Question

Find the equation of a plane which bisects perpendicularly the line joining the points $A(2,3,4)$ and $B(4,5,8)$ at right angles.

Answer 1

The given points are $A(2,3,4)$ and $B(4,5,8)$

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

$=(4-2),(5-3),(8-4)=(2,2,4)=(1,1,2)$

Since the plane bisects that right angle $\overrightarrow{AB}$ is normal is the plane which is $\overrightarrow{x}=\hat{i}+\hat{j}+2\hat{k}$

Let $C$ be the midpoint of $AB$

$\left(\frac{x_1+x_2}{2}\right),\left(\frac{y_1+y_2}{2}\right),\left(\frac{z_1+z_2}{2}\right)or,(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})=C(\frac{2+4}{2},\frac{3+5}{2},\frac{4+8}{2})=C(\frac{6}{2},\frac{8}{2},\frac{12}{2})=C(3,4,6)$
Let this be $\overrightarrow{a}=3\hat{i}+4\hat{j}+6\hat{k}$

Hence the vector equation of the line passing through $C$ and $AB$ is

$(\overrightarrow{r}-(3\hat{i}+4\hat{j}+6\hat{k}\left)\right).(\hat{i}+\hat{j}+2\hat{k})=0$

We have $\overrightarrow{r}=x\hat{i}+y\hat{j}+2\hat{k}\Rightarrow \left[\right(x\hat{i}+y\hat{j}+z\hat{k}).(3\hat{i}+4\hat{j}+6\hat{k}\left)\right](\hat{i}+\hat{j}+2\hat{k})=0$

On simplifying we get

$(x-3)\hat{i}+(y-4)\hat{j}+(z-6)\hat{k}.(\hat{i}+\hat{j}+2\hat{k})=0$

Applying product rule

$(x-3)+(y-4)+2(z-6)=0$

On simplifying

$x+y+2z=19$

This is the required equation/