# Equation of a plane given 3 points

• MrB3nn
So it looks like the problem was with my understanding of vector multiplication. When you multiply a vector by a scalar, the result is always a vector, no matter what the original vector was. In this case, when you multiplied the y coordinate by 0, you were really multiplying by the vector y-coordinate of the origin, which is why it didn't work. Thanks for the help!

## 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.

You did it right so far. Your plane ABC is perpendicular to the xz plane, and the equation will be of the form ax+cz=d.

If a point has x and z coordinates that satisfy this equation ax+cz=d, then the y coordinate can be anything (as opposed to "nothing" which you seem to be thinking.

Billy Bob said:
You did it right so far. Your plane ABC is perpendicular to the xz plane, and the equation will be of the form ax+cz=d.

If a point has x and z coordinates that satisfy this equation ax+cz=d, then the y coordinate can be anything (as opposed to "nothing" which you seem to be thinking.

Thanks for your post dude, I see now. It worked, I checked with the divergence theorem and got same answer.