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

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

Conclusion

Class 12 Exercise 10.3 of NCERT Solutions for Maths Chapter 10 - Vector Algebra is crucial for board exam preparation. It focuses on the scalar (dot) product of two vectors and the projection of a vector on a line, key concepts in vector algebra. Pay close attention to the properties of the scalar product and their applications. Practice the formulas in class 12 exercise 10.3 for the dot product and vector projection thoroughly, as they are foundational for advanced topics. Solving these questions with Vedantu’s guidance will solidify your understanding and help you tackle complex problems in exams.

Class 12 Maths Chapter 10: Exercises Breakdown

CBSE Class 12 Maths Chapter 10 Other Study Materials

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.