Q24 / 69 · JEE Advanced 2011 (Paper 1) · Mathematics

Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ be three vectors. A vector $\mathbf{v}$ in the plane of $\mathbf{a}$ and $\mathbf{b}$ whose projection of $\mathbf{c}$ is $\frac{1}{\sqrt{3}}$, is given by

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

Solution

Let $\mathbf{v}=\mathbf{a}+\lambda \mathbf{b}$ $$ \mathbf{v}=(1+\lambda) \hat{\mathbf{i}}+(1-\lambda) \hat{\mathbf{j}}+(1+\lambda) \hat{\mathbf{k}} $$ Projection of $\mathbf{v}$ on $\mathbf{c}=\frac{1}{\sqrt{3}}$ $$ \begin{array}{lc} \Rightarrow & \frac{\mathbf{v} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}} \\ \Rightarrow & \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ \Rightarrow & 1+\lambda-1+\lambda-1-\lambda=1 \\ \Rightarrow & \lambda-1=1 \Rightarrow \lambda=2 \\ \therefore & \mathbf{v}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{array} $$

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

Relevant Questions