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Let $\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}, \vec{b}=6 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{c}=3 \hat{i}-4 \hat{j}-12 \hat{k}$ be three vectors. If $\vec{p}$ is the projection of $\vec{b}$ on $\vec{a}$ and $\vec{q}$ is the projection of $\overrightarrow{\mathrm{c}}$ on $\overrightarrow{\mathrm{a}}$, then $13 \overrightarrow{\mathrm{p}}=$
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Verified Answer
The correct answer is:
$4 \vec{q}$
Projection of $\vec{b}$ on $\vec{a}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}=\frac{-4}{3} \vec{a}$
$\vec{p}=\frac{-4}{3}(\hat{i}-2 \hat{j}+2 \hat{k})$
Projection of $\vec{c}$ on $\vec{a}=\left(\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$
$\begin{aligned}
& \vec{q}=\frac{-13}{3} \vec{a} \\
\therefore & 13 \vec{p}=4\left(\frac{-13}{3}\right) \vec{a}=4 \vec{q} .
\end{aligned}$
$\vec{p}=\frac{-4}{3}(\hat{i}-2 \hat{j}+2 \hat{k})$
Projection of $\vec{c}$ on $\vec{a}=\left(\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$
$\begin{aligned}
& \vec{q}=\frac{-13}{3} \vec{a} \\
\therefore & 13 \vec{p}=4\left(\frac{-13}{3}\right) \vec{a}=4 \vec{q} .
\end{aligned}$
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