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(D) Let $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$.The unit vector perpendicular to the plane containing $\v

Vedclass · Accepted Answer

$\frac{\hat{i} + \hat{j}}{\sqrt{2}}$

Vedclass · Answer

(D) Let $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$.The unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\hat{n} = \pm \frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|}$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ -1 & 1 & 1 \end{vmatrix}$$= \hat{i}(-1 - 1) - \hat{j}(1 - (-1)) + \hat{k}(1 - 1)$$= -2\hat{i} - 2\hat{j} + 0\hat{k} = -2\hat{i} - 2\hat{j}$.Now,find the magnitude $|\vec{a} 	imes \vec{b}| = \sqrt{(-2)^2 + (-2)^2 + 0^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.The unit vector is $\pm \frac{-2\hat{i} - 2\hat{j}}{2\sqrt{2}} = \pm \frac{-(\hat{i} + \hat{j})}{\sqrt{2}} = \pm \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.Comparing with the given options,the correct option is $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$.

$A$ unit vector perpendicular to the plane containing the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$ is

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