Given two vectors vec{A} = 3hat{i} + hat{j} a | vedclass

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(C) 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{

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$\frac{2}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{3}{7}\hat{k}$

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(C) 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 $\vec{A} 	imes \vec{B}$:$\vec{A} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 0 \ 0 & 1 & 2 \end{vmatrix} = \hat{i}(1 	imes 2 - 0 	imes 1) - \hat{j}(3 	imes 2 - 0 	imes 0) + \hat{k}(3 	imes 1 - 1 	imes 0) = 2\hat{i} - 6\hat{j} + 3\hat{k}$.Next,find the magnitude of the cross product:$|\vec{A} 	imes \vec{B}| = \sqrt{2^2 + (-6)^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$.Finally,the unit vector is $\hat{n} = \frac{2\hat{i} - 6\hat{j} + 3\hat{k}}{7} = \frac{2}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{3}{7}\hat{k}$.

Given two vectors $\vec{A} = 3\hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + 2\hat{k}$. Find the unit vector perpendicular to both $\vec{A}$ and $\vec{B}$.

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