# Get direction of normal without matrix inversion

I am building a 3D engine and I want it to calculate normals for triangles automatically.

The user creates a model that is made of triangles. Every triangle is made of three points in the space, and every model has its own central point. Here is some pseudo code:

class Triangle
{
point p[3];
Vector normal;
}

class Model
{
List<Triangle> triangles;
Point centralPoint;
}


I know that a normal can be evaluated with a cross product of the triangle vertices

normal = (p[0]-p[2])X(p[0]-p[1])


But I want the normal to point from the central point, out from the model. I can verify this by creating a base

B = ( (p[0]-p[2]) (p[0]-p[1]) normal )


Giving me the equation

B*x = centralPoint


And if the third element of vector x is negative, means that my normal is pointing to the direction that I want. The problem here is that I have to define a matrix and calculate its inverse, just to make sure that this number is negative! Seems like an unnecessary performance thief. Is there any shortcut that I can take here?

• I think testing the sign of the scalar product p[0]-central and normal should be enough. – Matthias Dec 2 '14 at 14:38

The question is not very clearly formulated, but it seems that you need to know on which side of the plane $\{\, x\mid x\cdot N=0 \}$ parallel to your triangle, and passing through the origin, is situated your central point$~p$ . Here $N$ is the normal vector to that plane. This is easy: just evaluate the scalar product $p\cdot N$ and if it is positive then $p$ is on the same side as $N$ is, and on the opposite side if $p\cdot N<0$.
Maybe you wanted the plane of the triangle itself (but that is not what your computation using the basis $B$ would give); in that case the equation is $\{\, x\mid x\cdot N=p[0]\cdot N \}$, and you should take the scalar product $(p-p[0])\cdot N$ to know on which side $p$ is situated.