Question

If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.

Answer 1

$$\begin{gathered} \text{The normal is passing through the points}\mathit{\text{A}}\text{(4, -1, 2) and}\mathit{\text{B}}\text{(-10, 5, 4). So,} \\ \overrightarrow{n}\text{=}\overrightarrow{AB}\text{=}\overrightarrow{OB}\text{-}\overrightarrow{OA}\text{=}\left(-10\hat{i}\text{+5}\hat{j}+4\hat{k}\right)-\left(4\hat{i}-\hat{j}+2\hat{k}\right)=-14\hat{i}+6\hat{j}+2\hat{k} \\ \text{Since the plane passes through (-10, 5, 4),}\overrightarrow{a}\text{= -10}\hat{i}\text{+5}\hat{j}+4\hat{k} \\ \text{We know that the vector equation of the plane passing through a point}\overrightarrow{a}\text{and normal to}\overrightarrow{n}\text{is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \text{Substituting}\overrightarrow{a}\text{=}-10\hat{i}\text{+5}\hat{j}+4\hat{k}\text{and}\overrightarrow{n}\text{=}-14\hat{i}+6\hat{j}+2\hat{k}\text{, we get} \\ \overrightarrow{r}.\left(-14\hat{i}+6\hat{j}+2\hat{k}\right)=\left(-10\hat{i}\text{+5}\hat{j}+4\hat{k}\right).\left(-14\hat{i}+6\hat{j}+2\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(-14\hat{i}+6\hat{j}+2\hat{k}\right)=140+30+8 \\ \Rightarrow \overrightarrow{r}.\left(-2\left(7\hat{i}-3\hat{j}-\hat{k}\right)\right)=178 \\ \Rightarrow \overrightarrow{r}.\left(7\hat{i}-3\hat{j}-\hat{k}\right)=-89 \\ \text{Substituting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\text{in the vector equation, we get} \\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(7\hat{i}-3\hat{j}-\hat{k}\right)=-89 \\ \Rightarrow 7x-3y-z=-89 \\ \Rightarrow 7x-3y-z+89=0 \end{gathered}$$