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 $P,Q,R$
$=\pm \frac{(\overrightarrow{PQ})\times (\overrightarrow{PR})}{|\overrightarrow{PQ}\times \overrightarrow{PR}|}$
$\therefore$ Unit vector $=\pm \frac{(\overrightarrow{PQ})\times (\overrightarrow{PR})}{|\overrightarrow{PQ}\times \overrightarrow{PR}|}$
where, $\overrightarrow{PQ}=\hat{i}+\hat{j}-3\hat{k}$ and $\overrightarrow{PR}=\hat{i}-3\hat{j}-\hat{k}$
$\therefore \overrightarrow{PQ}\times \overrightarrow{PR}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -3\\ -1& 3& -1\end{array}\right|$
$=\hat{i}(-1+9)-\hat{j}(-1-3)+\hat{k}(3+1)$
$=8i+4j+4\hat{k}$
$\Rightarrow |\overrightarrow{PQ}\times \overrightarrow{PR}|=4\sqrt{4+1+1}=4\sqrt{6}=\pm \frac{4(2\hat{i}+\hat{j}+\hat{k})}{4\sqrt{6}}$
$=\pm \frac{(2\hat{i}+\hat{j}+\hat{k})}{\sqrt{6}}$