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(D) Let $\vec{a} = \hat{i}+2\hat{j}+\hat{k}$ and $\vec{b} = -2\hat{i}+\hat{j}+3\hat{k}$.The vector perpendicular to the plane containing $\vec{a}$ and

Vedclass · Accepted Answer

$\frac{5\hat{i}-5\hat{j}+5\hat{k}}{5\sqrt{3}}$

Vedclass · Answer

(D) Let $\vec{a} = \hat{i}+2\hat{j}+\hat{k}$ and $\vec{b} = -2\hat{i}+\hat{j}+3\hat{k}$.The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\vec{n} = \vec{a} 	imes \vec{b}$.$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ -2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3 - (-2)) + \hat{k}(1 - (-4)) = 5\hat{i} - 5\hat{j} + 5\hat{k}$.The magnitude of this vector is $|\vec{a} 	imes \vec{b}| = \sqrt{5^2 + (-5)^2 + 5^2} = \sqrt{25+25+25} = \sqrt{75} = 5\sqrt{3}$.The unit vector perpendicular to the plane is $\hat{n} = \pm \frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|} = \pm \frac{5\hat{i}-5\hat{j}+5\hat{k}}{5\sqrt{3}} = \pm \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$.Comparing this with the given options,option $D$ represents $\frac{5\hat{i}-5\hat{j}+5\hat{k}}{5\sqrt{3}}$,which simplifies to $\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$.

$A$ unit vector perpendicular to the plane containing the vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-2\hat{i}+\hat{j}+3\hat{k}$ is

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