# Help in equation of a plane?

1. May 7, 2012

### hivesaeed4

An equation for the plane that contains the three points $${(0, 0, 0)}$$, $${(1, 1, 1)}$$ and $${(2, 3, 4)}$$.

A$${x}$$ - B$${y}$$ +C $${z}$$ =0

Now I evaluated it, C comes out to be eual to A and B=2A. But that just means that A=B=C=0 which is just wrong.

Then I used the http://www.ehow.com/how_8072475_equations-planes.html
and evaluated A,B,C to be 1, 2, 1 respectively. But where was I making a mistake initially?

2. May 7, 2012

### optics123

plane equation is :ax+by+cz=d
we substitute 3 points (0,0,0), (1,1,1) and (2,3,4) in the plane equation, therefore:
(0,0,0) -> 0=d (1)
(1,1,1) -> a+b+c=d (2)
(2,3,4) -> 2a+3b+4c=d (3)

by solving equations (1), (2) and (3) relative to a we have:
d=0
b=-2a
c=a

therefore the plane eq. is : ax-2ay+az=0 ->

plane equation is: " x-2y+z=0"

3. May 7, 2012

### hivesaeed4

I don't think you got what I was asking. I arrived at the same equation you got
ax-2ay+az=0 but how do we find a=1?