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Offline Centres

Find the unit vector perpendicular to the plane determined by the points $P (1, - 1, 2), Q (2, 0, - 1)$
and $R (0, 2, 1)$ .

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$\pm (\frac{2 \hat{i} + 2 \hat{j} + \hat{k}}{3})$

$\pm (\frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}})$

$\pm (\frac{\hat{i} + 2 \hat{j} + \hat{k}}{\sqrt{6}})$

$\pm (\frac{\hat{i} + \hat{j} + 2 \hat{k}}{\sqrt{6}})$

answer is B.

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Detailed Solution

$\vec{P Q} = \hat{i} + \hat{j} - 3 \hat{k}, \vec{P R} = - \hat{i} + 3 \hat{j} - \hat{k}$

$\begin{aligned} \vec{P Q} \times \vec{P R} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & - 3 \\ - 1 & 3 & - 1 \end{array}| = 8 \hat{i} + 4 \hat{j} + 4 \hat{k} \end{aligned}$

$∴$ Unit vector $= \pm \frac{(2 \hat{i} + \hat{j} + \hat{k})}{\sqrt{6}}$

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