Question

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

• A. $\frac{i+2j+k}{\sqrt{6}}$
• B. $\frac{i-2j+2k}{\sqrt{6}}$
• C. $\frac{2i+j+k}{\sqrt{6}}$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{i+2j+k}{\sqrt{6}}$

Here$\overrightarrow{PQ}=(2-1)\hat{i}+(0-1-1)\hat{j}+(-1-2)\hat{k}$

$\therefore PQ=\hat{i}+\hat{j}-3\hat{k}$

$\therefore \overrightarrow{PR}=(0-1)\hat{i}+(2-1-1)\hat{j}+(1-2)\hat{k}$

$\therefore PR=\widehat{-i}+3\hat{j}-\hat{k}$

$\therefore \overrightarrow{PQ}\times \overrightarrow{PR}=\hat{i}+\hat{j}-3\hat{k}\times$

$\widehat{-i}+3\hat{j}-\hat{k}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -3\\ -1& 3& -1\end{array}\right|$

$=\hat{i}(-1+9)-(-1-3)+\hat{k}(3+1)$

$=8\hat{i}+4\hat{j}+4\hat{k}$

$\therefore Unitvector=\frac{8\hat{i}+4\hat{j}+4\hat{k}}{\sqrt{8^2+4^2+4^2}}$

$=\frac{\hat{i}+2\hat{j}+\hat{k}}{\sqrt{6}}$