The magnitude of vectors OA, OB, and OC in the given figure are equal. The direction of OA + OB - OC with X-axis will be

In the given figure, the magnitude of vectors OA, OB and OC are equal. Let it be m.

From figure, we can write

$\mathrm{OA}=\mathrm{m} \left(\cos 30° \hat{\mathrm{i}}+\sin 30° \hat{\mathrm{j}}\right)=\mathrm{m} \left(\frac{\sqrt{3}}{2}\hat{\mathrm{i}}+\frac{1}{2}\hat{\mathrm{j}}\right)$

$\mathrm{OB}=\mathrm{m} \left(\cos 60° \hat{\mathrm{i}}-\sin 60° \hat{\mathrm{j}}\right)=\mathrm{m} \left(\frac{1}{2}\hat{\mathrm{i}}-\frac{\sqrt{3}}{2}\hat{\mathrm{j}}\right)$

$\mathrm{OC}=\mathrm{m} \left(\cos 45° \left(-\hat{\mathrm{i}}\right)+\sin 45° \hat{\mathrm{j}}\right)=\mathrm{m} \left\{\frac{-1}{\sqrt{2}}\hat{\mathrm{i}}+\frac{1}{\sqrt{2}}\hat{\mathrm{j}}\right\}$

According to the question,

$\mathrm{OA}+\mathrm{OB}-\mathrm{OC}=\mathrm{m} \left(\frac{\sqrt{3}}{2}\hat{\mathrm{i}}+\frac{1}{2}\hat{\mathrm{j}}\right)+\mathrm{m} \left(\frac{1}{2}\hat{\mathrm{i}}-\frac{\sqrt{3}}{2}\hat{\mathrm{j}}\right)-\mathrm{m} \left(-\frac{1}{\sqrt{2}}\hat{\mathrm{i}}+\frac{1}{\sqrt{2}}\hat{\mathrm{j}}\right)$

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

$\therefore$ Angle with X-axis is given by

$\begin{array}{l} \tan \mathrm{\theta }=\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{m}\left(\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\right)}{\mathrm{m}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt{2}}\right)}\\ \Rightarrow \tan \mathrm{\theta }=\frac{\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{\sqrt{2}}{2}}\\ \Rightarrow \tan \mathrm{\theta }=\frac{1-\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\sqrt{2}}\\ \Rightarrow \mathrm{\theta }={\tan }^{-1} \left(\frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}\right)\end{array}$