Question

Find the equation of the plane bisecting the line segment joining the points $(-3,-2,1)and(1,6,-5)$ perpendicularly.

Answer 1

The given points are

$A\left(-3,-2,1\right)$ and $B\left(1,6,-5\right)$

The line segment $AB$ is given by $\left(x_2-x_1\right),\left(y_2-y_1\right)$ and $\left(z_2-z_1\right)$

$\therefore (1-(-3\left)\right),(6-(-2\left)\right),(-5-1)$

or, $(5,8,-6)$

$\because$ the plane bisects $AB$ at right angles, $\overrightarrow{AB}$ is the normal to the plane which is $\overrightarrow{n}$.

$\therefore \overrightarrow{n}=\hat{i}+\hat{j}+2\hat{k}$

Let $c$ be the mid point of $AB$,

then,

$\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)\\ =c\left(\frac{-3+1}{2},\frac{-2+6}{2},\frac{1+\left(-5\right)}{2}\right)\\ =c\left(\frac{-2}{2},\frac{4}{2},\frac{-4}{2}\right)=c\left(-1,2,-2\right)\end{array}$

Let this be $\overrightarrow{a}=-\hat{i}+2\hat{j}-2\hat{k}$

Hence,

The vector equation of the plane passing through $c$ and $\perp AB$ is

$\left(\overrightarrow{r}-\left(-\hat{i}+2\hat{j}-2\hat{k}\right)\right).\left(5\hat{i}+8\hat{j}-6\hat{k}\right)=0$

We know that,

$\begin{array}{c}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\\ \Rightarrow \left[\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(-\hat{i}+2\hat{j}-2\hat{k}\right)\right].\left(5\hat{i}+8\hat{j}-6\hat{k}\right)=0\end{array}$

Applying the product rule, we get,

$\left[\left(x+1\right)\hat{i}+\left(y-2\right)\hat{j}+\left(z+2\right)\hat{k}\right].\left(5\hat{i}+8\hat{j}-6\hat{k}\right)=0$

Applying product rule, we get,

$5\left(x+1\right)+8\left(y-2\right)+\left(-6\right)\left(z+2\right)=0$

On simplifying we get,

$5x+8y-6z=13$

This is the required equation of plane.