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.