Question

A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point $3\hat{i}+\hat{j}-\hat{k}.$ Find the vector and Cartesian forms of the equation of the plane.

Answer 1

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