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Question:
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If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are the points with position vectors $\hat{i}+\hat{j}-\hat{k}, 2 \hat{i}-\hat{j}+3 \hat{k}, 2 \hat{i}-3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ respectively, then find the projection of $\overline{\mathrm{AB}}$ along $\overline{\mathrm{CD}}$.
Solution:
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Verified Answer
$\overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+2 \hat{\mathrm{k}}$
$$
\overrightarrow{C D}=\hat{i}-2 \hat{j}+4 \hat{k}
$$
So, the projection of $\overrightarrow{\mathrm{AB}}$ along
$$
\begin{aligned}
&\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AB}} \cdot \frac{\overrightarrow{\mathrm{CD}}}{|\overrightarrow{\mathrm{CD}}|}=\frac{(\hat{\mathrm{i}}+2 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})}{\sqrt{1+4+16}} \\
&=\frac{1+8}{\sqrt{21}}=\frac{9}{\sqrt{3} \times \sqrt{7}}=\frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{7}}=3 \sqrt{\frac{3}{7}}
\end{aligned}
$$
$$
\overrightarrow{C D}=\hat{i}-2 \hat{j}+4 \hat{k}
$$
So, the projection of $\overrightarrow{\mathrm{AB}}$ along
$$
\begin{aligned}
&\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AB}} \cdot \frac{\overrightarrow{\mathrm{CD}}}{|\overrightarrow{\mathrm{CD}}|}=\frac{(\hat{\mathrm{i}}+2 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})}{\sqrt{1+4+16}} \\
&=\frac{1+8}{\sqrt{21}}=\frac{9}{\sqrt{3} \times \sqrt{7}}=\frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{7}}=3 \sqrt{\frac{3}{7}}
\end{aligned}
$$
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