Question

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

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

Answer 1

The correct option is C $\sqrt{155}/2sq.units$

Let the vertices of triangle are

$A(1,2,3),B(2,5,-1)$ and $C(-1,1,2)$

Then,

$\begin{array}{c}\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\hat{i}+3\hat{j}-4\hat{k}\\ \\ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=-2\hat{i}-\hat{j}-\hat{k}\end{array}$

Then,

$\begin{array}{c}\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& -4\\ -2& -1& -1\end{array}\right|\\ \\ =-7\hat{i}+9\hat{j}+5\hat{k}\\ \\ \left|\overrightarrow{AB}\times \overrightarrow{AC}\right|=\sqrt{{\left(-7\right)}^2+{\left(9\right)}^2+{\left(5\right)}^2}\\ \\ =\sqrt{49+8+25}\\ \\ =\sqrt{155}\end{array}$

Area of triangle $ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

$=\frac{1}{2}\times \sqrt{155}$

$=\frac{\sqrt{155}}{2}$ sq. units