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(D) Let $\overrightarrow{v} = \hat{i} + \hat{j} + \hat{k}$.First,find the cross product $\overrightarrow{b} 	imes \overrightarrow{c}$:$\overrightarro

Vedclass · Accepted Answer

$-\sqrt{6}$

Vedclass · Answer

(D) Let $\overrightarrow{v} = \hat{i} + \hat{j} + \hat{k}$.First,find the cross product $\overrightarrow{b} 	imes \overrightarrow{c}$:$\overrightarrow{b} 	imes \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2 - 9) - \hat{j}(-1 - 6) + \hat{k}(3 + 4) = -7\hat{i} + 7\hat{j} + 7\hat{k} = -7(\hat{i} - \hat{j} - \hat{k})$.Since $\hat{a}$ is a unit vector perpendicular to $\overrightarrow{b}$ and $\overrightarrow{c}$,$\hat{a} = \pm \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$.Given $\hat{a} \cdot \frac{\overrightarrow{v}}{|\overrightarrow{v}|} = \cos\left(\cos^{-1}\left(-\frac{1}{3}ight)ight) = -\frac{1}{3}$.If $\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$,then $\hat{a} \cdot \frac{\overrightarrow{v}}{\sqrt{3}} = \frac{1 - 1 - 1}{3} = -\frac{1}{3}$. This satisfies the condition.If $\hat{a} = \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$,then $\hat{a} \cdot \frac{\overrightarrow{v}}{\sqrt{3}} = \frac{-1 + 1 + 1}{3} = \frac{1}{3}$. This is rejected.So,$\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$.Now,$\hat{a}$ makes an angle $\frac{\pi}{3}$ with $\overrightarrow{u} = \hat{i} + \alpha\hat{j} + \hat{k}$.$\cos\left(\frac{\pi}{3}ight) = \frac{\hat{a} \cdot \overrightarrow{u}}{|\hat{a}| |\overrightarrow{u}|} \Rightarrow \frac{1}{2} = \frac{\frac{1 - \alpha - 1}{\sqrt{3}}}{\sqrt{1 + \alpha^2 + 1}} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$.$\frac{\sqrt{3}}{2} \sqrt{\alpha^2 + 2} = -\alpha$. Since $\alpha $3\alpha^2 + 6 = 4\alpha^2 \Rightarrow \alpha^2 = 6$. Since $\alpha < 0$,$\alpha = -\sqrt{6}$.

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