{\displaystyle \frac{2}{7}}\times {\displaystyle \frac{3}{7}}+\left({\displaystyle \frac{-3}{7}}\right)\times {\displaystyle \frac{6}{7}}={\displaystyle \frac{12}{49}}+{\displaystyle \frac{6}{49}}-{\displaystyle \frac{18}{49}}=0$

So, given vectors are mutually perpendicular to each other.

6. Find $|\overrightarrow{a}|$ and $|\overrightarrow{b}|$, if $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=8$ and $|\overrightarrow{a}|=8|\overrightarrow{b}|$.

Ans: $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=8$

$\Rightarrow \overrightarrow{a}\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{a}-\overrightarrow{b}\overrightarrow{b}=8$

$\Rightarrow |\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2=8$

$\Rightarrow (8|\overrightarrow{b}|{)}^2-|\overrightarrow{b}{|}^2=8$

$\Rightarrow 64|\overrightarrow{b}{|}^2-|\overrightarrow{b}{|}^2=8$

$\Rightarrow 63|\overrightarrow{b}{|}^2=8$

$\Rightarrow |\overrightarrow{b}{|}^2={\displaystyle \frac{8}{63}}$

$\Rightarrow |\overrightarrow{b}|=\sqrt{{\displaystyle \frac{8}{63}}}$

$\Rightarrow |\overrightarrow{b}|={\displaystyle \frac{2\sqrt{2}}{3\sqrt{7}}}$

$|\overrightarrow{a}|=8|\overrightarrow{b}|={\displaystyle \frac{8\times 2\sqrt{2}}{3\sqrt{7}}}={\displaystyle \frac{16\sqrt{2}}{3\sqrt{7}}}$

7. Evaluate the product $(3\overrightarrow{a}-5\overrightarrow{b})\cdot (2\overrightarrow{a}+7\overrightarrow{b})$

Ans: $(3\overrightarrow{a}-5\overrightarrow{b})\cdot (2\overrightarrow{a}+7\overrightarrow{b})$

$=3\overrightarrow{a}\cdot 2\overrightarrow{a}+3\overrightarrow{a}.7\overrightarrow{b}-5\overrightarrow{b}\cdot 2\overrightarrow{a}-5\overrightarrow{b}\cdot 7\overrightarrow{b}$

$=6\overrightarrow{a}\overrightarrow{a}+21\overrightarrow{a}\overrightarrow{b}-10\overrightarrow{a}\overrightarrow{b}-35\overrightarrow{b}\overrightarrow{b}$

$=6|\overrightarrow{a}{|}^2+11\overrightarrow{a}\overrightarrow{b}-35|\overrightarrow{b}{|}^2$

8. Find the magnitude of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, having the same magnitude and such that the angle between them is ${60}^{\circ }$ and their scalar product is ${\displaystyle \frac{1}{2}}$ 

Ans: Let $\theta$ be angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.

$|\overrightarrow{a}|=|\overrightarrow{b}|,\overrightarrow{a}.\overrightarrow{b}={\displaystyle \frac{1}{2}}$, and $\theta ={60}^{\circ }$

$\therefore {\displaystyle \frac{1}{2}}=|\overrightarrow{a}|\overrightarrow{b}\mid \cos {60}^{\circ }$

$\Rightarrow {\displaystyle \frac{1}{2}}=|\overrightarrow{a}{|}^2\times {\displaystyle \frac{1}{2}}$

$\Rightarrow |\overrightarrow{a}{|}^2=1$

$\Rightarrow |\overrightarrow{a}|=|\overrightarrow{b}|=1$

9. Find $|\overrightarrow{x}|$, if for a unit vector $\overrightarrow{a},(\overrightarrow{x}-\overrightarrow{a})\cdot (\overrightarrow{x}+\overrightarrow{a})=12$

Ans: $(\overrightarrow{x}-\overrightarrow{a})\cdot (\overrightarrow{x}+\overrightarrow{a})=12$

$\Rightarrow \overrightarrow{x}\cdot \overrightarrow{x}+\overrightarrow{x}\overrightarrow{a}-\overrightarrow{a}\overrightarrow{x}-\overrightarrow{a}\cdot \overrightarrow{a}=12$

$\Rightarrow |\overrightarrow{x}{|}^2-|\overrightarrow{a}{|}^2=12$

$\Rightarrow |\overrightarrow{x}{|}^2-1=12$

$\Rightarrow |\overrightarrow{x}{|}^2=13$

$\therefore |\overrightarrow{x}|=\sqrt{13}$

10. If $\overrightarrow{a}=2\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{b}=-\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{c}=3\hat{i}+\hat{j}$ are such that $\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$, then find the value of $\lambda$.

Ans: $\overrightarrow{a}+\lambda \overrightarrow{b}=(2\hat{i}+2\hat{j}+3\hat{k})+\lambda (-\hat{i}+2\hat{j}+\dot{k})=(2-\lambda )\hat{i}+(2+2\lambda )\hat{j}+(3+\lambda )\hat{k}$