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(B) To find a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$,we first calculate the cross product $\vec{a} 	imes \vec{b}$.$\vec{a} 	imes

Vedclass · Accepted Answer

$\frac{3 \hat{i} + 2 \hat{j} - 3 \hat{k}}{\sqrt{22}}$

Vedclass · Answer

(B) To find a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$,we first calculate the cross product $\vec{a} 	imes \vec{b}$.$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 0 & 3 & 2 \end{vmatrix} = \hat{i}(6 - 12) - \hat{j}(4 - 0) + \hat{k}(6 - 0) = -6 \hat{i} - 4 \hat{j} + 6 \hat{k}$.The magnitude of this vector is $|\vec{a} 	imes \vec{b}| = \sqrt{(-6)^2 + (-4)^2 + 6^2} = \sqrt{36 + 16 + 36} = \sqrt{88} = 2\sqrt{22}$.The unit vector is given by $\pm \frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|} = \pm \frac{-6 \hat{i} - 4 \hat{j} + 6 \hat{k}}{2\sqrt{22}} = \pm \frac{-3 \hat{i} - 2 \hat{j} + 3 \hat{k}}{\sqrt{22}}$.This is equivalent to $\pm \frac{3 \hat{i} + 2 \hat{j} - 3 \hat{k}}{\sqrt{22}}$. Comparing with the options,option $B$ is correct.

$A$ unit vector perpendicular to the vectors $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\vec{b} = 3 \hat{j} + 2 \hat{k}$ is

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