Question

A unit vector a makes an angle $\frac{\pi}{4}$ with z-axis. If $\mathbf{a} + \mathbf{i} + \mathbf{j}$ is a unit vector, then a is equal to

• A. $\frac{\mathbf{i}}{2} + \frac{\mathbf{j}}{2} + \frac{\mathbf{k}}{\sqrt{2}}$
• B. $\frac{\mathbf{i}}{2} + \frac{\mathbf{j}}{2} - \frac{\mathbf{k}}{\sqrt{2}}$
• C. $- \frac{\mathbf{i}}{2} - \frac{\mathbf{j}}{2} + \frac{\mathbf{k}}{\sqrt{2}}$
• D. None of these

Answer 1

Answer: (C)

$- \frac{\mathbf{i}}{2} - \frac{\mathbf{j}}{2} + \frac{\mathbf{k}}{\sqrt{2}}$

Answer 2

Let $\mathbf{a} = l\mathbf{i} + m\mathbf{j} + n\mathbf{k},$ where $l^{2} + m^{2} + n^{2} = 1.$

a makes an angle $\frac{\pi}{4}$ with $z -$axis.

$\therefore n = \frac{1}{\sqrt{2}},$ $l^{2} + m^{2} = \frac{1}{2}$ …..(i)

$\therefore\mathbf{a} = l\mathbf{i} + m\mathbf{j} + \frac{\mathbf{k}}{\sqrt{2}}$

$\mathbf{a} + \mathbf{i} + \mathbf{j} = (l + 1)\mathbf{i} + (m + 1)\mathbf{j} + \frac{\mathbf{k}}{\sqrt{2}}$

Its magnitude is 1, hence $(l + 1)^{2} + (m + 1)^{2} = \frac{1}{2}$ .....(ii)

From (i) and (ii), $2lm = \frac{1}{2} \Rightarrow l = m = - \frac{1}{2}$

Hence $\mathbf{a} = - \frac{\mathbf{i}}{2} - \frac{\mathbf{j}}{2} + \frac{\mathbf{k}}{\sqrt{2}}$.