Given three 3D points (A,B, & C) how do I calculate the normal vector? The three points define a plane and I want the vector perpendicular to this plane.
Can I get sample C# code that demonstrates this?
33
votes
Math has a lot to do with programming. Especially this math, if you're doing anything 3D.
- Todd Gamblin
Give him a break, it's marked C# and .net-3.5. This is a problem that comes up immediately when you start programming 3D apps. Look at the related questions, we usually allow this stuff.
- Frank Krueger
My decision line is: will the math that is being asked about be implemented in code. As it will, this gets to stay.
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3 Answers
44
votes
It depends on the order of the points. If the points are specified in a counter-clockwise order as seen from a direction opposing the normal, then it's simple to calculate:
Dir = (B - A) x (C - A)
Norm = Dir / len(Dir)
where x is the cross product.
If you're using OpenTK or XNA (have access to the Vector3 class), then it's simply a matter of:
class Triangle {
Vector3 a, b, c;
public Vector3 Normal {
get {
var dir = Vector3.Cross(b - a, c - a);
var norm = Vector3.Normalize(dir);
return norm;
}
}
}
5
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2
votes
You need to calculate the cross product of any two non-parallel vectors on the surface. Since you have three points, you can figure this out by taking the cross product of, say, vectors AB and AC.
When you do this, you're calculating a surface normal, of which Wikipedia has a pretty extensive explanation.
