If overrightarrow{P} = 3hat{i} + sqrt{3}hat{j | vedclass

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(D) First,calculate the cross product $\overrightarrow{P} 	imes \overrightarrow{Q}$ using the determinant method:$\overrightarrow{P} 	imes \overrigh

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) First,calculate the cross product $\overrightarrow{P} 	imes \overrightarrow{Q}$ using the determinant method:$\overrightarrow{P} 	imes \overrightarrow{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & \sqrt{3} & 2 \ 4 & \sqrt{3} & 2.5 \end{vmatrix}$$= \hat{i}(\sqrt{3} 	imes 2.5 - 2 	imes \sqrt{3}) - \hat{j}(3 	imes 2.5 - 2 	imes 4) + \hat{k}(3 	imes \sqrt{3} - 4 	imes \sqrt{3})$$= \hat{i}(2.5\sqrt{3} - 2\sqrt{3}) - \hat{j}(7.5 - 8) + \hat{k}(3\sqrt{3} - 4\sqrt{3})$$= 0.5\sqrt{3}\hat{i} + 0.5\hat{j} - \sqrt{3}\hat{k}$$= \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} - \sqrt{3}\hat{k}$Next,find the magnitude $|\overrightarrow{P} 	imes \overrightarrow{Q}| = \sqrt{(\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2 + (-\sqrt{3})^2} = \sqrt{\frac{3}{4} + \frac{1}{4} + 3} = \sqrt{1 + 3} = 2$.The unit vector is $\frac{\overrightarrow{P} 	imes \overrightarrow{Q}}{|\overrightarrow{P} 	imes \overrightarrow{Q}|} = \frac{1}{2}(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} - \sqrt{3}\hat{k}) = \frac{1}{4}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k})$.Comparing this with $\frac{1}{x}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k})$,we get $x = 4$.

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