Question

A non-zero vector $\overrightarrow{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i},\hat{i},\hat{j}$ and the plane determined by the vectors $\hat{i}-\hat{j},\hat{i}-\hat{k}$. The possible angle between $\overrightarrow{a}$ and $\hat{i}-2\hat{j}+2\hat{k}$ is

• A. $\frac{\pi }{3}$
• 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 $\frac{3\pi }{4}$

normal vector $n_1$ perpendicular to plane determining $\overrightarrow{i},\overrightarrow{j}+\overrightarrow{k}$ is
$n_1=i\times (\overrightarrow{i}-\overrightarrow{j})\times (\overrightarrow{i}-\overrightarrow{k})=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$
Now vector parallel to intersection,
$=\overrightarrow{n}\times \overrightarrow{n_1}=\overrightarrow{k}\times (\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})=-(\overrightarrow{j}-\overrightarrow{i}).$
and angle beteen $\overrightarrow{i}-\overrightarrow{j}$ and $\overrightarrow{i}-3\overrightarrow{j}+2\overrightarrow{k}$
$\Rightarrow \cos \theta =\left|\frac{1+2+2.0}{\sqrt{1+1}\sqrt{1+4+4}}\right|=\left|\frac{3}{\sqrt{2}.3}\right|=\left|\frac{1}{\sqrt{2}}\right|$
$\Rightarrow \theta =\frac{\pi }{4},\frac{3\pi }{4}$