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A vector a makes equal acute angles on the coordinate axis. Then the projection of vector $\mathbf{b}=5 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ on $\mathbf{a}$ is
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Verified Answer
The correct answer is:
$\frac{11}{\sqrt{3}}$
Let the equal angles of $\mathbf{a}$ with the coordinate axis be $\alpha$.
$\begin{aligned}
&l=\cos \alpha, m=\cos \alpha, n=\cos \alpha \\
&\because \quad l^{2}+m^{2}+n^{2} \equiv 1
\end{aligned}$
$\begin{aligned}
&\Rightarrow \quad 3 \cos ^{2} \alpha=1 \\
&\Rightarrow \quad \cos \alpha=\frac{1}{\sqrt{3}}
\end{aligned}$
Direction ratios are $1,1,1$.
$\therefore \quad \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Projection of $\mathbf{b}$ on $\mathbf{a}=\frac{\mathbf{b} \cdot \mathbf{a}}{|\mathbf{a}|}$
$\begin{aligned}
&=\frac{(5 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{1^{2}+1^{2}+1^{2}}} \\
&=\frac{5+7-1}{\sqrt{3}} \\
&=\frac{11}{\sqrt{3}}
\end{aligned}$
$\begin{aligned}
&l=\cos \alpha, m=\cos \alpha, n=\cos \alpha \\
&\because \quad l^{2}+m^{2}+n^{2} \equiv 1
\end{aligned}$
$\begin{aligned}
&\Rightarrow \quad 3 \cos ^{2} \alpha=1 \\
&\Rightarrow \quad \cos \alpha=\frac{1}{\sqrt{3}}
\end{aligned}$
Direction ratios are $1,1,1$.
$\therefore \quad \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Projection of $\mathbf{b}$ on $\mathbf{a}=\frac{\mathbf{b} \cdot \mathbf{a}}{|\mathbf{a}|}$
$\begin{aligned}
&=\frac{(5 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{1^{2}+1^{2}+1^{2}}} \\
&=\frac{5+7-1}{\sqrt{3}} \\
&=\frac{11}{\sqrt{3}}
\end{aligned}$
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