Q77 / 160 · TS EAMCET 2007 · Mathematics

If $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=+\lambda \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and the orthogonal projection of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}}$ is $\frac{4}{3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$, then $\lambda$ is equal to

  • A
    $0$
  • Correct answer
    B
    $2$
  • C
    $12$
  • D
    $-1$

Solution

Given orthogonal projection of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}}$ is $\begin{array}{r}\frac{(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^2}=\frac{4}{3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ \Rightarrow \frac{\{(\lambda \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})\}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})}{(1+1+1)}\end{array}$ $=\frac{4}{3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$ $\begin{aligned} & \Rightarrow(\lambda+3-1)(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})=4(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & \Rightarrow \quad(\lambda+2)(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})=4(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\end{aligned}$ On equating the coefficient of $\hat{\mathbf{i}}$, we get $\begin{aligned} \lambda+2 & =4 \\ \lambda & =2\end{aligned}$

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

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