Question

The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.

Answer 1

$$\begin{gathered} \text{The normal is passing through the points}\mathit{\text{A}}\text{(0, 0, 0) and}\mathit{\text{B}}\text{(12, -4, 3). So,} \\ \overrightarrow{n}\text{=}\overrightarrow{AB}\text{=}\overrightarrow{OB}\text{-}\overrightarrow{OA}\text{=}\left(12\hat{i}-4\hat{j}+3\hat{k}\right)-\left(0\hat{i}+0\hat{j}+0\hat{k}\right)=12\hat{i}-4\hat{j}+3\hat{k} \\ \text{Since the plane passes through (12, -4, 3),}\overrightarrow{a}\text{=}12\hat{i}-4\hat{j}+3\hat{k} \\ \text{We know that the vector equation of the plane passing through a point}\overrightarrow{a}\text{and normal to}\overrightarrow{n}\text{is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \text{Substituting}\overrightarrow{a}\text{=}\hat{i}\text{-}\hat{j}\text{+}\hat{k}\text{and}\overrightarrow{n}\text{= 4}\hat{i}\text{+ 2}\hat{j}\text{- 3}\hat{k}\text{, we get} \\ \overrightarrow{r}.\left(12\hat{i}-4\hat{j}+3\hat{k}\right)=\left(12\hat{i}-4\hat{j}+3\hat{k}\right).\left(12\hat{i}-4\hat{j}+3\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(12\hat{i}-4\hat{j}+3\hat{k}\right)=144+16+9 \\ \Rightarrow \overrightarrow{r}.\left(12\hat{i}-4\hat{j}+3\hat{k}\right)=169 \\ \Rightarrow \overrightarrow{r}.\left(12\hat{i}-4\hat{j}+3\hat{k}\right)=169 \\ \text{Substituting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\text{in the vector equation, we get} \\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(12\hat{i}-4\hat{j}+3\hat{k}\right)=169 \\ \Rightarrow 12x-4y+3z=169 \end{gathered}$$