Question

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

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

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{\pi }{4}$
D $\frac{3\pi }{4}$

$P_1=ax+by+cz=0$

$P_1||2\hat{j}+3\hat{k},4\hat{j}-3\hat{k}$

So normal vector $a\hat{i}+b\hat{j}+c\hat{k}$

$2b+3c=0$

$4b-3c=0$

Hence $b=c=0$

Similarly $3a+3b=0$

$a=-b=-c$

$P_2=-x+y+z$

Intersection of $P_1,P_2$ gives $y+2=0$

Vector $(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})or(0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

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

$\cos \theta =\frac{\overrightarrow{A}\overrightarrow{B}}{|A||B|}=\frac{1+2}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\theta =45°=\frac{\pi }{4}$

Or $\cos \theta =\frac{\overrightarrow{A}\overrightarrow{B}}{|A||B|}=\frac{-3}{2\sqrt{3}}$

$\theta =135°=\frac{3\pi }{4}$

Hence B and D