Question

The magniude of vectors \(\overrightarrow {OA},\overrightarrow {OB}\) and \(\overrightarrow {OC}\) In the given figure are equal.\(\overrightarrow {OA}+\overrightarrow {OB}-\overrightarrow {OC}\) with x-axis will be

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

Answer 1

The correct option is C ${\tan }^{-1}\frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}$

Let's write the components of each vector first.
$\overrightarrow{OA}=A\cos {30}^{\circ }\hat{i}+A\sin {30}^{\circ }\hat{j}$

Similarly,
$\overrightarrow{OB}=A\cos {60}^{\circ }\hat{i}-A\sin {60}^{\circ }\hat{j}$

And, $\overrightarrow{OC}=-A\cos {45}^{\circ }\hat{i}+A\sin {45}^{\circ }\hat{j}$

So, $\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}=\frac{\sqrt{3}}{2}A\hat{i}+\frac{A}{2}\hat{j}+\frac{A}{2}\hat{i}-\frac{\sqrt{3}}{2}A\hat{j}+\frac{A}{\sqrt{2}}\hat{i}-\frac{A}{\sqrt{2}}\hat{j}$

Now, separating $\hat{i}$ and $\hat{j}$ components we get,

$\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}=(\frac{\sqrt{3}}{2}A+\frac{A}{2}+\frac{A}{\sqrt{2}})\hat{i}+(\frac{A}{2}-\frac{\sqrt{3}}{2}A-\frac{A}{2})\hat{j}$

I.e. $=\frac{A}{2}(\sqrt{3}+1+\sqrt{2})\hat{i}+\frac{A}{2}(1-\sqrt{3}-\sqrt{2})\hat{j}$

The direction for the given set of vectors is given by,

$tan\theta =\frac{\frac{A}{2}(1-\sqrt{3}-\sqrt{2})}{\frac{A}{2}(\sqrt{3}+1+\sqrt{2})}$

i.e. $\theta ={\tan }^{-1}\left(\frac{1-\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\sqrt{2}}\right)$