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

Answer:

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

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- Jun 29th 2008, 07:21 AMApprentice123Flat
Finding the equation of the plan. Mediator of the segment of extreme A(1,-2,6) and B(3,0,0)

Answer:

$\displaystyle x+y-3z+8=0$ - Jun 29th 2008, 07:28 AMflyingsquirrel
Hi

To find the answer you need to know two things :

- the mediating plane is orthogonal to the line segment $\displaystyle [AB]$ (this boils to saying that the plane is orthogonal to $\displaystyle \vec{AB}$)
- the midpoint of $\displaystyle [AB]$ is a point of the mediating plane.

- Jun 29th 2008, 07:36 AMMoo
Hello,

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

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

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

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

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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] : $\displaystyle \left(\frac{3+1}{2}~,~\frac{-2+0}{2}~,~\frac{6+0}2\right)$, that is to say $\displaystyle (2~,~-1~,~3)$.

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

After that, divide the equation by 2. - Jun 29th 2008, 09:20 AMApprentice123
Thank you