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Question: Answered & Verified by Expert
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
MathematicsVector AlgebraJEE AdvancedJEE Advanced 2006
Options:
  • A
    $\frac{\pi}{2}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{6}$
  • D
    $\frac{3 \pi}{4}$
Solution:
2428 Upvotes Verified Answer
The correct answers are:
$\frac{\pi}{4}$
,
$\frac{3 \pi}{4}$
Let vector $\mathbf{A O}$ be parallel to line of intersection of planes $P_1$ and $P_2$ through, i.e.
$$
\begin{aligned}
& {[(2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times(4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})] \times[(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \times(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})]=54(\hat{\mathbf{j}}-\hat{\mathbf{k}}) .} \\
& \therefore \text { Angle between } 54(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and }(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\
& \Rightarrow \cos \theta=\pm\left(\frac{54+108}{3.54 \cdot \sqrt{2}}\right)=\pm \frac{1}{\sqrt{2}} \\
& \therefore \quad \theta=\frac{\pi}{4}, \frac{3 \pi}{4}
\end{aligned}
$$

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