ok, quickly...

use PVector.cross() to give you the normal

A--B
|  |
C--D

A = (0, 0, 0);
B = (300, 0, 0);
C = (0, 450, 0);
D = (300, 450, 0);

choose two vectors, A to C and C to D. those work out as (0 - 0, 450 - 0, 0 - 0) and (300 - 0, 450 - 450, 0 - 0)

ie (0, 450, 0) and (300, 0, 0), which, crossed, gives (0, 0, -135), pointing along the -ve z-axis which is a vector perpendicular to the first two.

for the other side choose A to D and D to C. (300 - 0, 450 - 0, 0 - 0) and (0 - 300, 450 - 450, 0 - 0)

(300, 450, 0) crossed with (-300, 0, 0) gives (0, 0, 135), ie along +ve z-axis.

that these two vectors have opposite signs indicates they are facing away from each other.

IT IS IMPORTANT to consistently number the edges in order for this to work. A - C - D is anticlockwise looking from the front, A - D - C is anti-clockwise when looking from the back. mixing clockwise orderings and anti-clockwise orderings will mean this doesn't work.

1. PVector pA = new PVector(0, 0, 0);
2. PVector pB = new PVector(300, 0, 0);
3. PVector pC = new PVector(0, 450, 0);
4. PVector pD = new PVector(300, 450, 0);
5. 
   
6. PVector pAC = PVector.sub(pC, pA);      // A to C
   
7. PVector pCD = PVector.sub(pD, pC);      // C to D
   
8. PVector c1 = pAC.cross(pCD);
9. println(c1.x + " " + c1.y + " " + c1.z);
10. 
    
11. PVector pAD = PVector.sub(pD, pA);      // A to D
    
12. PVector pDC = PVector.sub(pC, pD);      // D to C
    
13. PVector c2 = pAD.cross(pDC);
14. println(c2.x + " " + c2.y + " " + c2.z);