Question

A unit vector perpendicular to the plane $ABC$, where $A,B$ and $C$ are respectively the points $(3,-1,2),(1,-1,-3)$ and $(4,-3,1)$, is

• A. $-\frac{1}{\sqrt{29}}(2\hat{i}+5\hat{k})$
• B. $\frac{1}{\sqrt{6}}(\hat{i}-2\hat{j}-\hat{k})$
• C. $\frac{1}{\sqrt{26}}(4\hat{i}-3\hat{j}+\hat{k})$
• D. $-\frac{1}{\sqrt{165}}(10\hat{i}+7\hat{j}-4\hat{k})$

Answer 1

Answer: (D)

Let $O$ be the origin. Then $\overrightarrow{OA}=3\hat{i}-\hat{j}+2\hat{k}$,
$\overrightarrow{OB}=\hat{i}-\hat{j}-3\hat{k}$ and $\overrightarrow{OC}=4\hat{i}-3\hat{j}+\hat{k}$
$\therefore \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(\hat{i}-\hat{j}-3\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})=-2\hat{i}+0\hat{j}-5\hat{k}$
$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(4\hat{i}-3\hat{j}+\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})=\hat{i}-2\hat{j}-\hat{k}$
Now, $\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}<br/><br/>\hat{i}& \hat{j}& \hat{k}\\ -2& 0& -5\\ 1& -2& -1\end{array}\right|$
$=(0-10)\hat{i}-(2+5)\hat{j}+(4-0)\hat{k}=-10\hat{i}-7\hat{j}+4\hat{k}$
$\Rightarrow |\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{(-10{)}^2+(-7{)}^2+(4{)}^2}=\sqrt{100+49+16}=\sqrt{165}$
A unit vector perpendicular to the plane of $\Delta ABC$ is perpendicular to both $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Hence, a unit vector perpendicular, to the plane of
$\Delta ABC=\frac{\overrightarrow{AB}\times \overrightarrow{AC}}{|\overrightarrow{AB}\times \overrightarrow{AC}|}=\frac{-10\hat{i}-7\hat{j}+4\hat{k}}{\sqrt{165}}=-\frac{1}{\sqrt{165}}(10\hat{i}+7\hat{j}-4\hat{k})$