Find unit vector perpendicular to the plane of a=2i^-2j^+k^ and b=i^+2j^+2k^.

a×b=|a||b| n^ ∴n^=a×b|a|.|b| =|i^j^k^2−21122|(4+4+11+4+4) =i^(−4−2)−j^(4−1)+k^(4+2)9 =−69i^−39j^+69k^=−23i^−13j^+23k^

Find unit vector perpendicular to the plane of a=2i^-2j^+k^ and b=i^+2j^+2k^.

a×b=|a||b| n^ ∴n^=a×b|a|.|b| =|i^j^k^2−21122|(4+4+11+4+4) =i^(−4−2)−j^(4−1)+k^(4+2)9 =−69i^−39j^+69k^=−23i^−13j^+23k^

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Find unit vector perpendicular to the plane of $a = 2 \hat{i} - 2 \hat{j} + \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 2 \hat{k} .$

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$\frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} - \frac{1}{3} \hat{k}$

$\frac{2}{3} \hat{i} + \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k}$

$- \frac{2}{3} \hat{i} - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k}$

$- \frac{2}{3} \hat{i} - \frac{1}{3} \hat{j} - \frac{2}{3} \hat{k}$

answer is C.

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Detailed Solution

$a \times b = | a | | b | \hat{n}$
$∴ \hat{n} = \frac{a \times b}{| a | . | b |} = \frac{| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 2 & 1 \\ 1 & 2 & 2 \end{matrix} |}{(\sqrt{4 + 4 + 1} \sqrt{1 + 4 + 4})}$
$= \frac{\hat{i} (- 4 - 2) - \hat{j} (4 - 1) + \hat{k} (4 + 2)}{9}$
$= - \frac{6}{9} \hat{i} - \frac{3}{9} \hat{j} + \frac{6}{9} \hat{k} = - \frac{2}{3} \hat{i} - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k}$