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The projection of $\overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ on $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
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Verified Answer
The correct answer is:
$\frac{8}{\sqrt{14}}$
Given, $\overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
The projection of $\overrightarrow{\mathbf{a}}$ on $\overrightarrow{\mathbf{b}}=\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|}$
$=\frac{(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{2^{2}+3^{2}+1^{2}}}$
$$
=\frac{6-3+5}{\sqrt{14}}=\frac{8}{\sqrt{14}}
$$
The projection of $\overrightarrow{\mathbf{a}}$ on $\overrightarrow{\mathbf{b}}=\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|}$
$=\frac{(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{2^{2}+3^{2}+1^{2}}}$
$$
=\frac{6-3+5}{\sqrt{14}}=\frac{8}{\sqrt{14}}
$$
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