# The magnitude of vectors

Question:

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

1. $\tan ^{-1} \frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}$

2. $\tan ^{-1} \frac{(\sqrt{3}-1+\sqrt{2})}{(1+\sqrt{3}-\sqrt{2})}$

3. $\tan ^{-1} \frac{(\sqrt{3}-1+\sqrt{2})}{(1-\sqrt{3}+\sqrt{2})}$

4. $\tan ^{-1} \frac{(1+\sqrt{3}-\sqrt{2})}{(1-\sqrt{3}-\sqrt{2})}$

Correct Option: 1

Solution:

Let magnitude be equal to $\lambda$.

$\overrightarrow{\mathrm{OA}}=\lambda\left[\cos 30^{\circ} \hat{\mathrm{i}}+\sin 30 \hat{\mathrm{j}}\right]=\lambda\left[\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\right]$

$\overrightarrow{\mathrm{OB}}=\lambda\left[\cos 60^{\circ} \hat{\mathrm{i}}-\sin 60 \hat{\mathrm{j}}\right]=\lambda\left[\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right]$

$\overrightarrow{\mathrm{OC}}=\lambda\left[\cos 45^{\circ}(-\hat{\mathrm{i}})+\sin 45 \hat{\mathrm{j}}\right]=\lambda\left[-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}\right]$

$\therefore \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OC}}$

$=\lambda\left[\left(\frac{\sqrt{3}+1}{2}+\frac{1}{\sqrt{2}}\right) \hat{\mathrm{i}}+\left(\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\right) \hat{\mathrm{j}}\right]$

$\therefore$ Angle with x-axis

$\tan ^{-1}\left[\frac{\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt{2}}}\right]=\tan ^{-1}\left[\frac{\sqrt{2}-\sqrt{6}-2}{\sqrt{6}+\sqrt{2}+2}\right]$

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

Hence option (1)