KCET · Maths · Vector Algebra
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
- A \(\frac{11}{15}\)
- B \(\frac{11}{\sqrt{3}}\)
- C \(\frac{4}{5}\)
- D \(\frac{3}{5 \sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{11}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
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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