Question

Let $\overrightarrow{A}$ be a vector parallel to the line of intersection of the planes $P_1$ and $P_2$ . The plane $P_1$ is parallel to the vectors $2\hat{j}+3\hat{k}$ and $4\hat{j}-3\hat{k}$ while plane $P_2$ is parallel to the vectors $\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}$ . The acute angle between $\overrightarrow{A}$ 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

Answer: (B)

Let, $\overrightarrow{a}=2\hat{j}+3\hat{k}$
$\overrightarrow{b}=4\hat{j}-3\hat{k}$
$\overrightarrow{c}=\hat{j}-\hat{k}$
$\overrightarrow{d}=\hat{i}+\hat{j}$
$\overrightarrow{e}=2\hat{i}+\hat{j}-2\hat{k}$
A normal vector perpendicular to the plane $P_1$ is $\overrightarrow{a}\times \overrightarrow{b}$
A normal vector perpendicular to the plane $P_2$ is $\overrightarrow{c}\times \overrightarrow{d}$
A vector parallel to the line of intersection is
$\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)=\left(\hat{k}-\hat{j}\right)$
Let, $\theta$ be the acute angle between them then,
$cos\theta =\left|\frac{\overrightarrow{A}\cdot \overrightarrow{e}}{\left|\overrightarrow{A}\right||\overrightarrow{e}|}\right|=\left|\frac{\left(-3\right)}{\sqrt{2}\times 3}\right|=\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =\frac{\pi }{4}$