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# NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra Exercise 10.2

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## NCERT Solutions for Maths Class 12 Chapter 10 Exercise 10.2 Vector Algebra - FREE PDF Download

Access NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2  – Vector Algebra is available here in a downloadable PDF format. This exercise focuses on the basic operations of vectors, including vector addition and scalar multiplication. These foundational concepts are essential for understanding more complex topics in vector algebra.

Table of Content
1. NCERT Solutions for Maths Class 12 Chapter 10 Exercise 10.2 Vector Algebra - FREE PDF Download
2. Glance on NCERT Solutions Class 12 Maths Chapter 10 Exercise 10.2 | Vedantu
3. Formulas Used in Class 12 Chapter 10 Exercise 10.2
4. Access NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra Exercise 10.2
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

Vedantu’s expert teachers have carefully solved these questions, keeping the NCERT ex 10.2 class 12 guidelines in mind. It is important to pay attention to the methods of vector addition and scalar multiplication, as mastering these will help you tackle more advanced problems in the chapter. Use these solutions of ex 10.2 class 12 to strengthen your understanding and boost your exam preparation.

## Glance on NCERT Solutions Class 12 Maths Chapter 10 Exercise 10.2 | Vedantu

• Exercise 10.2 Class 12 covers Addition of Vectors, Multiplication of a Vector by a Scalar, Components of a vector, Vector joining two points, Section formula.

• Addition of vectors involves combining two or more vectors to form a resultant vector.

• Multiplication of a vector by a scalar changes the magnitude of the vector without affecting its direction.

• Components of a vector are the projections of the vector along the coordinate axes.

• A vector joining two points is represented by the difference of their position vectors.

• The section formula helps in finding the coordinates of a point dividing a line segment in a given ratio.

• There are 19 questions in Maths class 12 exercise 10.2 solutions  which are fully solved by experts at Vedantu.

## Formulas Used in Class 12 Chapter 10 Exercise 10.2

• Addition of Vectors: $\vec{A}+\vec{B} = \left ( A_{x} + B_{x}\right )\hat{i} + \left ( A_{y} + By\right )\hat{j} + \left ( A_{z} + B_{z}\right )\hat{k}$

• Multiplication of a Vector by a Scalar: $k\vec{A} = k\left ( A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k} \right )=\left ( kA_{x} \right )\hat{i}+\left ( kA_{y} \right )\hat{j}+\left ( kA_{z} \right )\hat{k}$

• Components of a Vector: $\vec{A}=A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k}$

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra Exercise 10.2

1. Compute the magnitude of the following vectors:

$\overrightarrow {\text{a}} {\text{ = }}\widehat {\text{i}}\,{\text{ + }}\widehat {\text{j}}\,{\text{ + }}\widehat {\text{k}}{\text{;}}\,\overrightarrow {\text{b}} {\text{ = 2}}\widehat {\text{i}}\,{\text{ - 7}}\widehat {\text{j}}\,{\text{ - 3}}\widehat {\text{k}}{\text{;}}\,\overrightarrow {\text{c}} {\text{ = }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{i}}\,{\text{ + }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{j}}\,{\text{ - }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{k}}\,$

Ans: The given vectors are,

$\overrightarrow {\text{a}} {\text{ = }}\widehat {\text{i}}\,{\text{ + }}\widehat {\text{j}}\,{\text{ + }}\widehat {\text{k}}{\text{;}}\,\overrightarrow {\text{b}} {\text{ = 2}}\widehat {\text{i}}\,{\text{ - 7}}\widehat {\text{j}}\,{\text{ - 3}}\widehat {\text{k}}{\text{;}}\,\overrightarrow {\text{c}} {\text{ = }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{i}}\,{\text{ + }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{j}}\,{\text{ - }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\widehat {\text{k}}\,$

Magnitude of a vector say,$\overrightarrow {{{\text{a}}_{\text{1}}}} {\text{ = }}\widehat {{{\text{i}}_{\text{1}}}}{\text{ + }}\widehat {{{\text{j}}_{\text{1}}}}{\text{ + }}\widehat {{{\text{k}}_{\text{1}}}}$ is,

$\left| {\overrightarrow {{{\text{a}}_{\text{1}}}} } \right|{\text{ = }}\sqrt {{{\left( {{{\text{i}}_{\text{1}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{j}}_{\text{1}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{k}}_{\text{1}}}} \right)}^{\text{2}}}}$

Therefore, the magnitude of the following vectors are,

$\left| {\overrightarrow {\text{a}} } \right|{\text{ = }}\sqrt {{{\left( {\text{1}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\text{1}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\text{1}} \right)}^{\text{2}}}} {\text{ = }}\sqrt {\text{3}}$

Now, for $\overrightarrow b$,

$\left| {\overrightarrow {\text{b}} } \right|{\text{ = }}\sqrt {{{\left( {\text{2}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{ - 7}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{ - 3}}} \right)}^{\text{2}}}} {\text{ = }}\sqrt {{\text{4 + 49 + 9}}} {\text{ = }}\sqrt {{\text{62}}}$

Again, for $\overrightarrow c$,

$\left| {\overrightarrow {\text{c}} } \right|{\text{ = }}\sqrt {{{\left( {\frac{{\text{1}}}{{\sqrt {\text{3}} }}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\frac{{\text{1}}}{{\sqrt {\text{3}} }}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\frac{{{\text{ - 1}}}}{{\sqrt {\text{3}} }}} \right)}^{\text{2}}}} {\text{ = 1}}$

2. Write two different vectors having the same magnitude.

Ans: Let us assume the two vectors to be,

and $\overrightarrow {\text{b}} {\text{ = 2}}\widehat {\text{i}}\,{\text{ - }}\widehat {\text{j}}\,{\text{ - 3}}\widehat {\text{k}}\,$

Magnitude of a vector say, $\overrightarrow {{{\text{a}}_{\text{1}}}} {\text{ = }}\widehat {{{\text{i}}_{\text{1}}}}{\text{ + }}\widehat {{{\text{j}}_{\text{1}}}}{\text{ + }}\widehat {{{\text{k}}_{\text{1}}}}$ is,

$\left| {\overrightarrow {{{\text{a}}_{\text{1}}}} } \right|{\text{ = }}\sqrt {{{\left( {{{\text{i}}_{\text{1}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{j}}_{\text{1}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{k}}_{\text{1}}}} \right)}^{\text{2}}}} \overrightarrow {\text{a}} {\text{ = }}\widehat {\text{i}}\,{\text{ + 2}}\widehat {\text{j}}\,{\text{ + 3}}\widehat {\text{k}}\,$

Therefore, the magnitude of the following vectors are,

$\left| {\overrightarrow {\text{a}} } \right|{\text{ = }}\sqrt {{{\left( {\text{1}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\text{2}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\text{3}} \right)}^{\text{2}}}} {\text{ = }}\,\sqrt {{\text{1 + 4 + 9}}} {\text{ = }}\sqrt {{\text{14}}}$

Now, for $\overrightarrow b$,

$\left| {\overrightarrow {\text{b}} } \right|{\text{ = }}\sqrt {{{\left( {\text{2}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{ - 1}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{ - 3}}} \right)}^{\text{2}}}} \,{\text{ = }}\sqrt {{\text{4 + 1 + 9}}} {\text{ = }}\sqrt {{\text{14}}}$

Both, vectors have got an equal magnitude that is $\sqrt {14}$ units but are completely different and not equal vectors since they are going in different directions.

3. Write two different vectors having the same direction.

Ans: Let us assume the two vectors as,

$\overrightarrow p {\text{ = }}\widehat {\text{i}}\,{\text{ + }}\widehat {\text{j}}\,{\text{ + }}\widehat {\text{k}}\,$and $\overrightarrow q = 2\widehat i\, + 2\widehat j\, + 2\widehat k\,$

The direction cosines of the two vectors can be calculated as,

In case of $\overrightarrow {\text{p}}$,

$l = \frac{1}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }} = \frac{1}{{\sqrt 3 }}$

$m = \frac{1}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }} = \frac{1}{{\sqrt 3 }}$

$n = \frac{1}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }} = \frac{1}{{\sqrt 3 }}$

Now, for $\overrightarrow {\text{q}}$,

$l = \frac{2}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{2}{{2\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

$m = \frac{2}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{2}{{2\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

$n = \frac{2}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{2}{{2\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

It can be clearly observed that the direction cosines obtained in case of both the vectors that is, $\overrightarrow p$ and $\overrightarrow {\text{q}}$ are same and therefore they have got the same direction but different vectors.

4. Find the values of x and y so that the vectors ${\text{2}}\widehat {\text{i}}\,{\text{ + 3}}\widehat {\text{j}}\,$and ${\text{x}}\widehat {\text{i}}\,{\text{ + y}}\widehat {\text{j}}$are equal.

Ans: Observe that it is required for the corresponding coefficients of $\widehat i$ and $\widehat j$to be equal in order to have two equal vectors.

Therefore,

${\text{x = 2,}}\,{\text{y = 3}}$

5. Find the scalar and vector components of the vector with initial point $\left( {{\text{2,}}\,{\text{1}}} \right)$ and terminal point $\left( {{\text{ - 5,}}\,{\text{7}}} \right)$.

Ans: The given initial and terminal point are respectively, $\left( {{\text{2,}}\,{\text{1}}} \right)$ and $\left( {{\text{ - 5,}}\,{\text{7}}} \right)$.

Therefore, the vector is,

$\overrightarrow {PQ} = \left( {{\text{ - }}5{\text{ - }}2} \right)\widehat i{\text{ + }}\left( {{\text{7 - 1}}} \right)\widehat j{\text{ = - 7}}\widehat i + 6\widehat {\text{j}}$

6. Find the sum of the vectors with initial point $\overrightarrow {\text{a}} {\text{ = }}\widehat {\text{i}}{\text{ - 2}}\widehat {\text{j}}{\text{ + }}\widehat {\text{k}}{\text{,}}\,\overrightarrow {\text{b}} {\text{ = - 2}}\widehat {\text{i}}{\text{ + 4}}\widehat {\text{j}}{\text{ + 5}}\widehat {\text{k}}\,$ and $\overrightarrow {\text{c}} {\text{ = }}\widehat {\text{i}}{\text{ - 6}}\widehat {\text{j}}{\text{ - 7}}\widehat {\text{k}}$.

Ans: The given vectors are respectively, $\overrightarrow a = \widehat i - 2\widehat j + \widehat k,\,\overrightarrow b = - 2\widehat i + 4\widehat j + 5\widehat k\,$ and $\overrightarrow c = \widehat i - 6\widehat j - 7\widehat k$.

Therefore, the sum of the following vectors can be calculated as,  $\,\,\,\,\,\,\,\overrightarrow a + \overrightarrow b + \overrightarrow c = \widehat i - 2\widehat j + \widehat k + ( - 2\widehat i) + 4\widehat j + 5\widehat k\, + \widehat i - 6\widehat j - 7\widehat k$

$\Rightarrow \overrightarrow a + \overrightarrow b + \overrightarrow c = \left( {1 - 2 + 1} \right)\widehat i + \left( { - 2 + 4 - 6} \right)\widehat j + \left( {1 + 5 - 7} \right)\widehat k$

$\Rightarrow \overrightarrow a + \overrightarrow b + \overrightarrow c = - 4\widehat j - \widehat k$

7. Find the unit vector in the direction of the vector $\overrightarrow {\text{a}} {\text{ = }}\widehat {\text{i}}{\text{ + }}\widehat {\text{j}}{\text{ + 2}}\widehat {\text{k}}$.

Ans: The given vectors is, $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$.

Therefore, the unit vector in the direction of the vector, that is, $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$ is represented as $\widehat a$ and can be calculated as,

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

$\therefore \,\widehat a = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} = \frac{{\widehat i + \widehat j + 2\widehat k}}{{\sqrt 6 }} = \frac{1}{{\sqrt 6 }}\widehat i + \frac{1}{{\sqrt 6 }}\widehat j + \frac{2}{{\sqrt 6 }}\widehat k$

8. Find the unit vector in the direction of the vector, $\overrightarrow {{\text{PQ}}}$, where ${\text{P }}$and ${\text{Q}}$ are the points $\left( {{\text{1,}}\,{\text{2,}}\,{\text{3}}} \right)$and $\left( {{\text{4,}}\,{\text{5,}}\,{\text{6}}} \right)$respectively.

Ans: The given points are$P{\text{ }}$and $Q$ which are $\left( {1,\,2,\,3} \right)$and $\left( {4,\,5,\,6} \right)$ respectively.

Now, the vector $\overrightarrow {PQ}$ can be calculated as,

$\overrightarrow {PQ} = \left( {4 - 1} \right)\widehat i + \left( {5 - 2} \right)\widehat j + \left( {6 - 3} \right)\widehat k = 3\widehat i + 3\widehat j + 3\widehat k$

Therefore, the unit vector in the direction of the vector, that is, $\overrightarrow {PQ} = 3\widehat i + 3\widehat j + 3\widehat k$ is represented as $\widehat {PQ}$ and can be calculated as,

$\left| {\overrightarrow {PQ} } \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 3 \right)}^2}} = \sqrt {27} \, = 3\sqrt 3 \,\,\,\,$

$\therefore \,\widehat {PQ} = \frac{{\overrightarrow {PQ} }}{{\left| {\overrightarrow {PQ} } \right|}} = \frac{{3\widehat i + 3\widehat j + 3\widehat k}}{{3\sqrt 3 }} = \frac{1}{{\sqrt 3 }}\widehat i + \frac{1}{{\sqrt 3 }}\widehat j + \frac{1}{{\sqrt 3 }}\widehat k$

9. For the given vectors $\overrightarrow {\text{a}} {\text{ = 2}}\widehat {\text{i}}{\text{ - }}\widehat {\text{j}}{\text{ + 2}}\widehat {\text{k}}$and $\overrightarrow {\text{b}} {\text{ = - }}\widehat {\text{i}}{\text{ + }}\widehat {\text{j}}{\text{ + }}\widehat {\text{k}}$, find the unit vector in the direction of the vector, $\overrightarrow {{\text{a + b}}}$.

Ans: It is given that the vectors are $\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$and $\overrightarrow b = - \widehat i + \widehat j + \widehat k$respectively.

Now, the vector $\overrightarrow {a + b}$ can be calculated as,

$\therefore \overrightarrow {a + b} = \left( {2 - 1} \right)\widehat i + \left( { - 1 + 1} \right)\widehat j + \left( {2 - 1} \right)\widehat k = \widehat i + \widehat k$

Therefore, the unit vector in the direction of the vector, that is, $\overrightarrow {a + b} = \widehat i + \widehat k$ is represented as $\widehat {a + b}$ and can be calculated as,

$\left| {\overrightarrow {a + b} } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 2 \, = \sqrt 2 \,\,\,\,$

$\therefore \,\widehat {a + b} = \frac{{\overrightarrow {a + b} }}{{\left| {\overrightarrow {a + b} } \right|}} = \frac{{\widehat i + \widehat k}}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\widehat i + \frac{1}{{\sqrt 2 }}\widehat k$

10. Find a vector in the direction of the vector ${\text{5}}\widehat {\text{i}}{\text{ - }}\widehat {\text{j}}{\text{ + 2}}\widehat {\text{k}}$which has magnitude ${\text{8}}$ units.

Ans: Assume the given vector as, $\overrightarrow a = 5\widehat i - \widehat j + 2\widehat k$.

Therefore, the unit vector in the direction of the vector, that is, $\overrightarrow a = 5\widehat i - \widehat j + 2\widehat k$ is represented as $\widehat a$ and can be calculated as,

$\left| {\overrightarrow a } \right| = \sqrt {{{\left( 5 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} = \sqrt {30} \,\,\,\,\,$

$\therefore \,\widehat a = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} = \frac{{5\widehat i - \widehat j + 2\widehat k}}{{\sqrt {10} }} = \frac{5}{{\sqrt {10} }}\widehat i - \frac{1}{{\sqrt {10} }}\widehat j + \frac{2}{{\sqrt {10} }}\widehat k$

Now, a vector in the direction of the vector $5\widehat i - \widehat j + 2\widehat k$which has magnitude $8$ units can be calculated as,

$\therefore \,8\widehat a = 8\left( {\frac{5}{{\sqrt {10} }}\widehat i - \frac{1}{{\sqrt {10} }}\widehat j + \frac{2}{{\sqrt {10} }}\widehat k} \right) = \frac{{40}}{{\sqrt {10} }}\widehat i - \frac{8}{{\sqrt {10} }}\widehat j + \dfrac{{16}}{{\sqrt {10} }}\widehat k$

11. Show that the vectors ${\text{2}}\widehat {\text{i}}{\text{ - 3}}\widehat {\text{j}}{\text{ + 4}}\widehat {\text{k}}$and ${\text{ - 4}}\widehat {\text{i}}{\text{ + 6}}\widehat {\text{j}}{\text{ - 8}}\widehat {\text{k}}$are collinear.

Ans: Assume the given vector as, $\overrightarrow a = 2\widehat i - 3\widehat j + 4\widehat k$ and $\overrightarrow b = - 4\widehat i + 6\widehat j - 8\widehat k$respectively.

Therefore, it can be observed that $\overrightarrow b$ can be deduced in terms of$\overrightarrow a$as,

$\overrightarrow b = - 4\widehat i + 6\widehat j - 8\widehat k = - 2\left( {2\widehat i - 3\widehat j + 4\widehat k} \right) = - 2\overrightarrow a$

$\therefore \overrightarrow b = \lambda \overrightarrow a$

The value of $\lambda = - 2$ in this case and therefore the given two vectors collinear.

12. Find the direction cosines of the vector $\widehat {\text{i}}{\text{ + 2}}\widehat {\text{j}}{\text{ + 3}}\widehat {\text{k}}$.

Ans: Assume the given vector as, $\overrightarrow a = \widehat i + 2\widehat j + 3\widehat k$.

Therefore, the magnitude of the vector, that is, $\overrightarrow a = \widehat i + 2\widehat j + 3\widehat k$ can be calculated as,

$\left| {\overrightarrow a } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}} = \sqrt {14}$

Therefore, the direction cosines of the vector, that is, $\overrightarrow a = \widehat i + 2\widehat j + 3\widehat k$ can be calculated as, $\left( {\dfrac{1}{{\sqrt {14} }},\,\frac{2}{{\sqrt {14} }},\,\frac{3}{{\sqrt {14} }}} \right)$ .

13. Find the direction cosines of the vector joining the points ${\text{A}}\left( {{\text{1,}}\,{\text{2,}}\,{\text{ - 3}}} \right)$and ${\text{B}}\left( {{\text{ - 1,}}\,{\text{ - 2,}}\,{\text{1}}} \right)$directed from A to B.

Ans: Observe that the given points are,$A\left( {1,\,2,\, - 3} \right)$and $B\left( { - 1,\, - 2,\,1} \right)$.

Hence, the vector $\overrightarrow {AB}$ can be calculated as,

$\therefore \overrightarrow {AB} = \left( { - 1 - 1} \right)\widehat i + \left( { - 2 - 2} \right)\widehat j + \left( {1 - \left( { - 3} \right)} \right)\widehat k = - 2\widehat i - 4\widehat j + 4\widehat k$

Therefore, the magnitude of the vector, that is, $\overrightarrow {AB} = - 2\widehat i - 4\widehat j + 4\widehat k$ can be calculated as,

$\left| {\overrightarrow {AB} } \right| = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 4 \right)}^2}} = 6$

Therefore, the direction cosines of the vector, that is, $\overrightarrow {AB} = - 2\widehat i - 4\widehat j + 4\widehat k$ can be calculated as, $\left( {\frac{-1}{3},\,\frac{-2}{3},\,\frac{2}{3}} \right)$.

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

Ans: Consider the given vector as,$\overrightarrow a = \widehat i + \widehat j + \widehat k$.

Therefore, the magnitude and the direction cosines of the vector, that is, $\overrightarrow a = \widehat i + \widehat j + \widehat k$ can be calculated as,

$\left| {\overrightarrow a } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 3$

$\therefore \left( {\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }}} \right)$.

Assume, $\alpha ,\,\beta$ and $\gamma$ to be the angles formed by $\overrightarrow a$ with the axes $OX,\,OY$ and$OZ$.$\therefore \cos \alpha = \frac{1}{{\sqrt 3 }},\,\cos \beta = \frac{1}{{\sqrt 3 }},\,\cos \gamma = \frac{1}{{\sqrt 3 }}$

This shows that the given vector is equally inclined to the  the positive direction of all the three axes.

15. Find the position vector of a point ${\text{R}}$which divides the line segment joining two points ${\text{P}}$ and ${\text{Q}}$ whose position vectors are $\widehat {\text{i}}{\text{ + 2}}\widehat {\text{j}}{\text{ - }}\widehat {\text{k}}$ and ${\text{ - }}\widehat {\text{i}}{\text{ + }}\widehat {\text{j}}{\text{ + }}\widehat {\text{k}}$ respectively in the ratio ${\text{2:1}}$

(i) internally

Ans: Observe that the given position vectors of the two points, $P$ and $Q$are,$\overrightarrow {OP} = \widehat i + 2\widehat j - \widehat k$and $\overrightarrow {OQ} = - \widehat i + \widehat j + \widehat k$.

Therefore, the position vector of $R$ dividing the line joining the two given points through an internal division in the ratio of $2:1$ can be calculated using the formula, $\frac{{m\overrightarrow b + n\overrightarrow a }}{{m + n}}$ (where the ratio is represented by $m:n$), as shown below,

$\overrightarrow {OR} = \frac{{2\left( { - \widehat i + \widehat j + \widehat k} \right) + 1\left( {\widehat i + 2\widehat j - \widehat k} \right)}}{{2 + 1}} = \frac{{ - \widehat i + 4\widehat j + \widehat k}}{3} = \frac{{ - 1}}{3}\widehat i + \frac{4}{3}\widehat j + \frac{1}{3}\widehat k$

(ii) externally

Ans: Observe that the given position vectors of the two points, $P$ and $Q$are,$\overrightarrow {OP} = \widehat i + 2\widehat j - \widehat k$and $\overrightarrow {OQ} = - \widehat i + \widehat j + \widehat k$.

Therefore, the position vector of $R$ dividing the line joining the two given points through an external division in the ratio of $2:1$ can be calculated using the formula, $\frac{{m\overrightarrow b - n\overrightarrow a }}{{m - n}}$ (where the ratio is represented by $m:n$), as shown below,

$\overrightarrow {OR} = \frac{{2\left( { - \widehat i + \widehat j + \widehat k} \right) - 1\left( {\widehat i + 2\widehat j - \widehat k} \right)}}{{2 - 1}} = \frac{{ - 3\widehat i - \widehat k}}{1} = - 3\widehat i + 3\widehat k$.

16. Find the position vector of the midpoint of the vector joining the points ${\text{P}}\left( {{\text{2,}}\,{\text{3,}}\,{\text{4}}} \right)$ and ${\text{Q}}\left( {{\text{4,}}\,{\text{1,}}\,{\text{ - 2}}} \right)$.

Ans: Consider the midpoint of the vector joining the position vectors of the two points,$\overrightarrow {OP} = 2\widehat i + 3\widehat j + 4\widehat k$and $\overrightarrow {OQ} = 4\widehat i + \widehat j - 2\widehat k$ to be $R$.

Therefore, the position vector of $R$ dividing the line joining the two given points through an internal division in the ratio of $1:1$ can be calculated as,

$\overrightarrow {OR} = \frac{{\left( {2\widehat i + 3\widehat j + 4\widehat k} \right) + \left( {4\widehat i + \widehat j - 2\widehat k} \right)}}{{1 + 1}} = \frac{{6\widehat i + 4\widehat j + 2\widehat k}}{2} = 3\widehat i + 2\widehat j + \widehat k$

17. Show that the positions ${\text{A,}}\,{\text{B}}$ and ${\text{C}}$ with position vectors $\overrightarrow {\text{a}} {\text{ = 3}}\widehat {\text{i}}{\text{ - 4}}\widehat {\text{j}}{\text{ - 4}}\widehat {\text{k}}{\text{,}}\,\overrightarrow {\text{b}} {\text{ = 2}}\widehat {\text{i}}{\text{ - }}\widehat {\text{j}}{\text{ + }}\widehat {\text{k}}$ and $\overrightarrow {\text{c}} {\text{ = }}\widehat {\text{i}}{\text{ - 3}}\widehat {\text{j}}{\text{ - 5}}\widehat {\text{k}}$ respectively form the vertices of a right-angled triangle.

Ans: Observe that the position vectors of the points$A,\,B$ and$C$are,$\overrightarrow a =3 \widehat i - 4\widehat j - 4\widehat k,\,\overrightarrow b = 2\widehat i - \widehat j + \widehat k$ and$\overrightarrow c = \widehat i - 3\widehat j - 5\widehat k$.

Now, $\overrightarrow {AB} ,\,\overrightarrow {BC}$and $\overrightarrow {CA}$can be calculated as,

$\therefore \overrightarrow {AB} = \overrightarrow b - \overrightarrow a = \left( {2 - 3} \right)\widehat i + \left( { - 1 + 4} \right)\widehat j + \left( {1 + 4} \right)\widehat k = - \widehat i + 3\widehat j + 5\widehat k$

$\overrightarrow {BC} = \overrightarrow c - \overrightarrow b = \left( {1 - 2} \right)\widehat i + \left( { - 3 + 1} \right)\widehat j + \left( { - 5 - 1} \right)\widehat k = - \widehat i - 2\widehat j - 6\widehat k$

$\overrightarrow {CA} = \overrightarrow a - \overrightarrow c = \left( {3 - 1} \right)\widehat i + \left( { - 4 + 3} \right)\widehat j + \left( { - 4 + 5} \right)\widehat k = 2\widehat i - \widehat j + \widehat k$

Again, calculate the squares of magnitudesas shown below,

$\therefore {\left| {AB} \right|^2} = {\left( { - 1} \right)^2} + {\left( 3 \right)^2} + {\left( 5 \right)^2} = 35$

$\,\,\therefore \,{\left| {BC} \right|^2} = {\left( { - 1} \right)^2} + {\left( { - 2} \right)^2} + {\left( { - 6} \right)^2} = 41$

$\therefore \,\,{\left| {CA} \right|^2} = {\left( 2 \right)^2} + {\left( { - 1} \right)^2} + {\left( 1 \right)^2} = 6$

Now, use Pythagoras theorem it can be shown that this triangle is right-angled triangle as shown below,

${\left| {AB} \right|^2} + {\left| {CA} \right|^2} = 35 + 6 = 41 = {\left| {BC} \right|^2}$

18. In triangle ${\text{ABC}}$ which of the following is not true:

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

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

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

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

Ans:

On considering the triangle law of addition for this triangle it can be observed that,

$\,\,\,\,\,\,\,\,\therefore \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}$

$\Rightarrow \overrightarrow {AB} + \overrightarrow {BC} = - \overrightarrow {CA}$

$\Rightarrow \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = \overrightarrow 0$

Therefore, the equation in (A) is true.

Again,

$\,\,\,\,\,\,\,\,\therefore \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}$

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

Therefore, the equation in (B) is true.

Again,

$\,\,\,\,\,\,\,\,\therefore \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}$

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

Therefore, the equation in (D) is true.

Now, for the equation in (C) also,

$\,\,\,\,\,\,\,\,\therefore \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}$

$\Rightarrow \overrightarrow {AB} + \overrightarrow {BC} = - \overrightarrow {CA}$

$\Rightarrow \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = \overrightarrow 0$

$\Rightarrow \overrightarrow {AB} + \overrightarrow {BC} - \overrightarrow {CA} \ne \overrightarrow 0$

Therefore, the equation in (C) is incorrect and henceforth, it is the correct answer.

19. If $\overrightarrow {\text{a}}$ and $\overrightarrow {\text{b}}$ are two collinear vectors, then which of the following are incorrect. A.A.

A. $\overrightarrow b = \lambda \overrightarrow a$, for some scalar $\lambda$.

B. $\overrightarrow a = \pm \overrightarrow b$

C. the respective components of $\overrightarrow {\text{a}}$ and$\overrightarrow {\text{b}}$are not proportional.

D. both the vectors $\overrightarrow {\text{a}}$ and$\overrightarrow {\text{b}}$have same direction, but different magnitudes.

Ans: It can be observed that if $\overrightarrow a$and$\overrightarrow b$ are collinear vectors then, they are parallel.

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

Again, if $\lambda = \pm 1$, then $\overrightarrow a = \pm \overrightarrow b$.

Again if is considered that, $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ and $\,\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$, then $\overrightarrow b = \lambda \overrightarrow a$.

$\,\,\,\,\,\,\,{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = \lambda \left( {{a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k} \right)$

$\Rightarrow {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = \lambda {a_1}\widehat i + \lambda {a_2}\widehat j + \lambda {a_3}\widehat k$

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

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

This shows that the respective components of $\overrightarrow a$ and $\overrightarrow b$, are proportional but the vectors can have different directions.

Therefore, the statement in (D) is incorrect and henceforth, it is the correct answer.

## Conclusion

NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2  - Vector Algebra is essential for understanding basic vector operations. This exercise covers the addition of vectors and the multiplication of vectors by a scalar, which are foundational concepts. Focus on mastering vector addition and scalar multiplication, as these skills are vital for solving more complex problems. Pay attention to the components of vectors and their manipulation. Vedantu’s detailed, step-by-step solutions in ex 10.2 class 12 help ensure a strong grasp of these concepts, aiding in effective exam preparation and boosting your confidence in handling vector algebra.

## Class 12 Maths Chapter 10: Exercises Breakdown

 Exercise Number of Questions Exercise 10.1 5 Question & Solutions Exercise 10.3 18 Questions & Solutions Exercise 10.4 12 Questions & Solutions

## CBSE Class 12 Maths Chapter 10 Other Study Materials

 S. No Important Links for Chapter 10 Vector Algebra 1 Class 12 Vector Algebra Important Questions 2 Class 12 Vector Algebra Revision Notes 3 Class 12 Vector Algebra Important Formulas 4 Class 12 Vector Algebra NCERT Exemplar Solution 5 Class 12 Vector Algebra RD Sharma 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 10.2

1. What does Class 12 Maths Chapter 10 Exercise 10.2 deal with?

Class 12 Maths Chapter 10 Exercise 10.2 of the NCERT textbook deals with Vector Algebra, which also contains the following topics:

1. Multiplication of a Vector by a Scalar

a. Components of a vector

b. Vector joining two points

c. Section formula

All these topics have been explained in a concise manner along with the solutions to the problems asked in this exercise. All the solutions are accurate as these are solved by the well qualified subject matter experts and teachers.

2. How many questions are there in Class 12 Maths Chapter 10 Exercise 10.2?

Class 12 Maths Chapter 10 Exercise 10.2 consists of 19 questions of the various patterns. Solutions to all these questions are also being provided in our NCERT Solutions which are created by the highly experienced subject matter experts from the respective industry.

3. How many total exercises are there in class 12 maths exercise 10.2?

There are a total of four exercises in the class 12 maths exercise 10.2 of the NCERT textbook. Have a look at the list of exercises given below, along with the number of questions contained in these exercises.

• Class 12 Maths Chapter 10 Exercise 10.1: 5 Questions

• Class 12 Maths Chapter 10 Exercise 10.2: 19 Questions

• Class 12 Maths Chapter 10 Exercise 10.3: 18 Questions

• Class 12 Maths Chapter 10 Exercise 10.4: 12 Questions

• Class 12 Maths Chapter 10 Miscellaneous Exercise: 19 Questions

4. How can I download the NCERT Solutions Class 12 Maths Chapter 10 Exercise 10.2 from Vedantu website?

It is very easy to download the NCERT Solutions Class 12 Maths Chapter 10 Exercise 10.2 from Vedantu website. You just have to visit the Vedantu official website and look for the Class 12 category under ‘NCERT Solutions’ section. That’s all! You can easily access those notes without any hassle. Download the NCERT solutions and study as per your convenience at the comfort of your home.

5. How many questions are there in Exercise 10.2 of Chapter 10 Vector Algebra Class 12 Maths?

The total number of questions given in Exercise 10.2 of Chapter 10 Vector Algebra Class 12 Maths is 19. These 19 questions contain questions of several types. The questions will allow you to test your understanding of a particular topic as well as the entire concept of the addition and multiplication of vectors. The questions asked include finding the magnitude of vectors, writing vectors having the same direction, finding values of unknown variables in two equal vectors, and also finding statements that are false by applying the knowledge of vectors.

6. How can I strongly prepare Exercise 10.2 of Chapter 10 Vector Algebra Class 12 Maths?

Exercise 10.2 of Chapter 10 Vector Algebra Class 12 Maths tests how well you have understood the concept of vectors, the addition of vectors, and the multiplication of vectors by a scalar quantity. To enhance your understanding of the exercise, you will find the resources available at the Vedantu website and application to be of extreme use. You can supplement your preparation of the exercise by smartly utilizing the Vedantu resources and understand the concept comprehensively.

7. Where can I find Exercise 10.2 of Chapter 10 Vector Algebra Class 12 Maths solutions?

The solutions for Exercise 10.2 of Chapter 10 Vector Algebra Class 12 Maths are available free of cost on the two platforms of Vedantu- website as well as mobile application. All the resources provided by Vedantu are free of cost. The solutions can be downloaded on your phones or your laptops. You can use the solutions to speed up your preparation and can go through them as many times as you want. The solutions are prepared by the faculty after extensively analyzing the previous years’ papers.

8. How can I use Vedantu solutions effectively?

The solutions that are provided by the platforms of Vedantu are prepared after going through the syllabus, analyzing patterns and trends of previous years’ question papers, and understanding the demand of questions asked. The solutions will allow you to understand how to approach a particular question. The solutions will also help you to test whether you have followed all required steps in your solutions. You can also improve your presentation skills by understanding how to answer the answer in steps.

9. Why is vector addition important in class 12 maths ex 10.2?

In class 12 maths ex 10.2 vector addition is important because it enables us to combine two or more vectors into a single resultant vector. This process is essential in fields like physics and engineering, where it helps in determining net forces, velocities, and other vector quantities.

10. How do I perform scalar multiplication of a vector in class 12 maths 10.2?

To perform scalar multiplication, multiply each component of the vector by the scalar. This operation scales in class 12 maths 10.2 the vector's magnitude by the scalar value while keeping its direction unchanged, which is useful in various mathematical and physical applications.