Question

Find the area of the triangle with vertices $A(1,1,2),B(2,3,5),and \space C(1,5,5).$

Answer 1

The vertices of triangle ABC are given as $A(1,1,2),B(2,3,5),and \space C(1,5,5).$
The adjacent sides $\overrightarrow{AB}$ and $\overrightarrow{BC}$ of $△ABC$ are given as:

\overrightarrow{AB}$$=(2-1)\hat{i}+(3-1)\hat{j}+(5-2)\hat{k}=\hat{i}+2\hat{j}+3\hat{k}
$\overrightarrow{BC}=(1-2)\hat{i}+(5-3)\hat{j}+(5-5)\hat{k}=-\hat{i}+2\hat{j}$
Area of $△ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{BC}|$
$\overrightarrow{AB}\times\overrightarrow{AC}$=$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ 1 & 2 & 3\\\\-1&2&0 \end{vmatrix}$=$\hat{i}(-6)-\hat{j}(3)+\hat{k}(2+2)=-6\hat{i}+3\hat{j}+4\hat{k}$
|$\overrightarrow{AB}\times\overrightarrow{AC}$|=\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}$$=\sqrt{36+9+16}=\sqrt{61}
Hence,the area of $△ABC$ is $\sqrt{\frac{61}{2}}$square units.