Question

A non-zero vector $\overrightarrow{a}$ is parallel to the line of intersection of the plane $P_1$ determined by $\hat{i}+\hat{j}$ and $\hat{i}-2\hat{j}$ and plane $P_2$ determined by vector $2\hat{i}+\hat{j}and3\hat{i}+2\hat{k}$, then angle between $\overrightarrow{a}$ and vector $\hat{i}-2\hat{j}+2\hat{k}$ is

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

Answer 1

The correct option is B $\frac{\pi }{2}$

The plane equations in vector form will be given by the direction of normal that is obtained by the cross product of vectors forming the plane.
For Plane $P_1$:
${\overrightarrow{n}}_1=(\hat{i}+\hat{j})\times (\hat{i}-2\hat{j})=-3\hat{k}$

For Plane $P_2$:
${\overrightarrow{n}}_2=(2\hat{i}+\hat{j})\times (3\hat{i}+2\hat{k})=2\hat{i}-4\hat{j}-3\hat{k}$

Now the cross product of these two normal vectors will give us a vector along the intersection of the planes.
${\overrightarrow{n}}_1\times {\overrightarrow{n}}_2=-3\hat{k}\times (2\hat{i}-4\hat{j}-3\hat{k})=-12\hat{i}-6\hat{j}$

$\overrightarrow{a}$ is parallel to the above vector.
Therefore angle between $({\overrightarrow{n}}_1\times {\overrightarrow{n}}_2)$ and $(\hat{i}-2\hat{j}+2\hat{k})$ will give us the angle betwee $\overrightarrow{a}$ and $(\hat{i}-2\hat{j}+2\hat{k})$.

Angle between $({\overrightarrow{n}}_1\times {\overrightarrow{n}}_2)$ and $(\hat{i}-2\hat{j}+2\hat{k})$ can be found out by:
$(-12\hat{i}-6\hat{j})\cdot (\hat{i}-2\hat{j}+2\hat{k})=0$

Angle between the vectors is therefore $\frac{\pi }{2}$.