Mathematics - Vector Question with Solution | TestHub

A vector perpendicular to the plane is
$\begin{array}{rl}& (\hat{i}-2\hat{j}-\hat{k})\times (3\hat{i}-2\hat{j}-\hat{k})\\ & =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& -1\\ 3& -2& -1\end{array}\right|=-2\hat{j}+4\hat{k}\\ & \Rightarrow \text{ unit vector }\hat{a}=\frac{-2\hat{j}+4\hat{k}}{\sqrt{4+16}}=\frac{-2\hat{j}+4\hat{k}}{2\sqrt{5}}\end{array}$
Angle between the unit vector and $\vec{r}=\hat{i}+\hat{j}+\hat{k}$
$={\cos }^{-1}\frac{\vec{r}\cdot \hat{a}}{|\vec{r}|\cdot |\hat{a}|}={\cos }^{-1}\frac{1}{\sqrt{15}}={\tan }^{-1}\sqrt{14}$