Question

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

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

Answer 1

Answer: (B), (D)

Let vector $\overrightarrow{AO}$ be parallel to line of intersection of planes $P_1$ and $P_2$ through origin.
Normal to plane $p_1$ is
$\overrightarrow{n_1}=[(2\hat{j}+3\hat{k})\times (4\hat{j}-3\hat{k})]=-18\hat{i}$
Normal to plane $p_2$ is
${\overrightarrow{n}}_2=(\hat{j}-\hat{k})\times (3\hat{i}+3\hat{j})=3\hat{i}-3\hat{j}-3\hat{k}$
So, $\overrightarrow{OA}$ is parallel to $\pm \left(\overrightarrow{n_1}\times \overrightarrow{n_2}\right)=54\hat{j}-54\hat{k}$.
$\therefore$ Angle between $54(\hat{j}-\hat{k})$ and $(2\hat{i}+\hat{j}-2\hat{k})$ is
$\cos \theta =\pm \left(\frac{54+108}{3\cdot 54\cdot \sqrt{2}}\right)=\pm \frac{1}{\sqrt{2}}$
$\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4}$
Hence, (b) and (d) are correct answers.