The magnitude of vectors overrightarrow{ OA } | vedclass

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Question

(A) Let the magnitude of each vector be $\lambda$.$\overrightarrow{ OA } = \lambda (\cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j}) = \lambda (\fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the magnitude of each vector be $\lambda$.$\overrightarrow{ OA } = \lambda (\cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j}) = \lambda (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j})$$\overrightarrow{ OB } = \lambda (\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j}) = \lambda (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})$$\overrightarrow{ OC } = \lambda (\cos 45^{\circ} (-\hat{i}) + \sin 45^{\circ} \hat{j}) = \lambda (-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$Now,calculate $\overrightarrow{ R } = \overrightarrow{ OA } + \overrightarrow{ OB } - \overrightarrow{ OC }$.$\overrightarrow{ R } = \lambda [(\frac{\sqrt{3}}{2} + \frac{1}{2} - (-\frac{1}{\sqrt{2}})) \hat{i} + (\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}) \hat{j}]$$\overrightarrow{ R } = \lambda [(\frac{\sqrt{3}+1+\sqrt{2}}{2}) \hat{i} + (\frac{1-\sqrt{3}-\sqrt{2}}{2}) \hat{j}]$The angle $	heta$ with the $x$-axis is given by $	an 	heta = \frac{R_y}{R_x}$.$	an 	heta = \frac{\frac{1-\sqrt{3}-\sqrt{2}}{2}}{\frac{\sqrt{3}+1+\sqrt{2}}{2}} = \frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}$Therefore,$	heta = 	an ^{-1} \frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}$.

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 the $x$-axis will be

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