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

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-2/

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

10.2 Class 12 Question 1.
Compute the magnitude of the following vectors:
$$\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { 2i } -\hat { 7j } -\hat { 3k }$$
$$\overrightarrow { c } =\frac { 1 }{ \sqrt { 3 } } \hat { i } +\frac { 1 }{ \sqrt { 3 } } \hat { j } -\frac { 1 }{ \sqrt { 3 } } \hat { k }$$
Solution:

Question 2.
Write two different vectors having same magnitude.
Solution:
Let $$\overrightarrow { a } =\hat { i } +\hat { 2j } +\hat { 3k } ,\overrightarrow { b } =\hat { 3i } +\hat { 2j } +\hat { k }$$
$$|\vec{a}|=\sqrt{1+1+9}=\sqrt{11}$$
$$|\vec{b}|=\sqrt{9+1+1}=\sqrt{11}$$
∴ $$\overline{a}$$ and $$\overline{b}$$ are examples for two vectors having the same magnitude.
There are infinitely many Vectors having same magnitude.

Question 3.
Write two different vectors having same direction.
Solution:
$$\hat{i}+\hat{j}+\hat{k}$$ and $$\hat{3i}+\hat{3j}+\hat{3k}$$ are examples for two vectors having the same direction. Generally, if $$\vec{a}$$ is a nonzero vector, then $$\vec{a}$$ and λ$$\vec{a}$$ have the same direction whenever λ is positive.

Question 4.
Find the values of x and y so that the vectors $$2 \hat{i}+3 \hat{j} \text { and } x \hat{i}+y \hat{j}$$ are equal.
Solution:
We are given $$2 \hat{i}+3 \hat{j}=x \hat{i}+y \hat{j}$$
If vectors are equal, then their respective components are equal. Hence x = 2, y = 3.

Question 5.
Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5,7).
Solution:
Let A(2, 1) be the initial point and B(-5,7) be the terminal point $$\overrightarrow { AB } =\left( { x }_{ 2 }-{ x }_{ 1 } \right) \hat { i } +\left( { y }_{ 2 }-{ y }_{ 1 } \right) \hat { j } =-\hat { 7i } +\hat { 6j }$$
∴ The vector components are $$\vec{-7i}$$, $$\vec{6j}$$ and scalar components are – 7 and 6.

Question 6.
Find the sum of three vectors:
$$\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k } ,$$
Solution:
$$\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k }$$

Question 7.
Find the unit vector in the direction of the vector $$\overrightarrow { a } =\hat { i } +\hat { j } +\hat { 2k }$$.
Solution:

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

Question 9.
For given vectors $$\overrightarrow { a } =2\hat { i } -\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =-\hat { i } +\hat { j } -\hat { k }$$ find the unit vector in the direction of the vector $$\overrightarrow { a } +\overrightarrow { b }$$
Solution:

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

Question 11.
Show that the vector $$2 \hat{i}-3 \hat{j}+4 \hat{k}$$ and $$-4 \hat{i}+6 \hat{j}-8 \hat{k}$$ are collinear.
Solution:
$$\overrightarrow { a } =2\hat { i } -3\hat { j } +4\hat { k } \quad and\quad \overrightarrow { b } =-4\hat { i } +6\hat { j } -8\hat { k }$$
$$=-2(2\hat { i } -3\hat { j } +4\hat { k } )$$
vector $$\vec{a}$$ and $$\vec{b}$$ have the same direction they are collinear.

Question 12.
Find the direction cosines of the vector $$\hat { i } +2\hat { j } +3\hat { k }$$
Solution:

Question 13.
Find the direction cosines of the vector joining the points A (1, 2, – 3) and B(- 1, – 2, 1), directed from A to B.
Solution:

∴ Direction cosines of $$\overline{AB}$$ = Scalar components of $$\overline{AB}$$
= $$\frac{-1}{3}, \frac{-2}{3}, \frac{2}{3}$$

Question 14.
Show that the vector $$\hat { i } +\hat { j } +\hat { k }$$ are equally inclined to the axes OX, OY, OZ.
Solution:
Let $$\vec{r}$$ = $$\hat { i } +\hat { j } +\hat { k }$$
The direction ratios of $$\vec{r}$$ are 1, 1, 1.
The direction cosines of $$\vec{r}$$ are
$$\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}, \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}, \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}$$
$$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$$
∴ $$\vec{r}$$ is equally inclined to the axes.

Question 15.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $$\hat{i}+2 \hat{j}-\hat{k}$$ and $$-\hat{i}+\hat{j}+\hat{k}$$ respectively, in the ratio 2:1
i. internally
ii. externally
Solution:
i. Internal division
Let $$\vec{a}$$ = position vectorof P = $$\hat{i}+2 \hat{j}-\hat{k}$$
Let $$\vec{b}$$ = position vectorof Q = $$-\hat{i}+\hat{j}+\hat{k}$$
Let R divides PQ internally in the ratio 2 : 1

ii. External division
Let S divides PQ externally in the ratio 2 : 1

Question 16.
Find position vector of the mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).
Solution:
Let R be the midpoint of PQ

Question 17.
Show that the points A, B and C with position vector $$\overrightarrow { a } =3\hat { i } -4\hat { j } -4\hat { k } ,\overrightarrow { b } =2\hat { i } -\hat { j } +\hat { k } and\quad \overrightarrow { c } =\hat { i } -3\hat { j } -5\hat { k }$$ respectively form the vertices of a right angled triangle.
Solution:

Question 18.
In triangle ABC (fig.), which of the following is not

(a) $$\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 }$$
(b) $$\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 }$$
(c) $$\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { CA } =\overrightarrow { 0 }$$
(d) $$\overrightarrow { AB } -\overrightarrow { CB } +\overrightarrow { CA } =\overrightarrow { 0 }$$
Solution:
By the triangle law of vector addition,
$$\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 }$$
$$\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 }$$, is not true.

Question 19.
If $$\overrightarrow { a } ,\overrightarrow { b }$$ are two collinear vectors then which of the following are incorrect:
(a) $$\overrightarrow { b } =\lambda \overrightarrow { a }$$, for some scalar λ.
(b) $$\overrightarrow { a } =\pm \overrightarrow { b }$$
(c) the respective components of $$\overrightarrow { a } ,\overrightarrow { b }$$ are proportional.
(d) both the vectors $$\overrightarrow { a } ,\overrightarrow { b }$$ have same direction, but different magnitudes.
Solution:
If $$\vec{a}$$ and $$\vec{b}$$ are collinear, then they need not be in the same direction, d and b may have opposite directions.

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