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 PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

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