A non-zero vector $\vec{a}$ is parallel yo the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i} + \hat{j}$ and the plane determined by the vectors $\hat{i} - \hat{j}, \hat{i} + \hat{j}$. The angle between $\vec{a}$ and $\hat{i} -2 \hat{j}+ 2\hat{k}$, is
Answer & explanation
Correct answer: option 2
Clearly, $\vec{a}$ is perpendicular to the normals to the two planes determined by the given pairs of vectors.
We have,
$\vec{n_1}$ = Normal vector to the plane determined by $\hat{i}$ and $\hat{i}+\hat{j}$
$⇒\vec{n_1} = \hat{i} × (\hat{i} + \hat{j})= \hat{k}$
$\vec{n_2}$ = Normal vector to the plane determined by $\hat{i} - \hat{j}$ and $\hat{i}+\hat{k}$
$⇒\vec{n_2}= (\hat{i} - \hat{j})×(\hat{i} + \hat{k})= -\hat{i} - \hat{j}+\hat{k}$
Since $\vec{a}$ is perpendicular to $\vec{n_1}$ and $\vec{n_2}$. Therefore,
$\vec{a}= λ(\vec{n_1}×\vec{n_2})=λ \begin{Bmatrix} \hat{k}× (-\hat{i}-\hat{j}+\hat{k})\end{Bmatrix}=λ(-\hat{j}+\hat{i})$
Let θ be the angle between $\vec{a}$ and $\vec{i} - 2\hat{j}+2\hat{k}$. Then,
$ cos θ =\frac{λ(1+2+0)}{λ\sqrt{2}\sqrt{1+4+4}}=\frac{1}{\sqrt{2}}⇒ θ = \frac{\pi}{4}$