Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}\overrightarrow{a,}\overrightarrow{b}\mathrm{and}\overrightarrow{c}\mathrm{be}\mathrm{the}\mathrm{position}\mathrm{vector}\mathrm{of}A,B\mathrm{and}C,\mathrm{respectively}. \\ \therefore \mathit{}\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k} \\ \overrightarrow{b}=2\hat{i}+3\hat{j}+5\hat{k} \\ \mathit{}\stackrel{}{\overrightarrow{c}}=\hat{i}+5\hat{j}+5\hat{k} \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right| \\ \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a} \\ =\left(2\hat{i}+3\hat{j}+5\hat{k}\right)-\left(\hat{i}+\hat{j}+2\hat{k}\right) \\ =\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a} \\ =\left(i+5j+5k\right)-\left(\hat{i}+\hat{j}+2\hat{k}\right) \\ =0\hat{i}+4\hat{j}+3\hat{k} \\ \therefore \overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 0& 4& 3\end{array}\right| \\ =-6\hat{i}-3\hat{j}+4\hat{k} \\ \Rightarrow \left|\overrightarrow{AB}\times \overrightarrow{AC}\right|=\sqrt{36+9+16} \\ =\sqrt{61} \end{gathered}$$

$$\begin{gathered} \text{Area of the triangle =}\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right| \\ \text{=}\frac{\sqrt{61}}{2}\text{sq. units} \end{gathered}$$