Let $\vec{a} = \bar{i} - 2 \vec{j} + 2 \vec{k}, \vec{b} = 6 \bar{i} + 2 \vec{j} - 3 \vec{k}$ and $\vec{c} = 3 \bar{i} - 4 \vec{j} - 12 \vec{k}$ be three vectors. If $\vec{P}$ is the projection vector of $\vec{b}$ on $\vec{a}$ and $\vec{q}$ is the projection vector of $\vec{c}$ on $\vec{a}$ then $13 \vec{p} =$

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$6 \vec{q}$

$7 \vec{q}$

$4 \vec{q}$

$5 \vec{q}$

answer is A.

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Detailed Solution

$\vec{p} = \frac{(\vec{a} . \vec{b}) \vec{a}}{{| \vec{a} |}^{2}} = \frac{- 4}{3} \vec{a}$
$\begin{array}{l} \vec{q} = \frac{(\vec{c} \cdot \vec{a}) \vec{a}}{| \vec{a} |^{2}} = \frac{- 13}{3} \vec{a} \\ 13 \vec{q} = 13 (\frac{- 4}{3} \vec{a}) \\ 13 \vec{p} = 4 (- \frac{13}{3} \vec{a}) \\ 13 \vec{p} = 4 \vec{q} \end{array}$

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