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The vector projection of $\overline{P Q}$ on $\overline{A B}$, where $P \equiv(-2,1,3), Q \equiv(3,2,5), A \equiv(4,-3,5)$ and $B \equiv(7,-5,-1)$ is
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Verified Answer
The correct answer is:
$\frac{1}{49}(3 \hat{i}-2 \hat{j}-6 \hat{k})$
$\begin{aligned}
& \overline{P Q}=(3+2) \hat{i}+(2-1) \hat{j}+(5-3) \hat{k}=5 \hat{i}+\hat{j}+2 \widehat{k} \\
& \overline{A B}=(7-4) \hat{i}+(-5+3) \hat{j}+(-1-5) \hat{k}=3 \hat{i}-2 \hat{j}-6 \hat{k}
\end{aligned}$
now vector projection of $\overrightarrow{P Q}$ on $\overrightarrow{A B}$ is $\frac{(\overrightarrow{P Q} \cdot \overrightarrow{A B}) \overrightarrow{A B}}{|\overrightarrow{A B}|^2}$
$=\frac{\{5 \times 3+1 \times(-1)+2 \times(-6)\}(3 \hat{i}-2 \hat{j}-6 \hat{k})}{3^2+(-2)^2+(-6)^2}=\frac{1}{49}(3 \hat{i}-2 \hat{j}-6 \hat{k})$
& \overline{P Q}=(3+2) \hat{i}+(2-1) \hat{j}+(5-3) \hat{k}=5 \hat{i}+\hat{j}+2 \widehat{k} \\
& \overline{A B}=(7-4) \hat{i}+(-5+3) \hat{j}+(-1-5) \hat{k}=3 \hat{i}-2 \hat{j}-6 \hat{k}
\end{aligned}$
now vector projection of $\overrightarrow{P Q}$ on $\overrightarrow{A B}$ is $\frac{(\overrightarrow{P Q} \cdot \overrightarrow{A B}) \overrightarrow{A B}}{|\overrightarrow{A B}|^2}$
$=\frac{\{5 \times 3+1 \times(-1)+2 \times(-6)\}(3 \hat{i}-2 \hat{j}-6 \hat{k})}{3^2+(-2)^2+(-6)^2}=\frac{1}{49}(3 \hat{i}-2 \hat{j}-6 \hat{k})$
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