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 angle between $\overrightarrow{a}$ and $\hat{i}-2\hat{j}+2\hat{k}$

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

Answer 1

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

Clearly $\overrightarrow{a}$ is perpendicular to the normals to the two planes determined by the given pairs of vectors
$\overrightarrow{n_1}=$ Normal vector to the plane determined by
$\hat{i}$ and $\hat{i}+\hat{j}=\hat{i}\times (\hat{i}+\hat{j})=\hat{k}$
$\overrightarrow{n_2}=$ Normal vector to the plane determined by
$\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}=(\hat{i}-\hat{j})\times (\hat{i}+\hat{k})=-\hat{i}-\hat{j}+\hat{k}$
since $\overrightarrow{a}$ is $\perp$ to $\overrightarrow{n_1}$ and $\overrightarrow{n_2}$
$\therefore \overrightarrow{a}=\lambda (\overrightarrow{n_1}\times \overrightarrow{n_2})=\lambda \left(\hat{k}(-\hat{i}-\hat{j}+\hat{k})\right)=\lambda (-\hat{j}+\hat{i})$
let $\theta$ be the angle between $\overrightarrow{a}$ and $\hat{i}-2\hat{j}+2\hat{k}$
Then $\cos \theta =\frac{\lambda (1+2+0)}{\lambda \sqrt{2}\sqrt{1+4+4}}=\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4}$