# Equation of a plane given 3 points

## Homework Statement

Hello, as part of a flux integral question in 3-d, I need to work out the equation of one side of a pyramid, of which the centre of the base is at the origin of the co-ordinate system.
The points setting out the side I am considering are B(L/2,L/2,0) A(L/2, -L/2,0) C(0,0,L) in cartesian co-ordinates. Basically, the pyramid has sides of its base of length L and height L.

## The Attempt at a Solution

I worked out the 2 vectors corresponding to the edges of the plane, that converge at point C:
C->B=(L/2)i+(L/2)j-Lk
C->A=(L/2)i-(L/2)j-Lk
I did the cross product to get the normal vector:
n=L2i+(L2/2)k
I then tried using the method in which you multiply the vector component from the normal vector by the corresponding cartesian variable, minus the co-ordinate of which ever point you are looking at.
It looks something like this:
For a normal vector V= Vxi + Vyj + Vzk
and a point (A,B,C) the surface would be
Vx(x-A)+Vy(y-B)+Vz(z-C)=0
This is where I am stuck. If I follow this method, this leads me to believe that I multiply the y component of the plane equation by zero (since the vector y-component is non-existant). This is obviously incorrect since there are many points on this plane with y co-ordinates. Can anyone tell me what I am doing wrong? I could work out a unit normal vector but using the above method it shouldn't make a difference since all variables will be scaled by the same 1/magnitude. I tried doing this all again for another point (A) and found the same problem. Please help.