Question

Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).

Answer 1

$$\begin{gathered} \text{Since the given plane passes through the point (1, -1, 1) and is normal to the line joining}\mathit{\text{A}}\text{(1, 2, 5) and}\mathit{\text{B}}\text{(-1, 3, 1),} \\ \overrightarrow{n}\text{=}\overrightarrow{AB}\text{=}\overrightarrow{OB}\text{-}\overrightarrow{OA}\text{=}\left(-\hat{i}+3\hat{j}+\hat{k}\right)-\left(\hat{i}+2\hat{j}+5\hat{k}\right)=-2\hat{i}+\hat{j}-4\hat{k} \\ \text{We know that the vector equation of the plane passing through a point}\overrightarrow{a}\text{and normal to}\overrightarrow{n}\text{is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \text{Substituting}\overrightarrow{a}\text{=}\hat{i}\text{-}\hat{j}\text{+}\hat{k}\text{and}\overrightarrow{n}\text{= - 2}\hat{i}\text{+}\hat{j}\text{- 4}\hat{k}\text{, we get} \\ \overrightarrow{r}.\left(-2\hat{i}+\hat{j}-4\hat{k}\right)=\left(\hat{i}\text{-}\hat{j}\text{+}\hat{k}\right).\left(-2\hat{i}+\hat{j}-4\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(-2\hat{i}+\hat{j}-4\hat{k}\right)=-2-1-4 \\ \Rightarrow \overrightarrow{r}.\left[-\left(2\hat{i}-\hat{j}+4\hat{k}\right)\right]=-7 \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}-\hat{j}+4\hat{k}\right)=7 \\ \text{For Cartesian form, we need to substitute}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\text{in the vector equation.} \\ \text{Then, we get} \\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(2\hat{i}-\hat{j}+4\hat{k}\right)=7 \\ \Rightarrow 2x-y+4z=7 \end{gathered}$$