Question

find the equation of the right bisector plane of the line segment joining the points (a,a,a) and (-a,-a,-a). (a# 0).

Answer 1

Answer:

x+y+z = 0

Step-by-step explanation:

Equation of the plane when dr's  (a, b, c) of normal to the plane are given s

  ax + by +cz = d(constant)

Let the given points be A(a,a,a) and B(-a,-a,-a)

DR's of line AB are (2a,2a,2a)

But AB is perpendicular to the required plane.

Hence , equation of the required plane will be of form 2ax+2ay+2az = d

But Midpoint of AB i.e., (0,0,0) will lie on the plane(since it is right bisector)

Thus,d =0

=>x+y+z = 0 is the required plane