Question

The magnitude of vectors $\overrightarrow{OA},\overrightarrow{OB}$ and $\overrightarrow{OC}$ in the given figure are equal. The direction of $\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}$ with $x$-axis will be :-

• A. ${\tan }^{-1}\frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}$
• B. ${\tan }^{-1}\frac{(\sqrt{3}-1+\sqrt{2})}{(1+\sqrt{3}-\sqrt{2})}$
• C. ${\tan }^{-1}\frac{(\sqrt{3}-1+\sqrt{2})}{(1-\sqrt{3}+\sqrt{2})}$
• D. ${\tan }^{-1}\frac{(1+\sqrt{3}-\sqrt{2})}{(1-\sqrt{3}-\sqrt{2})}$

Answer 1

Answer: (A)

Let magnitude be equal to $\lambda$.
$\overrightarrow{OA}=\lambda \left[\cos 30^{\circ }\hat{i}+\sin 30\hat{j}\right]=\lambda \left[\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right]$
$\overrightarrow{OB}=\lambda \left[\cos 60^{\circ }\hat{i}-\sin 60\hat{j}\right]=\lambda \left[\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j}\right]$
$\overrightarrow{OC}=\lambda \left[\cos 45^{\circ }(-\hat{i})+\sin 45\hat{j}\right]=\lambda \left[-\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}\right]$
$\therefore \overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}$
$=\lambda \left[\left(\frac{\sqrt{3}+1}{2}+\frac{1}{\sqrt{2}}\right)\hat{i}+\left(\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\right)\hat{j}\right]$
$\therefore$ Angle with $x$-axis
${\tan }^{-1}\left[\frac{\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt{2}}}\right]={\tan }^{-1}\left[\frac{\sqrt{2}-\sqrt{6}-2}{\sqrt{6}+\sqrt{2}+2}\right]$
$={\tan }^{-1}\left[\frac{1-\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\sqrt{2}}\right]$