Q79 / 150 · MHT CET 2007 · Mathematics

Let $\quad \overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \quad$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. A vector in the plane of $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ whose projection on $\overrightarrow{\mathbf{a}}$ is of magnitude $\sqrt{\frac{2}{3}}$, is

  • Correct answer
    A
    $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
  • B
    $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
  • C
    $2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
  • D
    $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}$

Solution

Any vector in the plane of $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ is $\overrightarrow{\mathbf{r}}=m \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}$ $$ =(m+1) \hat{\mathbf{i}}+(2 m+1) \hat{\mathbf{j}}+(-m-2) \hat{\mathbf{k}} $$ Projection of $\overrightarrow{\mathbf{r}}$ on $\overrightarrow{\mathbf{a}}=\frac{\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|}=\left|\sqrt{\frac{2}{3}}\right|$ $$ \begin{array}{l} \therefore \frac{2(m+1)-(2 m+1)+(-m-2)}{\sqrt{6}}=\pm \sqrt{\frac{2}{3}} \\ \Rightarrow-m-1=\pm 2 \end{array} $$ $\Rightarrow$ $m=-3$ and 1 Hence, $\overrightarrow{\mathbf{r}}=-2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$

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

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