Question

Using vectors find the area of the triangle with vertices, A (2, 3, 5), B (3, 5, 8) and C (2, 7, 8).

Answer 1

$$\begin{gathered} \mathrm{Let}\overrightarrow{a},\overrightarrow{b}\mathrm{and}\stackrel{}{\overrightarrow{c}}\mathrm{be}\mathrm{the}\mathrm{position}\mathrm{vector}s\mathrm{of}A,B\mathrm{and}C,\mathrm{respectively}.\mathrm{Then}, \\ \overrightarrow{a}=2\hat{i}+3\hat{j}+5\hat{k} \\ \overrightarrow{b}=3\hat{i}+5\hat{j}+8\hat{k} \\ \overrightarrow{c}=2\hat{i}+7\hat{j}+8\hat{k} \\ \mathrm{Now}, \\ \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a} \\ =\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a} \\ =0\hat{i}+4\hat{j}+3\hat{k} \\ \therefore \overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 0& 4& 3\end{array}\right| \\ =-6\hat{i}-3\hat{j}+4\hat{k} \\ \Rightarrow \left|\overrightarrow{AB}\times \overrightarrow{AC}\right|=\sqrt{36+9+16} \\ =\sqrt{61} \\ \text{Area of triangle}\mathit{\text{ABC}}\text{=}\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right| \\ \text{=}\frac{\sqrt{61}}{2}\text{sq. units} \end{gathered}$$