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 $2\overrightarrow{i}+\overrightarrow{j}-2\hat{k}$ is

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

Answer 1

Answer: (B), (D)

$\frac{3\pi }{4}$

Let vector $\overrightarrow{AO}$ be parallel to line of planes $P_1$ and $P_2$ through origin.
Normal to plane $p_1$ is
$\overrightarrow{n_1}=\left[\right(2\overrightarrow{j}+3\overrightarrow{k})\times (4\hat{j}-3\hat{k}\left)\right]=-18\hat{i}$
Normal to plane $p_2$ is
$\overrightarrow{n_2}=(\hat{j}-\hat{k})\times (3\hat{i}+3\hat{j})=3\hat{i}-3\hat{j}-3\hat{k}So,\overrightarrow{OA}isparallelto\pm (\overrightarrow{n_1}\times \overrightarrow{n_2})=54\hat{j}-54\hat{k}$
$\therefore \text{Angle between}54(\hat{j}-\hat{k})and(2\hat{i}+\hat{j}-2\hat{k})iscos\theta =\pm \left(\frac{54+108}{\mathrm{3.54.}\sqrt{2}}\right)=\pm \frac{1}{\sqrt{2}}\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4}$
Hence, B. and D. are correct answers.