Question

If the position vectors of the three points $ABC$, are $\hat{i}+\hat{j}+\hat{k},2\hat{i}+3\hat{j}-4\hat{k}$ and $7\hat{i}+4\hat{j}+9\hat{k}$, then the unit vector perpendicular to the plane of the triangle $ABC$ is

• A. $\frac{31\hat{i}-38\hat{j}-9\hat{k}}{2486}$
• B. $\frac{31\hat{i}-38\hat{j}+9\hat{k}}{2486}$
• C. $\frac{31\hat{i}+38\hat{j}-9\hat{k}}{2486}$
• D. $\frac{31\hat{i}+38\hat{j}+9\hat{k}}{2486}$

Answer 1

Answer: (C)

$\frac{31\hat{i}+38\hat{j}-9\hat{k}}{2486}$

$\overrightarrow{AB}=\hat{i}+2\hat{j}-5\hat{k}$
$\overrightarrow{AC}=\hat{6}i+3\hat{j}+8\hat{k}$
$\hat{V}=\frac{(\overrightarrow{AB}\times \overrightarrow{AC})}{|\overrightarrow{AB}\times \overrightarrow{AC}|}D$
$\overline{AB}\times \overline{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -5\\ 6& 3& 8\end{array}\right|=31\hat{i}-38\hat{j}-9\hat{k}$
$|\overline{AB}\times \overline{AC}|=\sqrt{2486}$