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 $\vec a$ and $\vec d$ are co-initial.

ii. Equal   

Ans: Vectors $\vec b$ and $\vec d$ are equal.

iii. Collinear but not equal 

Ans: Vectors $\vec a$ and $\vec c$ are collinear but not equal.

5. Answer the following as true or false.

i.$\vec a$ and $ - \vec 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:-

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

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

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

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

$ = \sqrt {62} $

$|\vec c| = \sqrt {{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} - {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}} $

$ = \sqrt {\dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}}  = 1$

2. Write two different vectors having the same magnitude. 

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

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

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

But $\vec a \ne \vec b$

3. Write two different vectors having same direction. 

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

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

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

The Direction Cosines of $\vec p$ and $\vec q$ are equal but $\vec p \ne \vec q$.

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 $\vec a = \hat i - 2\hat j + \hat k\;,\;\vec b =  - 2\hat i + 4\hat j + 5\hat k$ and $\vec c = \hat i - 6\hat j - 7\hat k$.

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

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

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

$\therefore \hat a = \dfrac{{\vec a}}{{|\vec a|}} = \dfrac{{\hat i + \hat j + 2\hat k}}{{\sqrt 6 }} = \dfrac{1}{{\sqrt 6 }}\hat i + \dfrac{1}{{\sqrt 6 }}\hat j + \dfrac{2}{{\sqrt 6 }}\hat k$

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 $ = \dfrac{{\overrightarrow {PQ} }}{{|\overrightarrow {PQ} |}} = \dfrac{{3\hat i + 3\hat j + 3\hat k}}{{3\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}\hat i + \dfrac{1}{{\sqrt 3 }}\hat j + \dfrac{1}{{\sqrt 3 }}\hat k$

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

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

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

So, unit vector $ = \dfrac{{(\vec a + \vec b)}}{{|\vec a + \vec b|}} = \dfrac{{\hat i + \hat k}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\hat i + \dfrac{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: $\vec a = 5\hat i - \hat j + 2\hat k$

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

$\therefore \hat a = \dfrac{{\vec a}}{{|\vec a|}} = \dfrac{{5\hat i - \hat j + 2\hat k}}{{\sqrt {30} }}$

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

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:  $\vec a = 2\hat i - 3\hat j + 4\hat k$ and $\vec b =  - 4\hat i + 6\hat j - 8\hat k$

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

$\therefore \vec b = \lambda \vec a,\;\;\lambda  =  - 2$

So, the given vectors are collinear.

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

Ans: $\vec a = \hat i + 2\hat j + 3\hat k$.

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

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

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 $\left( { - \dfrac{2}{6}, - \dfrac{4}{6},\dfrac{4}{6}} \right) = \left( { - \dfrac{1}{3}, - \dfrac{2}{3},\dfrac{2}{3}} \right)$

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

Ans: $\vec a = \hat i + \hat j + \hat k$

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

So, the Direction Cosines of $\vec a$ are $\left( {\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}} \right)$

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

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

$ = \dfrac{{ - \;\hat i + 4\hat j + \hat k}}{3} =  - \dfrac{1}{3}\hat i + \dfrac{4}{3}\hat j + \dfrac{1}{3}\hat k$

ii. Externally.

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

$ =  - 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

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

$ = \dfrac{{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, $\vec a = 3\hat i - 4\hat j - 4\hat k\;,\;\vec b = 2\hat i - \hat j + \hat k$ and $\vec c = \hat i - 3\hat j - 5\hat k$, respectively form the vertices of a right angled triangle. 

Ans: 

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

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

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

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

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

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

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

So, ABC is a right angled triangle.

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

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

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

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

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

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}}}  = \vec 0$

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

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

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

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

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

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

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

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

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

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

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

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

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

a. $\vec b = \lambda \vec a$, for some scalar $\lambda $

b. $\vec a =  \pm \vec b$

c. the respective components of $\vec a$ and $\vec b$ are proportional

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

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

If $\lambda  =  \pm 1$, then $\vec a =  \pm \vec b$

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

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

$ \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 \dfrac{{{b_1}}}{{{a_1}}} = \dfrac{{{b_2}}}{{{a_2}}} = \dfrac{{{b_3}}}{{{a_3}}} = \lambda $

Thus, respective components of $\vec a$ and $\vec b$ are proportional. 

But, $\vec a$ and $\vec b$ 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 $\vec a$ and $\vec b$ with magnitudes $\sqrt 3 $ and 2, respectively having $\vec a \cdot \vec b = \sqrt 6 $

Ans: $|\vec a| = \sqrt 3 \;,\;|\vec b| = 2\;,\;\vec a\;.\;\vec b = \sqrt 6 $

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

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

$ \Rightarrow \cos \theta  = \dfrac{1}{{\sqrt 2 }}$

$ \Rightarrow \theta  = \dfrac{\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: $|\vec a| = \sqrt {{1^2} + {{( - 2)}^2} + {3^2}}  = \sqrt {1 + 4 + 9}  = \sqrt {14} $

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

$\vec a\;.\;\vec b = (i - 2\hat j + 3\hat k)(3\hat i - 2\hat j + \hat k)$

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

$ = 3 + 4 + 3$

$ = 10$

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

$ \Rightarrow \cos \theta  = \dfrac{{10}}{{14}}$

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

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

Ans: $\vec a = \hat i - \hat j$ and $\vec b = \hat i + \hat j$

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

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: $\vec a = \hat i + 3\hat j + 7\hat k$ and $\vec b = 7\hat i - \hat j + 8\hat k$

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

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

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

$\vec b = \dfrac{1}{7}(3\bar i - 6\hat j + 2\hat k) = \dfrac{3}{7}\hat i - \dfrac{6}{7}\bar j + \dfrac{2}{7}\hat k$

$\vec c = \dfrac{1}{7}(6\hat i + 2\hat j - 3\hat k) = \dfrac{6}{7}\hat i + \dfrac{2}{7}\hat j - \dfrac{3}{7}\hat k$

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

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

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

So, each of the vectors is a unit vector.

$\vec a\;.\;\vec b = \dfrac{2}{7} \times \dfrac{3}{7} + \dfrac{3}{7} \times \left( {\dfrac{{ - 6}}{7}} \right) + \dfrac{6}{7} \times \dfrac{2}{7} = \dfrac{6}{{49}} - \dfrac{{18}}{{49}} + \dfrac{{12}}{{49}} = 0$

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

$\vec c\;.\;\vec a = \dfrac{6}{7} \times \dfrac{2}{7} + \dfrac{2}{7} \times \dfrac{3}{7} + \left( {\dfrac{{ - 3}}{7}} \right) \times \dfrac{6}{7} = \dfrac{{12}}{{49}} + \dfrac{6}{{49}} - \dfrac{{18}}{{49}} = 0$

So, given vectors are mutually perpendicular to each other.

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

Ans: $(\vec a + \vec b) \cdot (\vec a - \vec b) = 8$

$ \Rightarrow \vec a\vec a - \vec a\;.\;\vec b + \vec b \cdot \vec a - \vec b\vec b = 8$

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

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

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

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

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

$ \Rightarrow |\vec b| = \sqrt {\dfrac{8}{{63}}} $

$ \Rightarrow |\vec b| = \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }}$

$|\vec a| = 8|\vec b| = \dfrac{{8 \times 2\sqrt 2 }}{{3\sqrt 7 }} = \dfrac{{16\sqrt 2 }}{{3\sqrt 7 }}$

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

Ans: $(3\vec a - 5\vec b) \cdot (2\vec a + 7\vec b)$

$ = 3\vec a \cdot 2\vec a + 3\vec a.7\vec b - 5\vec b \cdot 2\vec a - 5\vec b \cdot 7\vec b$

$ = 6\vec a\vec a + 21\vec a\vec b - 10\vec a\vec b - 35\vec b\vec b$

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

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

Ans: Let $\theta $ be angle between $\vec a$ and $\vec b$.

$|\vec a| = |\vec b|,\vec a\;.\;\vec b = \dfrac{1}{2}$, and $\theta  = {60^\circ }$

$\therefore \dfrac{1}{2} = |\vec a|\vec b\mid \cos {60^\circ }$

$ \Rightarrow \dfrac{1}{2} = |\vec a{|^2} \times \dfrac{1}{2}$

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

$ \Rightarrow |\vec a| = |\vec b| = 1$

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

Ans: $(\vec x - \vec a) \cdot (\vec x + \vec a) = 12$

$ \Rightarrow \vec x \cdot \vec x + \vec x\vec a - \vec a\vec x - \vec a \cdot \vec a = 12$

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

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

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

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

10. 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 perpendicular to $\vec c$, then find the value of $\lambda $.

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

$(\vec a + \lambda \vec b) \cdot \vec c = 0$

$ \Rightarrow [(2 +  - )\hat i + (2 + 2\lambda )\hat j + (3 + \lambda )\hat k] \cdot (3\hat i + j) = 0$

$ \Rightarrow 3(2 - \lambda ) + (2 + 2\lambda ) + 0(3 + \lambda ) = 0$

$ \Rightarrow 6 - 3\lambda  + 2 + 2\lambda  = 0$

$ \Rightarrow  - \lambda  + 8 = 0$

$ \Rightarrow \lambda  = 8$

11. Show that $|\vec a|\vec b + |\vec b|\vec a$ is perpendicular to $|\vec a|\vec b - |\vec b|\vec a$ , For any two nonzero vectors $\vec a$ and $\vec b$.

Ans: $(|\vec a|\vec b + |\vec b|\vec a) \cdot (|\vec a|\vec b - |\vec b|\vec a)$

$ = |\vec a{|^2}\vec b\vec b - |\vec a||\vec b|\vec b\vec a + |\overrightarrow b ||a|\vec a\vec b - |\vec b{|^2}\vec a \cdot \vec a$

$ = |\vec a{|^2}|\overrightarrow b {|^2} - |\overrightarrow b {|^2}|\overrightarrow a {|^2}\;\; = \;\;0$

12. If $\vec a \cdot \vec a = 0$ and $\vec a\vec b = 0$, then what can be concluded above the vector $\vec b$ ? 

Ans: $\vec a \cdot \vec a = 0 \Rightarrow |\vec a{|^2} = 0 \Rightarrow |\vec a| = 0$

$\therefore \vec a$ is the zero vector. Thus, any vector $\vec b$ can satisfy $\vec a\;.\;\vec b = 0$.

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

Ans: $|\vec a + \vec b + \vec c{|^2} = (\vec a + \vec b + \vec c) \cdot (\vec a + \vec b + \vec c) = |\vec a{|^2} + |\vec b{|^2} + |\vec c{|^2} + 2(\vec a\vec b + \vec b\vec c + \vec c \cdot \vec a)$

$ \Rightarrow 0 = 1 + 1 + 1 + 2(\vec a\vec b + \vec b \cdot \vec c + \vec c \cdot \vec a)$

$ \Rightarrow (\vec a\;.\;\vec b + \vec b \cdot \vec c + \vec c\;.\;\vec a) = \dfrac{{ - 3}}{2}$

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

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

$\vec a \cdot \vec b = 2.3 + 4.3 + 3( - 6) = 6 + 12 - 18 = 0$

$|{\text{a}}| = \sqrt {{2^2} + {4^2} + {3^2}}  = \sqrt {29} $

$\therefore \vec a \ne \vec 0$

$|\vec b| = \sqrt {{3^2} + {3^2} + {{( - 6)}^2}}  = 54$

$\therefore \vec b \ne \bar 0$

So, it is clear from the above example that the converse of the given statement need not be true.

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

Ans: $\overrightarrow {{\text{BA}}}  = \{ 1 - ( - 1)\hat i + (2 - 0)\hat j + (3 - 0)\hat k = 2\hat i + 2\hat j + 3\hat k$

$\overrightarrow {{\text{BC}}}  = \{ 0 - ( - 1)\} \hat i + (1 - 0)\hat j + (2 - 0)\hat k = \hat i + \hat j + 2\hat k$

$\overrightarrow {{\text{BA}}}  \cdot \overrightarrow {{\text{BC}}}  = (2\hat i + 2\hat \jmath  + 3\hat k) \cdot (\hat i + \hat \jmath  + 2\hat k) = 2 \times 1 + 2 \times 1 + 3 \times 2 = 2 + 2 + 6 = 10$

$|\overrightarrow {{\text{BA}}} | = \sqrt {{2^2} + {2^2} + {3^2}}  = \sqrt {4 + 4 + 9}  = \sqrt {17} $

$|\overrightarrow {BC} | = \sqrt {1 + 1 + {2^2}}  = \sqrt 6 $

$\therefore 10 = \sqrt {17}  \times \sqrt 6 \cos (\angle {\text{ABC}})$

$ \Rightarrow \cos (\angle {\text{ABC}}) = \dfrac{{10}}{{\sqrt {17}  \times \sqrt 6 }}$

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

16. Show that the points ${\text{A}}(1,2,7),\;{\text{B}}(2,6,3)$ and ${\text{C}}(3,10, - 1)$ are collinear. 

Ans: $\overrightarrow {{\text{AB}}}  = (2 - 1)\hat i + (6 - 2)\hat j + (3 - 7)\hat k = \hat i + 4\hat j - 4\hat k$

$\overrightarrow {{\text{BC}}}  = (3 - 2)\hat i + (10 - 6)\hat \jmath  + ( - 1 - 3)\hat k = \hat i + 4\hat j - 4\hat k$

$\overrightarrow {{\text{AC}}}  = (3 - 1)\hat i + (10 - 2)\hat j + ( - 1 - 7)\hat k = 2\hat i + 8\hat j - 8\hat k$

$|\overrightarrow {AB} | = \sqrt {{1^2} + {4^2} + {{( - 4)}^2}}  = \sqrt {1 + 16 + 16}  = \sqrt {33} $

$|\overrightarrow {BC} | = \sqrt {{1^2} + {4^2} + {{( - 4)}^2}}  = \sqrt {1 + 16 + 16}  = \sqrt {33} $

$|\overrightarrow {AC} | = \sqrt {{2^2} + {8^2} + {8^2}}  = \sqrt {4 + 64 + 64}  = 2\sqrt {33} $

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

Hence, the given points are collinear.

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.

Ans: $\overrightarrow {{\text{OA}}}  = 2\hat i - \hat j + \hat k\;,\;\overrightarrow {{\text{OB}}}  = \hat i - 3\hat j - 5\hat k\;,\;\overrightarrow {{\text{OC}}}  = 3\hat i - 4\hat j - 4\hat k$

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

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

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

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

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

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

$\therefore |\overrightarrow {BC} {|^2} + |\overrightarrow {AC} {|^2} = 6 + 35 = 41 = |\overrightarrow {AB} {|^2}$

Hence, $\Delta {\text{ABC}}$ is a right triangle.

18. If $\vec a$ is a nonzero vector of magnitude 'a' and $\lambda $ a nonzero scalar. then $\lambda \vec a$ is unit vector if

a. $\lambda  = 1$

b. $\lambda  =  - 1$

c. $a = |\lambda |$

d. $a = \dfrac{1}{{|\lambda |}}$

Ans: $|\lambda \vec a| = 1$

$ \Rightarrow |\vec a| = \dfrac{1}{{|\lambda |}}$

$ \Rightarrow a = \dfrac{1}{{|\lambda |}}$

Exercise 10.4

1.  Find $|\vec a \times \vec b|$, if $\vec a = \hat i - 7\hat j + 7\hat k$ and $\vec b = 3\hat i - 2\hat j + 2\hat k$

Ans:  We have, $\vec a = \hat i - 7\hat j + 7\hat k$  and  $\vec b = 3\hat i - 2\hat j + 2\hat k$

$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&{ - 7}&7 \\ 3&{ - 2}&2 \end{array}} \right|$

$ = \hat i( - 14 + 14) - \hat j(2 - 21) + \hat k( - 2 + 21) = 19\hat j + 19\hat k$

$\therefore |\vec a \times \vec b| = \sqrt {{{(19)}^2} + {{(19)}^2}}  = \sqrt {2 \times {{(19)}^2}}  = 19\sqrt 2 $

2.  Find a unit vector perpendicular to each of the vector $\vec a + \vec b$ and $\vec a - \vec b$, where $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$.

Ans:  $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$

$\vec a + \vec b = 4\hat i + 4\hat j\;,\;\vec a - \vec b = 2\hat i + 4\hat k$

$(\vec a + \vec b) \times (\vec a - \vec b) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 4&4&0 \\ 2&0&4 \end{array}} \right| = \hat i(16) - \hat j(16) + \hat k( - 8) = 16\hat i - 16\hat j - 8\hat k$

$|(\vec a + \vec b) \times (\vec a - \vec b)| = \sqrt {{{16}^2} + {{( - 16)}^2} + {{( - 8)}^2}} $

$ = \sqrt {{2^2} \times {8^2} + {2^2} \times {8^2} + {8^2}} $

$ = 8\sqrt {{2^2} + {2^2} + 1}  = 8\sqrt 9  = 8 \times 3 = 24$

So, the unit vector is $ =  \pm \dfrac{{(\vec a + \vec b) \times (\vec a - \vec b)}}{{|(\vec a + \vec b) \times (\vec a - \vec b)\mid }} =  \pm \dfrac{{16\hat i - 16\hat j - 8\hat k}}{{24}}$

$ =  \pm \dfrac{{2\hat i - 2\hat j - \hat k}}{3} =  \pm \dfrac{2}{3}\hat i \mp \dfrac{2}{3}\hat j \mp \dfrac{1}{3}\hat k$

3.  If a unit vector $\vec a$ makes an angle $\dfrac{\pi }{3}$ with $\hat i,\dfrac{\pi }{4}$ with $\hat j$ and an acute angle $\theta $ with $\hat k$, then find $\theta $ and hence, the components of $\vec a$ 

Ans: $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$

$|\vec a|\; = 1$. $\cos \dfrac{\pi }{3} = \dfrac{{{a_1}}}{{|\vec a|}}$

$ \Rightarrow \dfrac{1}{2} = {a_1}$

$\cos \dfrac{\pi }{4} = \dfrac{{{a_2}}}{{|\vec a|}}$

$ \Rightarrow \dfrac{1}{{\sqrt 2 }} = {a_2}$

$\cos \theta  = \dfrac{{{a_3}}}{{|\vec a|}}$

$ \Rightarrow {a_3} = \cos \theta $

$ \Rightarrow \sqrt {a_1^2 + a_2^2 + a_3^2}  = 1$

$ \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} + {\cos ^2}\theta  = 1$

$ \Rightarrow \dfrac{1}{4} + \dfrac{1}{2} + {\cos ^2}\theta  = 1$

$ \Rightarrow \dfrac{3}{4} + {\cos ^2}\theta  = 1$

$ \Rightarrow {\cos ^2}\theta  = 1 - \dfrac{3}{4} = \dfrac{1}{4}$

$ \Rightarrow \cos \theta  = \dfrac{1}{2} \Rightarrow \theta  = \dfrac{\pi }{3}$

$\therefore {a_3} = \cos \dfrac{\pi }{3} = \dfrac{1}{2}$

So, $\theta  = \dfrac{\pi }{3}$ and components of $\vec a$ are $\left( {\dfrac{1}{2},\dfrac{1}{{\sqrt 2 }},\dfrac{1}{2}} \right)$

4.  Show that $(\vec a - \vec b) \times (\vec a + \vec b) = 2(\vec a \times \vec b)$

Ans: $(\vec a - \vec b) \times (\vec a + \vec b)$

$ = (a - b) \times \vec a + (a - b) \times b$

$ = a \times \vec a - \vec b \times \vec d + a \times \vec b - \vec b \times \vec b$

$ = 0 + \vec a \times \vec b + \vec a \times \vec b - 0$

$ = 2\;(\vec a \times \vec b)$

5. Find $\lambda $ and $\mu $ if $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda \hat j + \mu \hat k) = \vec 0$

Ans: $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda \hat j + \mu \hat k) = \vec 0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&6&{27} \\ 1&\lambda &\mu \end{array}} \right| = 0\hat i + 0\hat j + 0\hat k$

$ \Rightarrow \hat i(6\mu  - 27\lambda ) - \hat j(2\mu  - 27) + \hat k(2\lambda  - 6) = 0\hat i + 0\hat j + 0\hat k$

$6\mu  - 27\lambda  = 0$

$2\mu  - 27 = 0$

$2\lambda  - 6 = 0$

$\lambda  = 3$

$\mu  = \dfrac{{27}}{2}$

6.  Given that $\vec a\;.\;\vec b = 0$ and $\vec a \times \vec b = 0$. What can you conclude about $\vec a$ and $\vec b$ ? 

Ans: $\vec a\;.\;\vec b = 0$

i. $|\vec a| = 0$ or $|\vec b| = 0$ or $\vec a \bot \vec b\quad ($ if $|\vec a| \ne 0$ and $|\vec b| \ne 0)$

$\vec a \times \vec b = 0$

ii. $|\overrightarrow a |\; = \;0\;{\text{or}}\;|\vec b|\; = \;0$ or  (if $|\vec a| \ne 0$ and $|\vec b| \ne 0)$

But $\overrightarrow a $ and $\overrightarrow b $ cannot be parallel and perpendicular at same time.

So, $|\vec a| = 0$ or $|\vec b| = 0$

7. Let the vectors $\vec a,\vec b,\vec c$ given as ${a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;,\;{b_1}\hat i + {b_2}\hat j + {b_3}\hat k\;,\;{c_1}\hat i + {c_2}\hat j + {c_3}\hat k$.

Then show that $ = \vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c$

Ans: $(\vec b + \vec c) = \left( {{b_1} + {c_1}} \right)\hat i + \left( {{b_2} + {c_2}} \right)\hat j + \left( {{b_3} + {c_3}} \right)\hat k$

Now, $\vec a \times (\vec b + \vec c)\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1} + {c_1}}&{{b_2} + {c_2}}&{{b_3} + {c_3}} \end{array}} \right|$

$ = i\left[ {{a_2}\left( {{b_3} + {c_3}} \right) - {a_3}\left( {{b_2} + {c_2}} \right)} \right] - \hat j\left[ {{a_1}\left( {{b_3} + {c_3}} \right) - {a_3}\left( {{b_1} + {c_i}} \right)} \right] + \hat k\left[ {{a_1}\left( {{b_2} + {c_2}} \right) - {a_2}\left( {{b_1} + {c_1}} \right)} \right]$

$ = i\left[ {{a_2}{b_3} + {a_2}{c_3} - {a_3}{b_2} - {a_3}{c_2}} \right] + \hat j\left[ { - {a_1}{b_3} - {a_1}{c_3} + {a_3}{b_1} + {a_3}{c_1}} \right] + \hat k\left[ {{a_1}{b_2} + {a_1}{c_2} - {a_2}{b_1} - {a_2}{c_i}} \right]$

$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|$

$ = i\left[ {{a_2}{b_3} - {a_3}{b_2}} \right] + \hat j\left[ {{b_1}{a_3} - a{b_3}} \right] + \hat k\left[ {{a_1}{b_2} - {a_2}h} \right]$

$\vec a \times \vec c = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|$

$ = \hat i\left[ {{a_2}{c_3} - {a_3}{c_2}} \right] + \hat j\left[ {{a_3}{c_1} - {a_1}{c_3}} \right] + \hat k\left[ {a{b_2} - {a_2}b} \right]$

$\left. {(\vec a \times \vec b) + (\vec a \times \vec c) = \hat i\left[ {{a_2}{b_3} + {a_2}{c_3} - {a_3}{b_2} - {a_3}{c_2}} \right] + \hat j{{[{b_1}{a_3} + {a_3}{c_1} - {a_1}{b_3} - {a_i}\} }_3}} \right]$

$ + \hat k\left[ {a,b, + {a_1}c, - a,{b_1} - a,{c_1}} \right]$

$\vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c$

Hence, proved.

8. If either $\vec a = 0$ or $b = 0$, then $\vec a \times b = 0$.Is the converse true? Justify your answer with an example.

Ans: Let $\vec a = 2\hat i + 3\hat j + 4\hat k\;,\;\vec b = 4\hat i + 6\hat j + 8\hat k\;,\;\vec a \times \vec b = \vec 0$

$\vec a \times \vec b = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ 2&3&4 \\ 4&6&8 \end{array}} \right| = \hat i(24 - 24) - \hat j(16 - 16) + \hat k(12 - 12) = 0\hat i + 0\hat j + 0\hat k$

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

$\therefore \vec a \ne \vec 0$

$|\vec b| = \sqrt {{4^2} + {6^2} + {8^2}}  = \sqrt {116} $

$\therefore \vec b \ne \vec 0$

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

9. Find the area of the triangle with vertices A $(1,1,2)$, B $(2,3,5)$ and $C(1,5,5)$.

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

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

Area $ = \dfrac{1}{2}|\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{BC}}} |$

$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{BC}}} = \left| {\begin{array}{*{20}{l}} {\hat i}&{\hat j}&{\hat k} \\ 1&2&3 \\ { - 1}&2&0 \end{array}} \right| = \hat i( - 6) - \hat j(3) + \hat k(2 + 2) = - 6\hat i - 3\hat j + 4\hat k$

$|\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{BC}}} | = \sqrt {{{( - 6)}^2} + {{( - 3)}^2} + {4^2}}  = \sqrt {36 + 9 + 16}  = \sqrt {61} $

So, the area of $\Delta {\text{ABC}}$ is $\dfrac{{\sqrt {61} }}{2}$ sq units.

10.  Find the area of the parallelogram whose adjacent sides are determined by the vector $\vec a = \hat i - \hat j + 3\hat k$ and $\overrightarrow {\text{b}}  = 2\hat i - 7\hat j + \hat k$

Ans: $\vec a \times \overrightarrow {\text{b}} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&{ - 1}&3 \\ 2&{ - 7}&1 \end{array}} \right| = \hat i( - 1 + 21) - \hat j(1 - 6) + \hat k( - 7 + 2) = 20\hat i + 5\hat j - 5\hat k$

$\therefore |\vec a \times \vec b| = \sqrt {{{20}^2} + {5^2} + {5^2}}  = \sqrt {400 + 25 + 25}  = 15\sqrt 2 $

So, area of parallelogram is $15\sqrt 2 $ sq units

11. Let the vectors $\vec a$ and $\overrightarrow {\text{b}} $ be such that $|\vec a| = 3$ and $|\vec b| = \dfrac{{\sqrt 2 }}{3}$, then $\vec a \times \overrightarrow {\text{b}} $ is a unit vector, if the angle between $\vec a$ and $\vec b$is:

a. $\dfrac{\pi }{6}$

b. $\dfrac{\pi }{4}$

c. $\dfrac{\pi }{3}$

d. $\dfrac{\pi }{2}$

Ans:$|\vec a \times \vec b| = 1$

$ \Rightarrow ||a||\vec b|\sin \theta | = 1$

$ \Rightarrow |\vec a||\overrightarrow {\text{b}} ||\sin \theta | = 1$

$ \Rightarrow 3 \times \dfrac{{\sqrt 2 }}{3} \times \sin \theta  = 1$

$ \Rightarrow \sin \theta  = \dfrac{1}{{\sqrt 2 }}$

$ \Rightarrow \theta  = \dfrac{\pi }{4}$

12. Area of a rectangle having vertices ${\text{A}},{\text{B}},{\text{C}}$, and ${\text{D}}$ with position vectors $ - \hat i + \dfrac{1}{2}\hat j + 4\hat k,\hat i + \dfrac{1}{2}\hat j + 4\hat k,\hat i - \dfrac{1}{2}\hat j + 4\hat k$ and $ - \hat i - \dfrac{1}{2}\hat j + 4\hat k$ respectively is

a. $\dfrac{1}{2}$

b. 1

c. 2

d. 4

Ans: $\overrightarrow {{\text{AB}}}  = (1 + 1)\hat i + \left( {\dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + (4 - 4)\hat k = 2\hat i$

$\overrightarrow {{\text{BC}}}  = (1 - 1)\hat i + \left( { - \dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + (4 - 4)\hat k =  - \hat j$

$|\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{BC}}} | = \sqrt {{{( - 2)}^2}}  = 2$

So, the area of the required rectangle is 2 square units.

Miscellaneous Exercise

1. Write down a unit vector in XY-plane, making an angle of ${30^\circ }$ with the positive direction of x axis. 

Ans: Unit vector is $\vec r = \cos \theta \hat i + \sin \theta \hat j$, where $\theta $ is angle with positive ${\text{X}}$ axis. $\vec r = \cos {30^\circ }\hat i + \sin {30^\circ }\hat j = \dfrac{{\sqrt 3 }}{2}\hat i + \dfrac{1}{2}\hat j$

2. Find the scalar components and magnitude of the vector joining the points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$

Ans: $\overrightarrow {{\text{PQ}}}  = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k$

$|\overrightarrow {{\text{PQ}}} |\; = \;\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $

3. A girl walks 4 km towards west, then she walks 3 km in a direction ${30^\circ }$ east of north and stops. Determine the girl's displacement from her initial point of departure. 

Ans:  

$\overrightarrow {{\text{OA}}}  =  - 4\hat i$

$\overrightarrow {{\text{AB}}}  = \hat i|\overrightarrow {{\text{AB}}} |\cos {60^\circ } + \hat j|\overrightarrow {{\text{AB}}} |\sin {60^\circ }$

$ = \hat i3 \times \dfrac{1}{2} + \hat j3 \times \dfrac{{\sqrt 3 }}{2}$

$ = \dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$

$\overrightarrow {{\text{OB}}}  = \overrightarrow {{\text{OA}}}  + \overrightarrow {{\text{AB}}} $

$ = ( - 4\hat i) + \left( {\dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}j} \right)$

$ = \left( { - 4 + \dfrac{3}{2}} \right)\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$

$ = \left( {\dfrac{{ - 8 + 3}}{9}} \right)\hat i + \dfrac{{3\sqrt 3 }}{9}\hat j$

$ = \dfrac{{ - 5}}{2}\vec i + \dfrac{{3\sqrt 3 }}{2}\hat j$

4. If $\overrightarrow a  = \vec b + \vec c$, then is it true that $|\overrightarrow a | = |\vec b| + |\vec c|$ ? Justify your answer. 

Ans:  $\operatorname{In} \Delta {\text{ABC}},\overrightarrow {\;CB}  = \vec a,\;\overrightarrow {CA}  = b,\;\overrightarrow {AB}  = \vec c$

$\vec a = \vec b + \vec c$,  by triangle law of addition for vectors.

$|\vec a|\; < \;|\vec b| + |\vec c|$,  by triangle inequality law of lengths.

 Hence, it's not true that $|\vec a| = |\vec b| + |\vec c|$

5. Find the value of $x$ for which $x(\hat i + \hat j + \hat k)$ unit vector.

Ans: $|x(\hat i + \hat j + \hat k)| = 1$

$ \Rightarrow \sqrt {{x^2} + {x^2} + {x^2}}  = 1$

$ \Rightarrow \sqrt {3{x^2}}  = 1$

$ \Rightarrow \sqrt 3 x = 1$

$ \Rightarrow x =  \pm \dfrac{1}{{\sqrt 3 }}$

6.  Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec a = 2\hat i + 3\hat j - \hat k$ and $\vec b = \hat i - 2\hat j + \hat k$

Ans: $\vec c = \vec a + \vec b = (2 + 1)\hat i + (3 - 2)\hat j + ( - 1 + 1)\hat k = 3\hat i + \hat j$

$|\vec c| = \sqrt {{3^2} + {1^2}}  = \sqrt {9 + 1}  = \sqrt {10} $

$\therefore \hat c = \dfrac{{\vec c}}{{|\vec c|}} = \dfrac{{(3\hat i + \hat j)}}{{\sqrt {10} }}$

So, a vector of magnitude 5 and parallel to the resultant of $\vec a$ and $\vec b$ is $ \pm 5(\hat c) =  \pm 5\left( {\dfrac{1}{{\sqrt {10} }}(3\hat i + \hat j)} \right) =  \pm \dfrac{{3\sqrt {10} }}{2}\hat i \pm \dfrac{{\sqrt {10} }}{2}\hat j$

7.  If $\vec a = \hat i + \hat j + \hat k\;,\;\vec b = 2\hat i - \hat j + 3\hat k$ and $\vec c = \hat i - 2\hat j + \hat k$, find a unit vector parallel to the vector $2\vec a - \vec b + 3\vec c$.

Ans: $2\vec a - \vec b + 3\vec c = 2(\hat i + \hat j + \hat k) - (2\hat i - \hat j + 3\hat k) + 3(\hat i - 2\hat j + \hat k)$

$ = 2\hat i + 2\hat j + 2\hat k - 2\hat i + j - 3\vec k + 3\hat i - 6\hat j + 3\hat k$

$ = 3\hat i - 3\hat j + 2\hat k$

$|2\vec a - \vec b + 3c| = \sqrt {{3^2} + {{( - 3)}^2} + {2^2}}  = \sqrt {9 + 9 + 4}  = \sqrt {22} $

Thus ,required unit vector is $\dfrac{{2\vec a - \vec b + 3\vec c}}{{|2\vec a - \vec b + 3\vec c|}} = \dfrac{{3\hat i - 3\hat j + 2\hat k}}{{\sqrt {22} }} = \dfrac{3}{{\sqrt {22} }}\hat i - \dfrac{3}{{\sqrt {22} }}\hat j + \dfrac{2}{{\sqrt {22} }}\hat k$

8. Show that the points ${\text{A}}(1, - 2, - 8),{\text{B}}(5,0, - 2)$ and ${\text{C}}(11,3,7)$ are collinear, and find the ratio in which B divides Ac. 

Ans: $\overrightarrow {{\text{AB}}}  = (5 - 1)\hat i + (0 + 2)\hat j + ( - 2 + 8)\hat k = 4\hat i + 2\hat j + 6\hat k$

$\overrightarrow {{\text{BC}}}  = (11 - 5)\hat i + (3 - 0)\hat j + (7 + 2)\hat k = 6\hat i + 3\hat j + 9\hat k$

$\overrightarrow {AC}  = (11 - 1)\hat i + (3 + 2)\hat j + (7 + 8)\hat k = 10\hat i + 5\hat j + 15\hat k$

$|\overrightarrow {{\text{AB}}} | = \sqrt {{4^2} + {2^2} + {6^2}}  = \sqrt {16 + 4 + 36}  = \sqrt {56}  = 2\sqrt {14} $

$|\overrightarrow {{\text{BC}}} | = \sqrt {{6^2} + {3^2} + {9^2}}  = \sqrt {36 + 9 + 81}  = \sqrt {126}  = 3\sqrt {14} $

$|\overrightarrow {AC} | = \sqrt {{{10}^2} + {5^2} + {{15}^2}}  = \sqrt {100 + 25 + 225}  = \sqrt {350}  = 5\sqrt {14} $

$\therefore \;\;|\overrightarrow {{\text{AC}}} | = |\overrightarrow {{\text{AB}}} | + |\overrightarrow {{\text{BC}}} |$

So, the points are collinear.

Let B divide AC in ratio $\lambda :1$. $\overrightarrow {{\text{OB}}}  = \dfrac{{\lambda \overrightarrow {{\text{OC}}}  + \overrightarrow {{\text{OA}}} }}{{(\lambda  + 1)}}$

$ \Rightarrow 5\hat i - 2\hat k = \dfrac{{\lambda (11\hat i + 3\hat j + 7\hat k) + (\hat i - 2\hat j - 8\hat k)}}{{\lambda  + 1}}$

$ \Rightarrow (\lambda  + 1)(5\hat i - 2\hat k) = 11\lambda \hat i + 3\lambda \hat j + 7\lambda \hat k + \hat i - 2\hat j - 8\hat k$

$ \Rightarrow 5(\lambda  + 1)\hat i - 2(\lambda  + 1)\hat k = (11\lambda  + 1)\hat i + (3\lambda  - 2)\hat j + (7\lambda  - 8)\hat k$

$ \Rightarrow \lambda  = \dfrac{2}{3}$

So, the required ratio is 2:3

9. Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec a + \vec b)$ and $(\vec a - 3\vec b)$ externally in the ratio 1: 2. Also, show that $P$ is the mid point of the line segment RQ. 

Ans: $\overrightarrow {{\text{OP}}}  = 2\vec a + \vec b,\overrightarrow {{\text{OQ}}}  = \overrightarrow {\text{a}}  - 3\overrightarrow {\text{b}} $

$\overrightarrow {OR}  = \dfrac{{2(2\vec a + \vec b) - (\vec a - 3\vec b)}}{{2 - 1}} = \dfrac{{4\vec a + 2\vec b - \vec a - 3\vec b}}{1} = 3\vec a + 5\vec b$

So, the position vector of ${\text{R}}$ is $3\overrightarrow {\text{a}}  + 5\overrightarrow {\text{b}} $

Position vector of midpoint of ${\text{RQ}} = \dfrac{{\overrightarrow {{\text{OQ}}}  + \overrightarrow {{\text{OR}}} }}{2}$

$ = \dfrac{{(a\sqrt 6 ) + (3\vec a + 5\bar b)}}{2}$

$ = 2\vec a + \vec b$

$ = \overrightarrow {OP} $

Thus, ${\text{P}}$ is midpoint of line segment ${\text{RQ}}$

10. The two adjacent sides of a parallelogram are $2\hat i - 4\hat j + 5\hat k$ and $\hat i - 2\hat j - 3\hat k$. Find the unit vector parallel to its diagonal. Also, find its area. 

Ans:  Diagonal of a parallelogram is $\vec a + \vec b$ 

$\vec a + \vec b = (2 + 1)\hat i + ( - 4 - 2)\hat j + (5 - 3)\hat k = 3\hat i - 6\hat j + 2\hat k$

So, the unit vector parallel to diagonal is $\dfrac{{\vec a + \vec b}}{{|\vec a + \vec b|}} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {{3^2} + {{( - 6)}^2} + {2^2}} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {9 + 36 + 4} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{7} = \dfrac{3}{7}\hat i - \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k$

$\vec a \times \vec b = \left| {\begin{array}{*{20}{r}} {\hat i}&{\hat j}&k \\ 2&{ - 4}&3 \\ 1&{ - 2}&{ - 3} \end{array}} \right|$

$ = \hat i(12 + 10) - \hat j( - 6 - 5) + \bar k( - 4 + 4)$

$ = 22\hat i + 11\hat j$

$ = 11(2\hat i + \hat j)$

$\therefore |\vec a \times \vec b| = 11\sqrt {{2^2} + {1^2}}  = 11\sqrt 5 $

So, area of parallelogram is $11\sqrt 5 $ sq units

11. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ  are $\dfrac{1}{{\sqrt 3 }}$ $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.

Ans: Let a vector be equally inclined to ${\text{OX}},{\text{OY}}$, and ${\text{OZ}}$ at an angle $\alpha $.

So, the Direction Cosines of the vector are $\cos \alpha ,\cos \alpha $and $\cos \alpha $.

${\cos ^2}\alpha  + {\cos ^2}\alpha  + {\cos ^2}\alpha  = 1$

$ \Rightarrow 3{\cos ^2}\alpha  = 1$

$ \Rightarrow \cos \alpha  = \dfrac{1}{{\sqrt 3 }}$

So, the DCs of the vector are $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.

12. Let $\vec a = \hat i + 4\hat j + 2\hat k$ and $\vec b = 3\hat i - 2\hat j + 7\hat k$ and $\vec c = 2\hat i - \hat j + 4\hat k$. Find a vector $\vec d$ which is

perpendicular to both $\vec a$ and $\vec b$ and $\vec c \cdot \vec d = 15$ 

Ans: $\vec d = {d_1}\hat i + {d_2}\hat \jmath  + {d_3}\hat k$

$\vec d\;.\;\vec a = 0 \Rightarrow {d_1} + 4{d_2} + 2{d_3} = 0$

$\vec d\;.\;\vec b = 0 \Rightarrow 3{d_1} - 2{d_2} + 7{d_3} = 0$

$\vec c \cdot \vec d = 15 \Rightarrow 2{d_1} - {d_2} + 4{d_3} = 15$

Solving these equations, we get ${d_1} = \dfrac{{160}}{3},{d_2} =  - \dfrac{5}{3},{d_3} =  - \dfrac{{70}}{3}$

$\therefore \vec d = \dfrac{{160}}{3}\hat i + \dfrac{5}{3}\widehat j + \dfrac{{70}}{3}\hat k = \dfrac{1}{3}(160\hat i + 5\hat j + 70\hat k)$

13. The scalar product of the vector $\hat i + \hat j + \hat k$ with a unit vector along the sum of vectors $2\hat i + 4\hat j - 5\hat k$ änd $\lambda \hat i + 2\hat j + 3\hat k$ is equal to one. Find the value of $\lambda $. 

Ans: $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k) = (2 + \lambda )\hat i + 6\hat j - 2\hat k$

So, unit vector along $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k)$ is $\left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }}} \right)$

$(\hat i + \hat j + \hat k) \cdot \left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }}} \right) = 1$

$ \Rightarrow \dfrac{{(2 + \lambda ) + 6 - 2}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }} = 1$

$ \Rightarrow \sqrt {{\lambda ^2} + 4\lambda  + 44}  = \lambda  + 6$

$ \Rightarrow {\lambda ^2} + 4\lambda  + 44 = {(\lambda  + 6)^2}$

$ \Rightarrow {\lambda ^2} + 4\lambda  + 44 = {\lambda ^2} + 12\lambda  + 36$

$ \Rightarrow 8\lambda  = 8$

$ \Rightarrow \lambda  = 1$

14. If $\vec a,\vec b,\vec c$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec a + \vec b + \vec c$ is equally inclined to $\vec a\;,\;\vec b$ and $\vec c$. 

Ans: $\vec a\;.\;\vec b = \vec b\;.\;\vec c = \vec c\;.\;\vec a = 0$

$|\vec a| = |\vec b| = |\vec c|$

Let $\vec a + \vec b + \vec c$ be inclined to $\vec a,\vec b,\vec c$ at angles ${\theta _1},{\theta _2},{\theta _3}$ respectively. $\cos {\theta _1} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{\vec a \cdot \vec a + \vec b \cdot \vec a + \vec c \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{|\vec a{|^2}}}{{|\vec a + \vec b + \vec c||\vec a|}} = \dfrac{{|\vec a|}}{{|\vec a + \vec b + \vec c|}}$

$\cos {\theta _2} = \dfrac{{(\vec a + \vec b + \vec c)\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{\vec a\vec b + \vec b\vec b + \vec c\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{|\vec b{|^2}}}{{|\vec a + \vec b + \vec c||\vec b|}} = \dfrac{{|\vec b|}}{{|\vec a + \vec b + \vec c|}}$

$\operatorname{os} {\theta _3} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec c}}{{|\vec a + \vec b + \vec c||\bar c|}} = \dfrac{{\vec a\vec c + \vec b\vec c + \vec c\vec c}}{{|\vec a + \vec b + \vec c||\vec c\mid }} = \dfrac{{|\vec c{|^2}}}{{|\vec a + \vec b + \vec c||\vec c|}} = \dfrac{{|\vec c|}}{{|\vec a + \vec b + \vec c|}}$

Since, $|\vec a| = |\vec b| = |\vec c|\; \Rightarrow \;\cos {\theta _1} = \cos {\theta _2} = \cos {\theta _3}$, So, ${\theta _1} = {\theta _2} = {\theta _3}$

15. Prove that, $(\vec a + \vec b) \cdot (\vec a + \vec b) = \;|\vec a{|^2} + |\vec b{|^2}$ if and only if $\vec a,\vec b$ are perpendicular, given $\vec a \ne \vec 0,\vec b \ne 0$

Ans: $(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a{|^2} + |\vec b{|^2}$

$ \Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b \cdot \vec a + \vec b\;.\;\vec b = |\vec a{|^2} + |\vec b{|^2}$

$ \Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\vec b{|^2} = |a{|^2} + |\vec b{|^2}$

$ \Rightarrow 2\vec a\;.\;\vec b = 0$

$ \Rightarrow \vec a\;.\;\vec b = 0$

So $\vec a$ and $\vec b$ are perpendicular.

16. If $\theta $ is the angle between two vectors $\vec a$ and $\vec b$, then $\vec a\vec b \geqslant 0$ only when

a. $0 < \theta  < \dfrac{\pi }{2}$

b. $0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

c. $0 < \theta  < \pi $

d. $0 \leqslant \theta  \leqslant \pi $

Ans: $\therefore \vec a\vec b \geqslant 0$

$ \Rightarrow |\vec a||\vec b|\cos \theta  \geqslant 0$

$ \Rightarrow \cos \theta  \geqslant 0\quad \because [|\vec a| \geqslant 0$ and $|\vec b| \geqslant 0]$

$ \Rightarrow 0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

$\vec a \cdot \vec b \geqslant 0$ if $0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

So the right answer is B

17. Let $\vec a$ and $\vec b$ be two unit vectors and $\theta $ is the angle between them. Then $\vec a + \vec b$ is a unit vector if

a. $\theta  = \dfrac{\pi }{4}$

b. $\theta  = \dfrac{\pi }{3}$

c. $\theta  = \dfrac{\pi }{2}$

d. $\theta  = \dfrac{{2\pi }}{3}$

Ans: $|\vec a| = |\vec b| = 1$

$|\vec a + \vec b| = 1$

$ \Rightarrow (\vec a + \overrightarrow b )(\vec a + \overrightarrow b ) = 1$

$ \Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b\;.\;\vec a + \vec b\;.\;\vec b = 1$

$ \Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\overrightarrow b {|^2} = 1$

$ \Rightarrow {1^2} + 2|\vec a||\vec b|\cos \theta  + {1^2} = 1$

$ \Rightarrow 1 + 2.1.1\cos \theta  + 1 = 1$

$ \Rightarrow \cos \theta  =  - \dfrac{1}{2}$

$ \Rightarrow \theta  = \dfrac{{2\pi }}{3}$

So, $\vec a + \vec b$ is unit vector if $\theta  = \dfrac{{2\pi }}{3}$ 

The correct answer is D

18. The value of $\hat i.(\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$ is

$\begin{array}{*{20}{l}} {{\text{ a. }}0}&{{\text{ b. }} - 1}&{{\text{ c. }}1}&{{\text{ d. }}3} \end{array}$

Ans:

$\hat i \cdot (\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$

$ = \hat i\;.\;\hat i + \hat j \cdot ( - \hat j) + \hat k\;.\;\hat k$

$ = 1 - 1 + 1$

$ = 1$

The correct answer is C.

19. If $\theta $ is the angle between any two vectors $\vec a$ and $\vec b$, then $|\vec a\vec b| = |\vec a \times \vec b|$ when $\theta $ is equal to

a. 0 

b. $\dfrac{\pi }{4}$

c. $\dfrac{\pi }{2}$

d. n

Ans: $|\vec a\vec b| = |\vec a \times \vec b|$

$ \Rightarrow |\vec a|\vec b|\cos \theta  = |\vec a||\vec b\mid \sin \theta $

$ \Rightarrow \cos \theta  = \sin \theta $

$ \Rightarrow \tan \theta  = 1$

$ \Rightarrow \theta  = \dfrac{\pi }{4}$

The correct answer is B

Overview of Deleted Syllabus for CBSE Class 12 Maths Vector Algebra

Class 12 Maths Chapter 10: Exercises Breakdown

Conclusion

Vector Class 12 NCERT Solutions provides a comprehensive understanding of the fundamental concepts and applications of vectors. This chapter covers various essential topics such as vector addition, subtraction, scalar multiplication, dot product, cross product, and their geometric interpretations. Through well-structured examples and step-by-step solutions, students gain a solid foundation in vector algebra, which is crucial for higher studies in mathematics, physics, and engineering. The exercises are designed to enhance problem-solving skills and analytical thinking. Typically 1 to 2 short answer questions and 1 to 2 long answer questions appear from Ch 10 Maths Class 12 in exams.

Other Study Material for CBSE Class 12 Maths Chapter 10

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.