Question

The area of triangle whose vertices are $(1,2,3),(2,5,-1)$ and $(-1,1,2)$ is

• A. $\frac{145}{2}\text{sq. units}$
• B. $\frac{\sqrt{150}}{2}\text{sq. units}$
• C. $\frac{155}{2}\text{sq. units}$
• D. $\frac{\sqrt{155}}{2}\text{sq. units}$

Answer 1

The correct option is D $\frac{\sqrt{155}}{2}\text{sq. units}$

Given: $A(1,2,3),B(2,5,-1)$ and $C(-1,1,2)$ is
$\overrightarrow{AB}=\hat{i}+3\hat{j}-4\hat{k}$
$\overrightarrow{AC}=-2\hat{i}-\hat{j}-\hat{k}$
Area of triangle $=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$
$=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& -4\\ -2& -1& -1\end{array}\right|=\frac{1}{2}|-7\hat{i}+9\hat{j}+5\hat{k}|$
$=\frac{1}{2}\sqrt{49+81+25}=\frac{\sqrt{155}}{2}\text{sq. units}$