# Thread: How to find the equation of this plane?

1. ## How to find the equation of this plane?

Find the equation of the plane which consists of all points at the same distance from the points (3, 1, 5) and (5, -1, 3).

As far as I know, the equation of a plane passing through the point Po (xo, yo, zo) with normal vector <a, b, c> is:

a(x-xo) + b(y-yo) + c(z-zo) = 0.

Can anyone help me?

2. Originally Posted by essedra
Find the equation of the plane which consists of all points at the same distance from the points P(3, 1, 5) and Q(5, -1, 3).

As far as I know, the equation of a plane passing through the point Po (xo, yo, zo) with normal vector <a, b, c> is:

a(x-xo) + b(y-yo) + c(z-zo) = 0.

Can anyone help me?
1. The plane must pass through the midpoint of $\overline{PQ}$

2. The normal vector of the plane is the vector $\vec n = \overrightarrow{PQ}$

3. Let be $\vec p = \overrightarrow{OP}$ and $\vec q = \overrightarrow{OQ}$ then the equation of the plane is:

$(\vec p - \vec q) \cdot \left((x,y,z)-\frac{\vec q + \vec p}2 \right) = 0$