Question

Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

Answer 1

$$\begin{gathered} \text{The normal is passing through the points}\mathit{\text{A}}\text{(1, 2, 3) and}\mathit{\text{B}}\text{(3, 4, 5). So,} \\ \overrightarrow{n}\text{=}\overrightarrow{AB}\text{=}\overrightarrow{OB}\text{-}\overrightarrow{OA}\text{=}\left(3\hat{i}+4\hat{j}+5\hat{k}\right)-\left(\hat{i}+2\hat{j}+3\hat{k}\right)=2\hat{i}+2\hat{j}+2\hat{k} \\ \text{Mid-point of}\mathit{\text{AB}}\text{=}\left(\frac{1+3}{2},\frac{2+4}{2},\frac{3+5}{2}\right)\text{=}\left(2,3,4\right) \\ \text{Since the plane passes through}\left(2,3,4\right)\text{,}\overrightarrow{a}\text{=}2\hat{i}+3\hat{j}+4\hat{k} \\ \text{We know that the vector equation of the plane passing through a point}\overrightarrow{a}\text{and normal to}\overrightarrow{n}\text{is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \text{Substituting}\overrightarrow{a}\text{=}\hat{i}\text{-}\hat{j}\text{+}\hat{k}\text{and}\overrightarrow{n}\text{= 4}\hat{i}\text{+ 2}\hat{j}\text{- 3}\hat{k}\text{, we get} \\ \overrightarrow{r}.\left(2\hat{i}+2\hat{j}+2\hat{k}\right)=\left(2\hat{i}+3\hat{j}+4\hat{k}\right).\left(2\hat{i}+2\hat{j}+2\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+2\hat{j}+2\hat{k}\right)=4+6+8 \\ \Rightarrow \overrightarrow{r}.\left(2\hat{i}+2\hat{j}+2\hat{k}\right)=18 \\ \Rightarrow \overrightarrow{r}.\left[2\left(\hat{i}+\hat{j}+\hat{k}\right)\right]=18 \\ \Rightarrow \overrightarrow{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)=9 \\ \text{Substituting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\text{in the vector equation, we get} \\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(\hat{i}+\hat{j}+\hat{k}\right)=9 \\ \Rightarrow x+y+z=9 \end{gathered}$$