Question

The equation of the plane which bisects the line joining $(2,3,4)$ and $(6,7,8)$ at right angle is

• A. $x+y+z-15=0$
• B. $x-y+z-15=0$
• C. $x-y-z-15=0$
• D. $x+y+z+15=0$

Answer 1

The correct option is A $x+y+z-15=0$

Let the given points be $A\left(2,3,4\right)$ and $B\left(6,7,8\right)$ and C be the midpoint of AB.

Then,

$C\left(\frac{2+6}{2},\frac{3+7}{2},\frac{4+8}{2}\right)=C\left(4,5,6\right)$.

Since, the plane passes through this point, then the equation of line is,

$A\left(x-4\right)+B\left(y-5\right)+C\left(z-6\right)=0$ (1)

Direction ratio of AB is,

$\left(6-2\right),\left(7-3\right),\left(8-4\right)=4,4,4$.

The line with direction ratio $4,4,4$ is normal to the plane (1),

$4\left(x-4\right)+4\left(y-5\right)+4\left(z-6\right)=0$

$4x-16+4y-20+4z-24=0$

$4x+4y+4z-60=0$

$x+y+z-15=0$

This is the required equation of plane.