Question

A vector $\overrightarrow{A}=\hat{i}+\sqrt{3}\hat{j}$ is rotated clockwise through an angle ${30}^{\circ }$ about one of its ends. The unit vector along the new vector is

• A. $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$
• B. $\sqrt{3}\hat{i}+\hat{j}$
• C. $\sqrt{\frac{2}{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}$
• D. $\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$

Answer 1

The correct option is D $\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$

$\overrightarrow{A}=\hat{i}+\sqrt{3}\hat{j}\tan \theta =\frac{y}{x}=\frac{\sqrt{3}}{1}\Rightarrow \theta =\frac{\pi }{3}={60}^{\circ }$

Let, $\overrightarrow{P}$be the new vector
Then,$A=P=\sqrt{1^2+{\left(\sqrt{3}\right)}^2}=2\text{units}$

New vector
$\overrightarrow{P}=P\cos 30\hat{i}+P\sin 30\hat{i}\overrightarrow{P}=2\times \frac{\sqrt{3}}{2}\hat{i}+2\times \frac{1}{2}\hat{j}\overrightarrow{P}=\sqrt{3}\hat{i}+\hat{j}$

Unit Vector $=\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$