Q117 / 180 · KCET 2021 · Mathematics

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
    $\frac{11}{15}$
  • Correct answer
    B
    $\frac{11}{\sqrt{3}}$
  • C
    $\frac{4}{5}$
  • D
    $\frac{3}{5 \sqrt{3}}$

Solution

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 $\mathbf{a}=\frac{\mathbf{b} \cdot \mathbf{a}}{|\mathbf{a}|}$ $\begin{aligned} &=\frac{(5 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{1^{2}+1^{2}+1^{2}}} \\ &=\frac{5+7-1}{\sqrt{3}} \\ &=\frac{11}{\sqrt{3}} \end{aligned}$

Hence, the correct answer is option: 'B'.

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