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(D) Let $\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} - \hat{j} + 2\hat{k}$.The unit vector $\hat{n}$ perpendicular to both $\vec{A

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$\frac{1}{\sqrt{83}} (7\hat{i} - 3\hat{j} - 5\hat{k})$

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(D) Let $\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} - \hat{j} + 2\hat{k}$.The unit vector $\hat{n}$ perpendicular to both $\vec{A}$ and $\vec{B}$ is given by $\hat{n} = \frac{\vec{A} 	imes \vec{B}}{|\vec{A} 	imes \vec{B}|}$.First,calculate the cross product $\vec{A} 	imes \vec{B}$:$\vec{A} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(6 - (-1)) - \hat{j}(4 - 1) + \hat{k}(-2 - 3) = 7\hat{i} - 3\hat{j} - 5\hat{k}$.Next,calculate the magnitude of the cross product:$|\vec{A} 	imes \vec{B}| = \sqrt{7^2 + (-3)^2 + (-5)^2} = \sqrt{49 + 9 + 25} = \sqrt{83}$.Therefore,the unit vector is $\hat{n} = \frac{1}{\sqrt{83}} (7\hat{i} - 3\hat{j} - 5\hat{k})$.

The unit vector perpendicular to the two vectors $(2\hat{i} + 3\hat{j} + \hat{k})$ and $(\hat{i} - \hat{j} + 2\hat{k})$ is:

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