NCERT Solutions for Class 12 Maths Chapter 10 - Exercise

Exercise 10.3

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

Ans: It is given in the question that

$\left| \overrightarrow{a} \right|=\sqrt{3}$

$\left| \overrightarrow{b} \right|=2$

$\overrightarrow{a}.\overrightarrow{b}=\sqrt{6}$

Also we know that 

$\text{    }\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta $where $\theta $ is the angle between the vectors

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

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

Henc the angle between the vectors is $\dfrac{\pi }{4}$

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

Ans: Let the given vectors be 

$\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k}$

$\overrightarrow{b}=3\widehat{i}-2\widehat{j}+\widehat{k}$

We know that 

$\text{    }\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta $where $\theta $ is the angle between the vectors

$\Rightarrow \left( \widehat{i}-2\widehat{j}+3\widehat{k} \right).\left( 3\widehat{i}-2\widehat{j}+\widehat{k} \right)=\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}\sqrt{{{3}^{2}}+{{2}^{2}}+{{1}^{2}}}$

$\Rightarrow 10=14\cos \theta $

Hence, we found $\theta ={{\cos }^{-1}}\left( \dfrac{5}{7} \right)$

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

Ans: Given we have vectors

 $\overrightarrow{a}=\widehat{i}-\widehat{j}$ and 

$\overrightarrow{b}=\widehat{i}+\widehat{j}$ 

Now the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is given by

$\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{b} \right|}=\dfrac{\left( \widehat{i}-\widehat{j} \right).\left( \widehat{i}+\widehat{j} \right)}{\sqrt{2}}$

Hence, the projection of $\widehat{i}-\widehat{j}$ on $\widehat{i}+\widehat{j}$ is $0$

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

Ans: : Given we have vectors

 $\overrightarrow{a}=\widehat{i}+3\widehat{j}+7\widehat{k}$ and 

$\overrightarrow{b}=7\widehat{i}-\widehat{j}+8\widehat{k}$ 

Now the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is given by

$\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{b} \right|}=\dfrac{\left( \widehat{i}+3\widehat{j}+7\widehat{k} \right).\left( \widehat{i}+3\widehat{j}+7\widehat{k} \right)}{\sqrt{114}}$

Hence, the projection of $\widehat{i}-\widehat{j}$ on $\widehat{i}+\widehat{j}$ is $\dfrac{60}{\sqrt{114}}$

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

$\dfrac{1}{7}\left( 2\widehat{i}+3\widehat{j}+6\widehat{k} \right)$, $\dfrac{1}{7}\left( 3\widehat{i}-6\widehat{j}+2\widehat{k} \right)$, $\dfrac{1}{7}\left( 6\widehat{i}+2\widehat{j}-3\widehat{k} \right)$

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

Ans: let us have following notations 

 $7\overrightarrow{a}=2\widehat{i}+3\widehat{j}+6\widehat{k}$

$7\overrightarrow{b}=3\widehat{i}-6\widehat{j}+2\widehat{k}$

$7\overrightarrow{c}=6\widehat{i}+2\widehat{j}-3\widehat{k}$

Its magnitude is given by

$\text{   }7a=\sqrt{{{2}^{2}}+{{3}^{2}}+{{6}^{2}}}$

$\Rightarrow a=\dfrac{7}{7}=1$

Hence, the magnitude is $1$, it is a unit vector 

Similarly, magnitude of $7\overrightarrow{b}$ is given by

$\text{   }7b=\sqrt{{{3}^{2}}+{{6}^{2}}+{{2}^{2}}}$

$\Rightarrow b=\dfrac{7}{7}=1$

Hence, the magnitude is $1$, it is a unit vector 

Similarly, magnitude of $7\overrightarrow{c}$ is given by

$\text{   }7c=\sqrt{{{6}^{2}}+{{2}^{2}}+{{3}^{2}}}$

$\Rightarrow c=\dfrac{7}{7}=1$

Hence, the magnitude is $1$, it is a unit vector 

Now we will calculate the followings

$\overrightarrow{a}.\overrightarrow{b}=\dfrac{1}{7}\left( 2\widehat{i}+3\widehat{j}+6\widehat{k} \right).\left( 3\widehat{i}-6\widehat{j}+2\widehat{k} \right)=0$

$\overrightarrow{c}.\overrightarrow{b}=\dfrac{1}{7}\left( 6\widehat{i}+2\widehat{j}-3\widehat{k} \right).\left( 3\widehat{i}-6\widehat{j}+2\widehat{k} \right)=0$

$\overrightarrow{a}.\overrightarrow{c}=\dfrac{1}{7}\left( 2\widehat{i}+3\widehat{j}+6\widehat{k} \right).\left( 6\widehat{i}+2\widehat{j}-3\widehat{k} \right)=0$

Hence, all these vectors are mutually perpendicular

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

Ans: Given in the question that 

$\left( \overrightarrow{a}+\overrightarrow{b} \right).\left( \overrightarrow{a}-\overrightarrow{b} \right)=8$……($1$)

$\left| \overrightarrow{a} \right|=8\left| \overrightarrow{b} \right|$                       …… ($2$)

Now from ($1$) 

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

Also from ($2$)

${{\left| \overrightarrow{b} \right|}^{2}}\left( 64-1 \right)=8$

$\Rightarrow \left| \overrightarrow{b} \right|=\sqrt{\dfrac{8}{63}}$

Hence, $\left| \overrightarrow{a} \right|=8\sqrt{\dfrac{8}{63}}$

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

Ans: We are given with two vectors $3\overrightarrow{a}-5\overrightarrow{b}$ and $2\overrightarrow{a}+7\overrightarrow{b}$

Now $\left( 3\overrightarrow{a}-5\overrightarrow{b} \right).\left( 2\overrightarrow{a}+7\overrightarrow{b} \right)=6{{a}^{2}}+11\overrightarrow{a}.\overrightarrow{b}-35{{b}^{2}}$

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 $\dfrac{1}{2}$

Ans: It is given in the question 

$\theta ={{60}^{\circ }}$

$\overrightarrow{a}.\overrightarrow{b}=\dfrac{1}{2}$

According to the question 

$\dfrac{1}{2}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\dfrac{1}{2}$

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

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

Ans: It is given that $\left| \overrightarrow{a} \right|=1$

$\left( \overrightarrow{x}-\overrightarrow{a} \right).\left( \overrightarrow{x}+\overrightarrow{a} \right)=12$

$\text{        }\Rightarrow {{\left| \overrightarrow{x} \right|}^{2}}-1=12$  (since $\left| \overrightarrow{a} \right|=1$)

Hence, we found that $\left| \overrightarrow{x} \right|=\sqrt{13}$

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

Ans: The given vectors are 

$\text{           }\overrightarrow{a}=2\widehat{i}+2\widehat{j}+3\widehat{k}$

$\text{           }\overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k}$

$\therefore \overrightarrow{a}+\lambda \overrightarrow{b}=\left( 2-\lambda  \right)\widehat{i}+\left( 2+2\lambda  \right)\widehat{j}+\left( 3-\lambda  \right)\widehat{k}$

$\text{           }\overrightarrow{c}=3\widehat{i}+\widehat{j}$

According to the question 

$\text{                                          }\left( \overrightarrow{a}+\lambda \overrightarrow{b} \right).\left( 3\widehat{i}+\widehat{j} \right)=0$

$\Rightarrow \left( \left( 2-\lambda  \right)\widehat{i}+\left( 2+2\lambda  \right)\widehat{j}+\left( 3-\lambda  \right)\widehat{k} \right).\left( 3\widehat{i}+\widehat{j} \right)=0$

Hence, we found that $\lambda =8$

11. Show that:

$\overrightarrow{\left| a \right|}\overrightarrow{b}+\left| \overrightarrow{b} \right|\overrightarrow{a}$ is perpendicular to $\overrightarrow{\left| a \right|}\overrightarrow{b}-\left| \overrightarrow{b} \right|\overrightarrow{a}$ For any two nonzero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$

Ans: Let us suppose the two vectors as shown

$\overrightarrow{\eta }=\left| \overrightarrow{a} \right|\overrightarrow{b}+\left| \overrightarrow{b} \right|\overrightarrow{a}$

$\overrightarrow{\kappa }=\left| \overrightarrow{a} \right|\overrightarrow{b}-\left| \overrightarrow{b} \right|\overrightarrow{a}$

Now $\overrightarrow{\eta }.\overrightarrow{\kappa }=\left( \left| \overrightarrow{a} \right|\overrightarrow{b}+\left| \overrightarrow{b} \right|\overrightarrow{a} \right).\left( \left| \overrightarrow{a} \right|\overrightarrow{b}-\left| \overrightarrow{b} \right|\overrightarrow{a} \right)$

$\Rightarrow \overrightarrow{\eta }.\overrightarrow{\kappa }={{\left| \overrightarrow{a} \right|}^{2}}{{\left| \overrightarrow{b} \right|}^{2}}-{{\left| \overrightarrow{a} \right|}^{2}}{{\left| \overrightarrow{b} \right|}^{2}}=0$

Hence, proved

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

Ans: Given we have two equations 

$\overrightarrow{a}.\overrightarrow{a}=0$…….($1$)

$\overrightarrow{a}.\overrightarrow{b}=0$

It is clear from the equation ($1$) that ${{\left| a \right|}^{2}}=0$

Hence, $\overrightarrow{b}$ can be any vector 

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

Ans: Given we have three unit vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ such that 

$\text{                                      }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$

$\text{             }\Rightarrow \left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right).\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)=0$

$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a} \right)=0$

$\text{                       }\Rightarrow \left( \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a} \right)=\dfrac{-3}{2}$

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

Ans: Let us suppose two vectors as shown

$\overrightarrow{a}=2\widehat{i}+4\widehat{j}+3\widehat{k}$ 

$\overrightarrow{b}=3\widehat{i}+3\widehat{j}-6\widehat{k}$

Now $\overrightarrow{a}.\overrightarrow{b}=\left( 2\widehat{i}+4\widehat{j}+3\widehat{k} \right).\left( 3\widehat{i}+3\widehat{j}-6\widehat{k} \right)$

$\Rightarrow \overrightarrow{a}.\overrightarrow{b}=6+12-18=0$

But clearly neither $\overrightarrow{a}=0$nor $\overrightarrow{b}=0$

15. If the vertices A,B,C of a triangle ABC are $\left( 1,2,3 \right)$, $\left( -1,0,0 \right)$ and $\left( 0,1,2 \right)$ respectively then find $\angle ABC$ 

Ans: It is given that the vertices of $\Delta ABC$ are 

$A\left( 1,2,3 \right)$

$B\left( -1,0,0 \right)$

$C\left( 0,1,2 \right)$

Now $\angle ABC$ is the angle between $\overrightarrow{BA}$ and $\overrightarrow{BC}$

$\therefore \overrightarrow{BA}.\overrightarrow{BC}=\left( 2\widehat{i}+2\widehat{j}+3\widehat{k} \right).\left( \widehat{i}+\widehat{j}+2\widehat{k} \right)$

$\text{     }\Rightarrow 10=\sqrt{17}\sqrt{6}\cos \angle ABC$

Hence, we found that $\angle ABC={{\cos }^{-1}}\left( \dfrac{10}{\sqrt{17}\sqrt{6}} \right)$

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

Ans: Given we have points as shown

$A\left( 1,2,7 \right)$

$B\left( 2,6,3 \right)$

$C\left( 3,10,-1 \right)$

$\therefore \overrightarrow{AB}=\widehat{i}+4\widehat{j}-4\widehat{k}$

$\therefore \overrightarrow{BC}=\left( \widehat{i}+4\widehat{j}-4\widehat{k} \right)$

$\therefore \overrightarrow{AC}=2\left( \widehat{i}+4\widehat{j}-4\widehat{k} \right)$

Clearly, 

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

Hence, vectors are collinear

17. Show that the vectors $2\widehat{i}-\widehat{j}+\widehat{k}$, $\widehat{i}-3\widehat{j}-5\widehat{k}$, $3\widehat{i}-4\widehat{j}-4\widehat{k}$ forms the vertices of the right angled triangle 

Ans: It is given in the question that

$\overrightarrow{OA}=2\widehat{i}-\widehat{j}+\widehat{k}$

$\overrightarrow{OB}=\widehat{i}-3\widehat{j}-5\widehat{k}$

$\overrightarrow{OC}=3\widehat{i}-4\widehat{j}-4\widehat{k}$

Where $\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}$ are position vectors of A,B,C

Now 

$\overrightarrow{AB}=-\widehat{i}-2\widehat{j}-6\widehat{k}$

$\left| \overrightarrow{AB} \right|=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{41}$

$\overrightarrow{BC}=-2\widehat{i}-\widehat{j}+\widehat{k}$

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

$\left| \overrightarrow{AC} \right|=\sqrt{35}$

Clearly $\left| \overrightarrow{AC} \right|+\left| \overrightarrow{BC} \right|=\sqrt{41}=\left| \overrightarrow{AB} \right|$

Hence, proved

18. If $\overrightarrow{a}$ is a non-zero vector of magnitude ‘a’ and $\lambda $ a non-zero scalar, then $\lambda \overrightarrow{a}$ is unit vector for what value of $\lambda $

Ans: Given we have a vector $\overrightarrow{a}$ and a scalar $\lambda $

For $\lambda \overrightarrow{a}$ to be a unit vector 

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

Hence, the value is$\left| \lambda  \right|=\dfrac{1}{a}$

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3

Opting for the NCERT solutions for Ex 10.3 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.3 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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NCERT Solutions for Class 12 Math Chapter 10- Vector Algebra Exercise 10.3 is based on the following topics:

Product of Two Vectors

• Scalar or Dot Product of Two Vectors

• Projection In Number Line

What Is the Product of Two Vectors?

A vector in Mathematics incorporates both magnitude and direction. The multiplication of two or more vectors can be made either by using dot product or cross product. Here, we will study the scalar (or dot) product of two vectors.

What Is the Scalar (Or Dot) Product of Two Vectors?

The resultant of the scalar or dot product of the vectors is always a scalar value.

Dot Product of vectors is equivalent to:

• The product of the magnitudes of the two vectors, and

• The cosine of the angle between the two vectors. 

The outcomes of the dot product of two vectors lie in the same plane of the two vectors. The dot product may be:

• A Positive Real Number 

• A Negative Real Number.

NCERT Solutions for Class 12 Maths

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability