NCERT Solutions For Class 12 Maths Chapter 10 Vector Algebra Exercise 10.2 - 2025-26

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

CBSE Class 12 Maths Chapter 10 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.