Question

Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.

Answer 1

$$\begin{gathered} \text{The normal is passing through the points}\mathit{\text{A}}\text{(-1, 2, 3) and}\mathit{\text{B}}\text{(3, -5, 6). So,} \\ \overrightarrow{n}\text{=}\overrightarrow{AB}\text{=}\overrightarrow{OB}\text{-}\overrightarrow{OA}\text{=}\left(3\hat{i}-5\hat{j}+6\hat{k}\right)-\left(-\hat{i}+2\hat{j}+3\hat{k}\right)=4\hat{i}-7\hat{j}+3\hat{k} \\ \text{Mid-point of}\mathit{\text{AB}}\text{=}\left(\frac{-1+3}{2},\frac{2-5}{2},\frac{3+6}{2}\right)\text{=}\left(1,\frac{-3}{2},\frac{9}{2}\right) \\ \text{Since the plane passes through}\left(1,\frac{-3}{2},\frac{9}{2}\right)\text{,}\overrightarrow{a}\text{=}\hat{i}-\frac{3}{2}\hat{j}+\frac{9}{2}\hat{k} \\ \text{We know that the vector equation of the plane passing through a point}\overrightarrow{a}\text{and normal to}\overrightarrow{n}\text{is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \text{Substituting}\overrightarrow{a}\text{=}\hat{i}\text{-}\hat{j}\text{+}\hat{k}\text{and}\overrightarrow{n}\text{=}4\hat{i}-7\hat{j}+3\hat{k}\text{, we get} \\ \overrightarrow{r}.\left(4\hat{i}-7\hat{j}+3\hat{k}\right)=\left(\hat{i}-\frac{3}{2}\hat{j}+\frac{9}{2}\hat{k}\right).\left(4\hat{i}-7\hat{j}+3\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(4\hat{i}-7\hat{j}+3\hat{k}\right)=28 \\ \text{Substituting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\text{in the vector equation, we get} \\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(4\hat{i}-7\hat{j}+3\hat{k}\right)=28 \\ \Rightarrow 4x-7y+3z=28 \\ \Rightarrow 4x-7y+3x-28=0 \end{gathered}$$