A non-zero $\overline{a}$ is parallel to the line of intersection of the plane determined by the $vectos\hat{i},\hat{i}+\hat{j}$ and the plane determined by the

$\text{ vectors }\hat{i}-\hat{j},\hat{i}+\hat{k}$ . The angle between $\overline{a}$ and $\hat{i}-2\hat{j}+2\hat{k}$ is

 

Clearly $\overrightarrow{a}$ is perpendicular to the normals to the two planes determined by the given pairs of vectors We have,

${\overrightarrow{n}}_1=$ Normal vector to the plane determined by $\hat{i}$ and $\hat{i}+\hat{j}$

$\Rightarrow {\overrightarrow{n}}_1=\hat{i}\times (\hat{i}+\hat{j})=\hat{k}$

${\overrightarrow{n}}_2=$ = Normal vector to the plane 'determined by $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$

$\Rightarrow {\overrightarrow{n}}_2=(\hat{i}-\hat{j})\times (\hat{i}+\hat{k})=-\hat{i}-\hat{j}+\hat{k}$

Since $\overrightarrow{a}$ is perpendicular to ${\overrightarrow{n}}_1$ and ${\overrightarrow{n}}_2$.

$\therefore \overrightarrow{a}=\lambda \left({\overrightarrow{n}}_1\times {\overrightarrow{n}}_2\right)=\lambda \{\hat{k}\times (-\hat{i}-\hat{j}+\hat{k}\left)\right\}=\lambda (-\hat{j}+\hat{i})$

Let 0 be the angle between $\overrightarrow{a}$ and $\hat{i}-2\hat{j}+2\hat{k}$ then

$\cos \theta =\frac{\lambda (1+2+0)}{\lambda \sqrt{2}\sqrt{1+4+4}}=\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4}$