Question

Let $A$ be a vector parallel to line of intersection of planes $P_1:x=0$ and $P_2:x-y-z=0$ through origin. The acute angle between $A$ and $2\hat{i}+\hat{j}-2\hat{k}$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{5\pi }{12}$

Answer 1

The correct option is B $\frac{\pi }{4}$

Given planes are $P_1:x=0$ and $P_2:x-y-z=0$
Normal vector of the planes are
$\overrightarrow{n_1}=\hat{i}\overrightarrow{n_2}=\hat{i}-\hat{j}-\hat{k}$
The vector along the line of intersection is
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& 0\\ 1& -1& -1\end{array}\right|\Rightarrow \overrightarrow{n}=\hat{j}-\hat{k}$
Therefore,
$\overrightarrow{A}=a(\hat{j}-\hat{k})$
Where $a$ is a constant
Now, angle between $\overrightarrow{A}$ and $\overrightarrow{B}=2\hat{i}+\hat{j}-2\hat{k}$ is
$\cos \theta =\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}\Rightarrow \cos \theta =\frac{a(1+2)}{a\sqrt{2}\times 3}=\frac{1}{\sqrt{2}}\therefore \theta =\frac{\pi }{4}$