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] $
$=\tan ^{-1}\left[\frac{1-\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\sqrt{2}}\right]$