# Flat

• Jun 29th 2008, 08:21 AM
Apprentice123
Flat
Finding the equation of the plan. Mediator of the segment of extreme A(1,-2,6) and B(3,0,0)

$x+y-3z+8=0$
• Jun 29th 2008, 08:28 AM
flyingsquirrel
Hi
Quote:

Originally Posted by Apprentice123
Finding the equation of the plan. Mediator of the segment of extreme A(1,-2,6) and B(3,0,0)

$x+y-3z+8=0$

To find the answer you need to know two things :
• the mediating plane is orthogonal to the line segment $[AB]$ (this boils to saying that the plane is orthogonal to $\vec{AB}$)
• the midpoint of $[AB]$ is a point of the mediating plane.
• Jun 29th 2008, 08:36 AM
Moo
Hello,

Quote:

Originally Posted by Apprentice123
Finding the equation of the plan. Mediator of the segment of extreme A(1,-2,6) and B(3,0,0)

$x+y-3z+8=0$

A mediator is perpendicular to the segment and passes throught the midpoint.

The coordinates of $\overrightarrow{AB}$ are $(2~,~2~,~-6)$

So the equation of the plan, which has $AB$ as orthogonal vector, will be :

$2x+2y-6z+d=0$

-----------------
With d to determine. And for this, you will have to use the fact that it goes throught the midpoint of AB.

Coordinates of the midpoint of [AB] : $\left(\frac{3+1}{2}~,~\frac{-2+0}{2}~,~\frac{6+0}2\right)$, that is to say $(2~,~-1~,~3)$.

This point is on the plane, so substitute and get d.

After that, divide the equation by 2.
• Jun 29th 2008, 10:20 AM
Apprentice123
Thank you