Question

If the position vectors of three points $A,B,C$ are $\hat{i}+\hat{j}+\hat{k},2\hat{i}+3\hat{j}-4\hat{k}$ and $3\hat{i}+2\hat{j}+\hat{k}$ respectively, then the unit vector perpendicular to the plane of the triangle $ABC$ is

• A. $\frac{5\overline{i}-10\overline{j}+3\overline{k}}{\sqrt{134}}$
• B. $\frac{5\overline{i}-10\overline{j}-3\overline{k}}{\sqrt{134}}$
• C. $\frac{5\overline{i}+10\overline{j}-3\overline{k}}{\sqrt{134}}$
• D. $\frac{5\overline{i}+10\overline{j}+3\overline{k}}{\sqrt{134}}$

Answer 1

The correct option is B $\frac{5\overline{i}-10\overline{j}-3\overline{k}}{\sqrt{134}}$

$\overrightarrow{AB}=\hat{i}+2\hat{j}-5\hat{k}$
$\overrightarrow{AC}=2\hat{i}+\hat{j}$
$\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -5\\ 2& 1& 0\end{array}\right|=5\hat{j}-10\hat{j}-3\hat{k}$
$|\overrightarrow{n}|=\sqrt{134}$
$\hat{n}=\frac{\overrightarrow{n}}{|\overrightarrow{n}|}=\frac{5\hat{i}-10\hat{j}-3\hat{k}}{\sqrt{134}}$