Question

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

Answer 1

$\text{Here,}\overrightarrow{AB}=(2-1)\hat{i}+(-1-2)\hat{j}+(4-3)\hat{k}=\hat{i}-3\hat{j}+\hat{k}and\overrightarrow{AC}=(4-1)\hat{i}+(5-2)\hat{j}+(-1-3)\hat{k}=3\hat{i}+3\hat{j}-4\hat{k}$

$\therefore \overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 1\\ 3& 3& -4\end{array}\right|=\hat{i}(12-3)-\hat{j}(-4-3)+\hat{k}(3+9)=9\hat{i}+7\hat{j}+12\hat{k}and|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{9^2+7^2+{12}^2}=\sqrt{81+49+144}=\sqrt{274}\therefore \text{Area of}△ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\frac{1}{2}\sqrt{274}squnits$