# NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.3) Exercise 10.3

## NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.3) Exercise 10.3

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## Access NCERT Solutions for Maths Class 12 Chapter 10 - Vector Algebra

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 10  Vector Algebra All Exercises in PDF Format Exercise 10.1 5 Question & Solutions Exercise 10.2 19 Questions & Solutions Exercise 10.4 12 Questions & Solutions

## FAQs on NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.3) Exercise 10.3

1. What is the main concept given in Class 12 Maths Exercise 10.3?

It is primarily based on the scalar product of two vectors. It is possible to define the dot product as the product of the magnitudes of the two vectors and the angle between the two vectors' cosine values. Another definition is the product of a first-to-second projection and magnitude of the second vector. This exercise deals with different problems related to the dot product of 2 vectors.

2. Which is the most important question from Exercise 10.3 of Class 12 Maths?

There is no such thing as the most important question! As you can see, the simpler ones help you remember your ideas, whilst the more difficult ones test your problem-solving abilities and how well you can indicate your principles in practical situations. You must pay close attention to Chapter 10. Refer to all problems in NCERT Class 12 Maths Chapter 10.

3. What is the dot product of 2 Vectors?

This number (Scalar quantity) is generated by executing an operation on the vector components and it is termed the dot product or scalar product. Only vectors with the same number of dimensions can be combined using the dot product. The dot product has a heavy dot as its logo. It is a very crucial topic and must be learned well as it has multiple applications across domains of Math and Physics.

4. Is dot Product Scalar Class 12?

Often referred to as the Scalar Product, the Dot Product returns a Scalar (Normal Number) response. On the other hand, there's also the Cross Product, which delivers the solution as a Vector. Both are equally crucial and have their own significance in Math and Physics. Learn both the topics well with clear knowledge of their fundamentals and also practice the related exercises.

5. Is Cross Product Vector Class 12?

Using the right-hand thumb rule, you may calculate its magnitude by measuring the area of the parallelogram between them. When two vectors are crossed, the outcome of that cross is called a vector quantity. Hence yes, the cross product of any two vectors yields another vector as the answer. Practice the different problems as much as you can.