I would like to generate a random axis or unit vector in 3D. In 2D it would be easy, I could just pick an angle between 0 and 2*Pi and use the unit vector pointing in that direction.
But in 3D I don't know how can I pick a random point on a surface of a sphere.
If I pick two angles the distribution won't be uniform on the surface of the sphere. There would be more points at the poles and less points at the equator.
If I pick a random point in the (-1,-1,-1):(1,1,1) cube and normalise it, then there would be more chance that a point gets choosen along the diagonals than from the center of the sides. So thats not good either.
But then what's the good solution?
Best Answer
You need to use an equal-area projection of the sphere onto a rectangle. Such projections are widely used in cartography to draw maps of the earth that represent areas accurately.
One of the simplest such projections is the axial projection of a sphere onto the lateral surface of a cylinder, as illustrated in the following figure:
This projection is area-preserving, and was used by Archimedes to compute the surface area of a sphere.
The result is that you can pick a random point on the surface of a unit sphere using the following algorithm:
Choose a random value of $\theta$ between $0$ and $2\pi$.
Choose a random value of $z$ between $-1$ and $1$.
Compute the resulting point: $$ (x,y,z) \;=\; \left(\sqrt{1-z^2}\cos \theta,\; \sqrt{1-z^2}\sin \theta,\; z\right) $$