1. ## Calculating a Normal

I've done a lot of reading about how to find normals, but I just don't quite get it. My understanding so far is:

To find the normal of triangle ABC.
[a-b]x[b-c]

I'm not sure how you subtract vectors.
I'm don't understand how to find a cross product, mostly just because of the math symbols used.

I just need someone to clarify these things for me.

Thanks

2. So cross product, given triange a (x1, y1, z1) b (x2, y2, z2) c (x3, y3, z3), you take any two vectors, the edges, which compose it

Either

ab and bc
ac and ab
ca and cb

And take the cross product

So, a, b, and c represent points in space (must be space, or in 2d, z is some constant)

Cross product of ab and bc for example is ab x bc

ab = <x2 - x1, y2 - y1, z2 - z1>
bc = <x2 - x3, y2 - y3, z2 - z3>

ab x bc = <(y2 - y1)(z2 - z3), (z2 - z1)(y2 - y3), (x2 - x1)(z2 - z3) - (z2 - z1)(x2 - x3), (x2 - x1)(y2 - y3) - (y2 - y1)(x2 - x3)>

This is my vector class implementing such operations
http://pokepong.svn.sourceforge.net/...pp?view=markup

3. And vector subtracting for a = b - c is just
[ ax, ay, az ] = [ bx, by, bz ] - [ cx, cy, cz ] with component-wise subtraction, so the result is
[ bx - cx, by - cy, bz - cz ]

4. Aw, it seems so simple now! Thanks guys
edit:
wait..
ab x bc = <(y2 - y1)(z2 - z3), (z2 - z1)(y2 - y3), (x2 - x1)(z2 - z3) - (z2 - z1)(x2 - x3), (x2 - x1)(y2 - y3) - (y2 - y1)(x2 - x3)>
was that subtraction I bolded a mistake?

5. I couldn't see what you posted but I missed a subtraction sign in the first vector components

ab x bc = <(y2 - y1)(z2 - z3) - (z2 - z1)(y2 - y3), (x2 - x1)(z2 - z3) - (z2 - z1)(x2 - x3), (x2 - x1)(y2 - y3) - (y2 - y1)(x2 - x3)>