Let vec{a}=hat{i}-2 hat{j}+2 hat{k}, vec{b}=6 | vedclass

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(A) The projection of a vector $\vec{v}$ on $\vec{a}$ is given by $\left(\frac{\vec{v} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$.First, calculate $|

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$4$

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(A) The projection of a vector $\vec{v}$ on $\vec{a}$ is given by $\left(\frac{\vec{v} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$.First, calculate $|\vec{a}|^2 = (1)^2 + (-2)^2 + (2)^2 = 1 + 4 + 4 = 9$.For $\vec{p}$ (projection of $\vec{b}$ on $\vec{a}$):$\vec{p} = \left(\frac{(6)(1) + (2)(-2) + (-3)(2)}{9}\right) \vec{a} = \left(\frac{6 - 4 - 6}{9}\right) \vec{a} = \frac{-4}{9} \vec{a}$.For $\vec{q}$ (projection of $\vec{c}$ on $\vec{a}$):$\vec{q} = \left(\frac{(3)(1) + (-4)(-2) + (-12)(2)}{9}\right) \vec{a} = \left(\frac{3 + 8 - 24}{9}\right) \vec{a} = \frac{-13}{9} \vec{a}$.Now, we need to find $13 \vec{p}$:$13 \vec{p} = 13 \left(\frac{-4}{9} \vec{a}\right) = \frac{-52}{9} \vec{a}$.Comparing this with $\vec{q} = \frac{-13}{9} \vec{a}$, we see that $4 \vec{q} = 4 \left(\frac{-13}{9} \vec{a}\right) = \frac{-52}{9} \vec{a}$.Therefore, $13 \vec{p} = 4 \vec{q}$.

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 $\vec{c}$ on $\vec{a}$, then $13 \vec{p}=$ (in $\vec{q}$)

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