If e is the unit vector perpendicular to the plane determined by the points 2i^+j^+k^,i^−j^+k^ and −i^+j^−k^ . If a=2i^−3j^+6k^ , then the projection vector of a on e is

Let e=2i+j+k−i−j+k×2i+j+k−−i+j−k =i+2j+0k×3i+oj+2k=4i−2j−6k ∴ e^=4i−2j−6k214=1142i−j−3k ∴Projection of a¯ on e =a¯⋅e¯=−1114 ∴ projection of vector of a¯ on e =a¯⋅e¯e¯=1114−2i+j+3k

If e is the unit vector perpendicular to the plane determined by the points 2i^+j^+k^,i^−j^+k^ and −i^+j^−k^ . If a=2i^−3j^+6k^ , then the projection vector of a on e is

Let e=2i+j+k−i−j+k×2i+j+k−−i+j−k =i+2j+0k×3i+oj+2k=4i−2j−6k ∴ e^=4i−2j−6k214=1142i−j−3k ∴Projection of a¯ on e =a¯⋅e¯=−1114 ∴ projection of vector of a¯ on e =a¯⋅e¯e¯=1114−2i+j+3k

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If $e$ is the unit vector perpendicular to the plane determined by the points $2 \hat{i} + \hat{j} + \hat{k}, \hat{i} - \hat{j} + \hat{k}$ and $- \hat{i} + \hat{j} - \hat{k}$ . If $a = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ , then the projection vector of $a$ on $e$ is

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$\frac{11}{14} (- 2 \hat{i} + \hat{j} + 3 \hat{k})$

$\frac{1}{\sqrt{14}} (2 \hat{i} - \hat{j} + 3 \hat{k})$

$\frac{1}{7} (2 \hat{i} - 3 \hat{j} + 6 \hat{k})$

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

answer is A.

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Detailed Solution

Let $e = [(2 i + j + k) - (i - j + k)] \times [(2 i + j + k) - (- i + j - k)]$
$= (i + 2 j + 0 k) \times (3 i + o j + 2 k) = 4 i - 2 j - 6 k ∴ \hat{e} = \frac{4 i - 2 j - 6 k}{2 \sqrt{14}} = \frac{1}{\sqrt{14}} (2 i - j - 3 k)$

$∴$ Projection of $\bar{a}$ on $e$ $= \bar{a} \cdot \bar{e} = \frac{- 11}{\sqrt{14}}$
$∴$ projection of vector of $\bar{a}$ on $e$ $= (\bar{a} \cdot \bar{e}) \bar{e} = \frac{11}{14} (- 2 i + j + 3 k)$