Question

Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.

Answer 1

$$\begin{gathered} \text{We know that the vector equation of the plane passing through a point}\overrightarrow{a}\text{and normal to}\overrightarrow{n}\text{is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \text{Substituting}\overrightarrow{a}\text{= 5}\hat{i}\text{+2}\hat{j}\text{-4}\hat{k}\text{and}\overrightarrow{n}\text{= 2}\hat{i}\text{+3}\hat{j}\text{-}\hat{k}\text{(because the direction ratios of}\overrightarrow{n}\text{are 2, 3, -1), we get} \\ \overrightarrow{r}.\left(2\hat{i}\text{+3}\hat{j}\text{-}\hat{k}\right)=\left(5\hat{i}\text{+2}\hat{j}\text{-4}\hat{k}\right).\left(2\hat{i}\text{+3}\hat{j}\text{-}\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}\text{+3}\hat{j}\text{-}\hat{k}\right)=10+6+4 \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}\text{+3}\hat{j}\text{-}\hat{k}\right)=20 \\ \text{For Cartesian form, we need to substitute}\overrightarrow{r}\text{=}x\hat{i}+y\hat{j}+z\hat{k}\text{in this equation. Then, we get} \\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(2\hat{i}\text{+3}\hat{j}\text{-}\hat{k}\right)=20 \\ \Rightarrow 2x+3y-z=20 \end{gathered}$$