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:
Answer: (B) $\tan^{-1}\frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}$
Let each vector have magnitude $a$.
$\overrightarrow{OA} = a\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right)$ (at $30^\circ$ above $+x$)
$\overrightarrow{OB} = a\left(\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right)$ (at $60^\circ$ below $+x$)
$\overrightarrow{OC} = a\left(-\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right)$ (at $45^\circ$ above $-x$)
$\overrightarrow{OA} + \overrightarrow{OB} - \overrightarrow{OC} = \dfrac{a}{2}\left[(\sqrt{3} + 1 + \sqrt{2})\hat{i} + (1 - \sqrt{3} - \sqrt{2})\hat{j}\right]$
$$\theta = \tan^{-1}\frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}}$$
Solution by Sreeraj P, M.Sc Physics