Question

The area of the triangle whose vertices are $A(1,-1,2),B(2,1,-1)$ and $C(3,-1,2)$ is

• A. $13$
• B. $\sqrt{13}$
• C. $6$
• D. $\sqrt{6}$

Answer 1

Answer: (B)

$\sqrt{13}$

Here

$\overrightarrow{OA}=\hat{i}-\hat{j}+2\hat{k}$

$\overrightarrow{OB}=\widehat{2i}+\hat{j}-\hat{k}$

$\overrightarrow{OC}=\widehat{3i}-\hat{j}+2\hat{k}$

This implies $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$$=\hat{i}+2\hat{j}-3\hat{k}$

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\widehat{2i}$

Thus required area is given by $\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

$\Rightarrow |\overrightarrow{AB}\times \overrightarrow{AC}|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -3\\ 2& 0& 0\end{array}\right|=-2(3\hat{j}+2\hat{k})$

So area = $\frac{1}{2}\times 2\left|(3\hat{j}+2\hat{k})\right|=\sqrt{13}$