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

Last updated date: 17th Sep 2024
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## NCERT Solutions for Maths Class 12 Chapter 10 Exercise 10.3 Vector Algebra - FREE PDF Download

Access Exercise 10.3 Class 12 of NCERT Solutions for Maths Chapter 10 – Vector Algebra is available here in a downloadable PDF format. This exercise focuses on key concepts of vector algebra, including vector addition, scalar multiplication, and the geometric interpretation of vectors. Numerous questions are designed to reinforce these fundamental principles and improve problem-solving skills. Vedantu’s subject experts have carefully prepared the NCERT Solutions for Class 12 Maths, observing the CBSE syllabus and guidelines. These solutions are an excellent resource for Class 12 students preparing for their board exams, ensuring a thorough understanding of vector algebra.

Table of Content
1. NCERT Solutions for Maths Class 12 Chapter 10 Exercise 10.3 Vector Algebra - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 10 Exercise 10.3 Class 12 | Vedantu
3. Formulas Used in Class 12 Maths Exercise 10.3
4. Access NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra Exercise 10.3
5. Class 12 Maths Chapter 10: Exercises Breakdown
6. CBSE Class 12 Maths Chapter 10 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 10 Exercise 10.3 Class 12 | Vedantu

• NCERT Solutions Maths exercise 10.3 class 12 covers Product of Two Vectors, Scalar (or dot) product of two vectors, Two important properties of scalar product, projection of a vector on a line.

• The product of two vectors includes both scalar (dot) products and vector (cross) products, each with unique properties and applications.

• The scalar (or dot) product of two vectors is a single number obtained by multiplying their magnitudes and the cosine of the angle between them.

• Two important properties of the scalar product are that it is commutative and distributive over vector addition.

• The projection of a vector on a line involves projecting one vector onto another, resulting in a scalar quantity that represents the component of one vector along the direction of the other in ex 10.3 class 12.

• The scalar product is useful in determining the angle between two vectors and calculating work done by a force.

• Projection helps in breaking down vectors into components, simplifying the analysis of vector quantities in physics and engineering.

• There are 18 questions in maths chapter 10 ex 10.3 class 12 solutions  which are fully solved by experts at Vedantu.

## Formulas Used in Class 12 Maths Exercise 10.3

• Scalar (or dot) product of two vectors: $\vec{A}\cdot \vec{B}=\left | \vec{A} \right | \left | \vec{B} \right | cos \theta$

• Projection of a vector on a line: The projection of vector $\vec{A}$ on vector $\vec{B}$ is given by $proj_{\vec{B}}\vec{A}=\left ( \frac{\vec{A}\cdot \vec{B}}{\left | B \right |^{2}} \right )\vec{B}$

Competitive Exams after 12th Science

## Access 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

 Exercise Number of Questions Exercise 10.1 5 Question & Solutions Exercise 10.2 19 Questions & Solutions Exercise 10.4 12 Questions & Solutions

## 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.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra 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.

6. How can I increase my knowledge of vector algebra while working on Exercise 10.3 from NCERT Class 12 Maths?

To improve the understanding of vector algebra in Exercise 10.3, first learn the fundamental operations of addition, subtraction, and scalar multiplication. Practice using these operations on each problem. Refer to textbook examples for problem-solving strategies. For clarification use online resources in Vedantu. Summarise key themes for a thorough comprehend.

7. Why is the scalar product of two vectors important in 10.3 class 12 maths?

In 10.3 class 12 maths the scalar product, or dot product, helps in determining the angle between two vectors and their projections. It is also used in various applications in physics and engineering, such as calculating work done by a force.

8. How can Vedantu’s NCERT Solutions help me with class 12 ex 10.3

Vedantu’s NCERT Solutions provide detailed, step-by-step explanations for each question in class 12 ex 10.3. These solutions are created by experts following the NCERT guidelines, ensuring you understand the concepts thoroughly and are well-prepared for your exams.