Q60 / 120 · JEE Advanced 2006 · Mathematics

Let $\mathbf{A}$ be vector parallel to line of intersection of planes $P_1$ and $P_2$ through origin. $P_1$ is parallel to the vectors $2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $P_2$ is parallel to $\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$, then the angle between vector $\mathbf{A}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ is

  • A
    $\frac{\pi}{2}$
  • Correct answer
    B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{6}$
  • Correct answer
    D
    $\frac{3 \pi}{4}$

Solution

Let vector $\mathbf{A O}$ be parallel to line of intersection of planes $P_1$ and $P_2$ through, i.e. $$ \begin{aligned} & {[(2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times(4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})] \times[(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \times(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})]=54(\hat{\mathbf{j}}-\hat{\mathbf{k}}) .} \\ & \therefore \text { Angle between } 54(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and }(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & \Rightarrow \cos \theta=\pm\left(\frac{54+108}{3.54 \cdot \sqrt{2}}\right)=\pm \frac{1}{\sqrt{2}} \\ & \therefore \quad \theta=\frac{\pi}{4}, \frac{3 \pi}{4} \end{aligned} $$

Hence, the correct answer is options: 'B', 'D'.

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