Question

Using vectors, find the area of the triangle with vertices: $A(1,2,3),B(2,-1,4)$ and $C(4,5,-1)$

Answer 1

Given$:A(1,2,3),B(2,-1,4)$ and $C(4,5,-1)$

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

We have $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(2-1)\hat{i}+(-1-2)\hat{j}+(4-3)\hat{k}=\hat{i}-3\hat{j}+\hat{k}$

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(4-1)\hat{i}+(5-2)\hat{j}+(-1-3)\hat{k}=3\hat{i}+3\hat{j}-4\hat{k}$

$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 1\\ 3& 3& -4\end{array}\right|$

$=(12-3)\hat{i}-(-4-3)\hat{j}+(3+9)\hat{k}$

$=9\hat{i}+7\hat{j}+12\hat{k}$

Magnitude of $\overrightarrow{AB}\times \overrightarrow{AC}=\sqrt{{\left(9\right)}^2+{\left(7\right)}^2+{12}^2}$

$|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{81+49+144}=\sqrt{274}$

Area of $△ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\frac{1}{2}\times \sqrt{274}=\frac{\sqrt{274}}{2}$sq.units

$\therefore$ the required area is $\frac{\sqrt{274}}{2}$sq.units