How to work out the angle of a line passing through a plane I have a triangular plane composed of three points. From this it it easy to deduce that the plane is in fact composed of two vectors which must touch at some point. because all of this is relative, one of the points can be taken as being at reference (0,0,0). How do I calculate the equation for a line that passes through the plane which is perpendicular to it? Its actual location is of no great importance as the angle is more use - it will be used to show how 'bright' the plane should be due to facing light.

My maths isn't great so an answer in plain English would be appreciated!
Hint: you have two vectors (in the plane), call them $\vec a$ and $\vec b$. Then their cross product $\vec p$ will be a vector perpendicular to both of them, which is also therefore perpendicular to the plane:
$$\vec p = \vec a \times \vec b$$
Why not find the x direction, y direction and z direction of the plane? Then, rotate it so that the plane is, say, the plane at y=0. You're then going to find the 'y-angle' of the vector hitting the plane. Do the same for x and z. This way probably isn't particularly elegant, but it's simple.