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# NCERT Solutions for Class 12 Maths Chapter 10 - Exercise Last updated date: 28th Nov 2023
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## NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.2) Exercise 10.2

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2 (Ex 10.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 10 Vector Algebra Exercise 10.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

### Concepts covered in Exercise 10.2:

• Position Vector

• Direction Cosines

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

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,

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

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, $\left( {\frac{1}{{\sqrt {14} }},\,\frac{2}{{\sqrt {14} }},\,\frac{3}{{\sqrt {14} }}} \right)$ .

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.

### Importance of Vector Algebra

In the fields of physics and mathematics, vector algebra has several applications. A vector is a quantity that has both a direction and a magnitude. Numerous quantities, such as velocity, acceleration, and force, must be represented as mathematical expressions and can be represented as vectors.

### Applications of Vector Algebra

• In the study of partial differential equations and differential geometry, vectors are extremely important.

• In Physics and engineering, vectors are particularly useful in the study of electromagnetic fields, gravitational fields, and fluid flow.

• When determining the component of a force in a specific direction, vector algebra is used.

• In Physics, vector algebra is used to find the interaction of two or more quantities.

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

Opting for the NCERT solutions for Ex 10.2 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 10.2 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.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Subject Vector Algebra textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 10 Exercise 10.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 10 Exercise 10.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 10 Exercise 10.2from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

### NCERT Solution Class 12 Maths of Chapter 10 All Exercises

 Chapter 10 - Vector Algebra Exercises in PDF Format Exercise 10.1 5 Questions & Solutions (5 Short Answers) Exercise 10.2 19 Questions & Solutions (5 Short Answers, 14 Long Answers) Exercise 10.3 18 Questions & Solutions (5 Short Answers, 13 Long Answers) Exercise 10.4 12 Questions & Solutions (4 Short Answers, 8 Long Answers)

## FAQs on NCERT Solutions for Class 12 Maths Chapter 10 - Exercise

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 Chapter 10?

There are a total of four exercises in the Class 12 Maths Chapter 10 of the NCERT textbook. Take 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.