Question

The unit vector perpendicular to the plane determined by $P(1,-1,2), Q(2,0,-1)$ and $R(0,2,1)$ is $\cdots$.

Answer 1

A unit vector perpendicular to the plane determined by $P, Q, R$
$=\pm \frac{(\overrightarrow{ PQ }) \times(\overrightarrow{ P R })}{|\overrightarrow{ PQ } \times \overrightarrow{ PR }|} $
$\therefore $ Unit vector $=\pm \frac{(\overrightarrow{ P Q }) \times(\overrightarrow{ P R })}{|\overrightarrow{ P Q } \times \overrightarrow{ PR }|}$
where, $ \overrightarrow{ P Q }=\hat{ i }+\hat{ j }-3 \hat{ k }$ and $\overrightarrow{ P R }=\hat{ i }-3 \hat{ j }-\hat{ k }$
$ \therefore \overrightarrow{ P Q } \times \overrightarrow{ P R } =\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 1 & -3 \\ -1 & 3 & -1\end{vmatrix} $
$=\hat{ i }(-1+9)-\hat{ j }(-1-3)+\hat{ k }(3+1) $
$=8 i +4 j +4 \hat{ k } $
$\Rightarrow |\overrightarrow{ P Q } \times \overrightarrow{ P R }| =4 \sqrt{4+1+1}=4 \sqrt{6}=\pm \frac{4(2 \hat{ i }+\hat{ j }+\hat{ k })}{4 \sqrt{6}} $
$=\pm \frac{(2 \hat{ i }+\hat{ j }+\hat{ k })}{\sqrt{6}}$