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

Since, the equation of a plane is bisecting perpendicular the line joining the points A(2,3,4) and B(4,5,8) at right angles.
So, mid-point of AB is $(\frac{2+4}{2},\frac{3+5}{2},\frac{4+8}{2})$i.e., (3,4,6).
Also, $\overrightarrow{N}=(4-2)\hat{i}+(5-3)\hat{j}+(8-4)\hat{k}=2\hat{i}+2\hat{j}+4\hat{k}$
So, the required equation of the plane is $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{N}=0\Rightarrow \left[\right(x-3)\hat{i}+(y-4)\hat{j}+(z-6\left)\hat{k}\right].(2\hat{i}+2\hat{j}+4\hat{k})=0[\because \overrightarrow{a}=3\hat{i}+4\hat{j}+6\hat{k}]\Rightarrow 2x-6+2y-8+4z-24=0\Rightarrow 2x+2y+4z=38\therefore x+y+2z=19$