Vector Algebra Exercise 10.2

Question 1
For the following vectors, calculate the magnitude of the following.
$\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}; \;\;\; \overrightarrow{b} = \hat{2i} – \hat{7j} – \hat{3k}; \;\;\; \overrightarrow{c} = \frac{1}{\sqrt{3}} \;\hat{i} + \frac{1}{\sqrt{3}}\; \hat{j} – \frac{1}{\sqrt{3}}\; \hat{k}$  
Solution
The magnitude are calculated as below
$\left | \overrightarrow{a} \right | = \sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}} = \sqrt{3} \\ \left | \overrightarrow{b} \right | = \sqrt{(2) ^{2} + (- 7) ^{2} + (- 3) ^{2}} = \sqrt{4 + 49 + 9} = \sqrt{62} \\ \left | \overrightarrow{c} \right | = \sqrt{(\frac{1}{\sqrt{3}}) ^{2} + (\frac{1}{\sqrt{3}}) ^{2} + (\frac{1}{\sqrt{3}}) ^{2}} = 1 \\$  

Question-2
Write two different vectors having same magnitude?
Solution
$\overrightarrow{p} = (2\hat{i} – \hat{j} + 3\hat{k}); \;and\; \overrightarrow{q} = (2\hat{1i} + 3\hat{j} – \hat{k}) \\ It\; can\; be\; observed\; that:\; \\ \left | \overrightarrow{p} \right | = \sqrt{(2) ^{2} + (1) ^{2} + (3) ^{2}} = \sqrt{17} \; and \\ \left | \overrightarrow{q} \right | = \sqrt{(2) ^{2} + (3) ^{2} + (1) ^{2}} = \sqrt{14}$ So, these are two different vectors $\overrightarrow{p} \;and\; \overrightarrow{q}$ having the similar magnitude. They are different as their direction is different.
 
 
Question 3
Write two different vectors having same direction?
Solution
Consider two vectors, $\overrightarrow{p} = \hat{i} + \hat{j} + \hat{k}; \;\;\;and\; \overrightarrow{q} = 3\hat{i} + 3\hat{j} + 3\hat{k} \\ The\; direction\; cosines\; of\; \overrightarrow{p}\; are\; given\; by, \\ l = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}, \; m = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}, \;and\; n = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}$
$The\; direction\; cosines\; of\; \overrightarrow{q}\; are\; given\; by, \\ l = \frac{3}{\sqrt{(3) ^{2} + (3) ^{2} + (3) ^{2}}} = \frac{1}{\sqrt{3}}, \;m = \frac{3}{\sqrt{(3) ^{2} + (3) ^{2} + (3) ^{2}}} = \frac{1}{\sqrt{3}}\; and\; n = \frac{3}{\sqrt{(3) ^{2} + (3) ^{2} + (3) ^{2}}} = \frac{1}{\sqrt{3}}$ $The\; direction\; cosines\; of\; \overrightarrow{p}\; and\; \overrightarrow{q}\; are\; similar.$
Thus, the direction of the two vectors is similar.

Question 4:
Find the values of x and y so that the vectors $2 \hat{i} + 3 \hat{j} \;and\; x \hat{i} + y \hat{j}$ are equal
Solution
Given, $2 \hat{i} + 3 \hat{j} \;and\; x \hat{i} + y \hat{j}$ are equal.
The equivalent components are equal.
So, the value of x = 2 and y = 3.

Question 5
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (-5, 7).
Solution
Given,
The initial point of the vector A (2, 1) and the terminal point of the vector is B (- 5, 7).
The vector $\overrightarrow{AB} = (- 5 – 2) \hat{i} + (7 – 1) \hat{j} \\ \overrightarrow{AB} = – 7 \hat{i} + 6 \hat{j}$
The vector components of the given vector are $– 7 \hat{i} \;and\; 6 \hat{j}$.
The scalar components of the given vector are -9 and 6.


Question 6
Find the sum of the vectors $\overrightarrow{a} = \hat{i} -2\hat{j} + \hat{k}, \;\;\; \overrightarrow{b} = -2\hat{i} + 4\hat{j} +5 \hat{k}, \;\;and \; \overrightarrow{c} = \hat{i} – 6 \hat{j} – 7\hat{k}$.
Solution
$\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = (1 – 2 + 1) \hat{i} + (-2 + 4 – 6) \hat{j} + (1 + 5 – 7) \hat{k} \\ = – 4 \hat{j} – \hat{k}$

Question 7
Find the unit vector in the direction of the vector $\overrightarrow{a}= \hat{i} + \hat{j} + 2\hat{k}$
Solution
First we need to find the magnitude
$\left | \overrightarrow{a} \right | = \sqrt{(1) ^{2} + (1) ^{2} + (2) ^{2}} = \sqrt{1 + 1 + 4} = \sqrt{6} \\ \hat{a} = \frac{\overrightarrow{a}}{\left | \overrightarrow{a} \right |} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{j} + 2\frac{1}{\sqrt{6}} \hat{k} \\ = \frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}$  
 

Question 8
Find the unit vector $\overrightarrow{PQ}$ in the direction of vector , where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
Solution
Given, points P (1, 2, 3) and Q (4, 5, 6).
$\overrightarrow{PQ} = (4 – 1) \hat{i} + (5 – 2) \hat{j} + (6 – 3) \hat{k} \\ \overrightarrow{PQ} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ \left | \overrightarrow{PQ} \right | = \sqrt{(3) ^{2} + (3) ^{2} + (3) ^{2}} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3 \sqrt{3} \\ The\; unit\; vector\; in\; the\; direction\; of\; \overrightarrow{PQ}\; is \\ \frac{\overrightarrow{PQ}}{\left | \overrightarrow{PQ} \right |} = \frac{3 \hat{i} + 3 \hat{j} + 3 \hat{k}}{3 \sqrt{3}} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k}$

Question 9
For given vectors,$\overrightarrow{a} = 2\hat{i} – \hat{j} + 2\hat{k} \; and\; \overrightarrow{b} = -\hat{i} + \hat{j} – \hat{k}$ , find the unit vector in the direction of the vector $\overrightarrow{a} + \overrightarrow{b}$ .
Solution
$\overrightarrow{a} + \overrightarrow{b} = (2 -1) \hat{i} + (- 1 + 1) \hat{j} + (2 – 1) \hat{k} \\ = \hat{i} + 1 \hat{k} \\ \left | \overrightarrow{a} + \overrightarrow{n} \right | = \sqrt{(1) ^{2} + (1) ^{2} } = \sqrt{1 +1} = \sqrt{2}$ Thus, in the direction of $\overrightarrow{a} + \overrightarrow{b}$, the vector is,
$\frac{(\overrightarrow{a} + \overrightarrow{b})}{\left | \overrightarrow{a} + \overrightarrow{b} \right |} = \frac{ \hat{i} + \hat{k}}{\sqrt{2}}$

 

Also Read

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• Notes
  • What is vector
  • Difference between Scalar and Vector
  • Type Of Vectors
  • Addition Of vector
  • Subtraction of vectors
  • Scalar multiplication of vectors
  • Components of the vector
  • Multiplication of two vectors
  • Dot product of vectors
  • Cross product of vectors
  • Triple product
  • How to solve vector algebra problems
• Video
  • vector addition and subtraction
  • Vectors and their properties
• NCERT Solutions
  • NCERT Solutions for Vector Algebra Exercise 10.1
  • NCERT Solutions for Vector Algebra Exercise 10.2

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