Question

If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.

Answer 1

$$\begin{gathered} \text{The normal is passing through the points}\mathit{\text{O}}\text{(0, 0, 0) and}\mathit{\text{P}}\text{(2, 3 ,-1). So,} \\ \overrightarrow{n}\text{=}\overrightarrow{OP}\text{=}\left(2\hat{i}+3\hat{j}-\hat{k}\right)-\left(0\hat{i}+0\hat{j}+0\hat{k}\right)=2\hat{i}+3\hat{j}-\hat{k} \\ \text{Since the plane passes through the point (2, 3 ,-1),}\overrightarrow{a}\text{=}2\hat{i}+3\hat{j}-\hat{k} \\ \text{We know that the vector equation of the plane passing through a point}\overrightarrow{a}\text{and normal to}\overrightarrow{n}\text{is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \text{Substituting}\overrightarrow{a}\text{=}2\hat{i}+3\hat{j}-\hat{k}\text{and}\overrightarrow{n}\text{=}2\hat{i}+3\hat{j}-\hat{k}\text{, we get} \\ \overrightarrow{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)=\left(2\hat{i}+3\hat{j}-\hat{k}\right).\left(2\hat{i}+3\hat{j}-\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)=4+9+1 \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)=14 \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)=14 \\ \text{Substituting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\text{in the vector equation, we get} \\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(2\hat{i}+3\hat{j}-\hat{k}\right)=14 \\ \Rightarrow 2x+3y-z=14 \end{gathered}$$