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Let $\vec{a}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-\hat{j}-2 \hat{k}$ be two vectors. If the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is $\vec{x}$ and orthogonal projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{y}$ then $|\vec{x}-\vec{y}|=$
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Verified Answer
The correct answer is:
$\frac{4}{9} \sqrt{10}$
Orthogonal projection of $\vec{a}$ on $\vec{b}=\frac{(\vec{a} \cdot \vec{b})}{|\vec{b}|^2} \vec{b}$
$\therefore \vec{x}=\frac{4}{9} \vec{b}$
Orthogonal projection of $\vec{b}$ on $\vec{a}=\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2} \vec{a}$
$\begin{aligned}
& \vec{y}=\frac{4}{9} \vec{a} \\
& |\vec{x}-\vec{y}|=\frac{4}{9}|\vec{b}-\vec{a}| \\
& =\frac{4}{9}|\hat{i}-3 \hat{j}|=\frac{4}{9} \sqrt{10}
\end{aligned}$
$\therefore \vec{x}=\frac{4}{9} \vec{b}$
Orthogonal projection of $\vec{b}$ on $\vec{a}=\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2} \vec{a}$
$\begin{aligned}
& \vec{y}=\frac{4}{9} \vec{a} \\
& |\vec{x}-\vec{y}|=\frac{4}{9}|\vec{b}-\vec{a}| \\
& =\frac{4}{9}|\hat{i}-3 \hat{j}|=\frac{4}{9} \sqrt{10}
\end{aligned}$
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