Question

Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B and C are A (3, −1, 2), B (1, −1, −3) and C (4, −3, 1).

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{vector}\overrightarrow{AB}\times \overrightarrow{AC}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{vectors}\overrightarrow{AB}\mathrm{and}\overrightarrow{AC}. \\ \therefore \mathrm{Required}\mathrm{unit}\mathrm{vector}=\frac{\overrightarrow{AB}\times \overrightarrow{AC}}{\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|} \\ \mathrm{Now}, \\ \overrightarrow{AB}=\mathrm{Position}\mathrm{vector}\mathrm{of}B-\mathrm{Position}\mathrm{vector}\mathrm{of}A \\ =\left(\hat{i}-\hat{j}-3\hat{k}\right)-\left(3\hat{i}-\hat{j}+2\hat{k}\right) \\ =-2\hat{i}+0\hat{j}-5k \\ \overrightarrow{AC}=\mathrm{Position}\mathrm{vector}\mathrm{of}C-\mathrm{Position}\mathrm{vector}\mathrm{of}A \\ =\left(4\hat{i}-3\hat{j}+\hat{k}\right)-\left(3\hat{i}-\hat{j}+2\hat{k}\right) \\ =\hat{i}-2\hat{j}-\hat{k} \\ \therefore \overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 0& -5\\ 1& -2& -1\end{array}\right| \\ =\left(0-10\right)i-\left(2+5\right)j+\left(4-0\right)\hat{k} \\ =-10\hat{i}-7\hat{j}+4\hat{k} \\ \left|\overrightarrow{AB}\times \overrightarrow{BC}\right|=\sqrt{{\left(-10\right)}^2+{\left(-7\right)}^2+4^2} \\ =\sqrt{165} \\ \text{Unit vector perpendicular to the plane ABC =}\frac{\overrightarrow{AB}\times \overrightarrow{AC}}{\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|}=\frac{-10\hat{i}-7\hat{j}+4\hat{k}}{\sqrt{165}} \end{gathered}$$