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Let $\mathbf{a}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ be two vectors. Then the projection vector of $\mathbf{b}$ on a vector perpendicular to $\mathbf{a}$ is
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Verified Answer
The correct answer is:
$\frac{31}{9} \hat{i}-\frac{20}{9} \hat{j}-\frac{41}{9} \hat{k}$
Given, $\mathbf{a}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$
The projection of vector $\mathbf{b}$ on a vector
$\begin{aligned}
\text { perpendicular to } \mathbf{a}=\mathbf{b}-\left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2}\right) \mathbf{a} \\
=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})-\left(\frac{-2}{9}\right)(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})+\left(\frac{4}{9} \hat{\mathbf{i}}-\frac{2}{9} \hat{\mathbf{j}}+\frac{4}{9} \hat{\mathbf{k}}\right) \\
=\frac{31}{9} \hat{\mathbf{i}}-\frac{20}{9} \hat{\mathbf{j}}-\frac{41}{9} \hat{\mathbf{k}}
\end{aligned}$
The projection of vector $\mathbf{b}$ on a vector
$\begin{aligned}
\text { perpendicular to } \mathbf{a}=\mathbf{b}-\left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2}\right) \mathbf{a} \\
=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})-\left(\frac{-2}{9}\right)(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})+\left(\frac{4}{9} \hat{\mathbf{i}}-\frac{2}{9} \hat{\mathbf{j}}+\frac{4}{9} \hat{\mathbf{k}}\right) \\
=\frac{31}{9} \hat{\mathbf{i}}-\frac{20}{9} \hat{\mathbf{j}}-\frac{41}{9} \hat{\mathbf{k}}
\end{aligned}$
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