21

I need to create a method to generate a unit vector in three dimensions that points in a random direction using a random number generator. The distribution of direction MUST be isotropic.
Here is how I am trying to generate a random unit vector:
v = randn(1,3);
v = v./sqrt(v*v');
But I don't know how to complete the isotropic part. Any ideas?

Aina
  • 653
  • 2
  • 9
  • 22
  • possible duplicate of [Uniform random (Monte-Carlo) distribution on unit sphere](http://stackoverflow.com/questions/1841014/uniform-random-monte-carlo-distribution-on-unit-sphere) – finnw Mar 22 '12 at 13:09
  • There are also some good answers [here](http://stackoverflow.com/questions/8839086/how-to-randomize-points-on-a-sphere-surface-evenly) and [here](http://stackoverflow.com/questions/6389984/how-to-randomly-select-a-point-on-the-surface-of-the-earth) – finnw Mar 22 '12 at 13:14
  • @finnw: The question is a duplicate, but not the answers. Maybe we could merge the questions? – Jonas Mar 24 '12 at 13:29

1 Answers1

22

You're doing it right. A random normal distribution of coordinates gives you a uniform distribution of directions.

To generate 10000 uniform points on the unit sphere, you run

v = randn(10000,3);
v = bsxfun(@rdivide,v,sqrt(sum(v.^2,2)));

plot3(v(:,1),v(:,2),v(:,3),'.')
axis equal

enter image description here

Jonas
  • 74,690
  • 10
  • 137
  • 177
  • wow, thanks, @Jonas! I didn't know that about normal distribution! – Aina Mar 17 '12 at 23:24
  • @Aina: The 2D normal distribution is rotationally symmetric. Thus, the 3D normal distribution has spherical symmetry. – Jonas Mar 17 '12 at 23:51
  • Impressive. Any intuitive explanation on why it happens? – Andrey Rubshtein Mar 18 '12 at 16:43
  • 3
    @Andrey: Why the normal distribution is rotationally symmetric, you mean? The 3D normal distribution is proportional to `exp(- (x^2+y^2+z^2))`. The symmetry becomes obvious if you transform to spherical coordinates, where that expression becomes `exp(-(r^2))`. In other words, the density is only a function of the radius, not the angle, which means the points are uniformly distributed among all angles. – Jonas Mar 24 '12 at 13:25
  • @Jonas As far as I can see, this is an excellent method. Would you be able to produce a version for R? Thanks. – JASC Jan 14 '19 at 03:29
  • 1
    @JASC: Sadly no, as I'm just barely literate in R. However, the algorithm is really (1) draw a number of points from a 3D normal distribution (i.e. a list of coordinates where x, y, and z are drawn from independent normal distributions with mean 0 and standard deviation 1), and then divide each coordinate triplet by its norm so that x^2+y^2+z^2 is 1 for each of them. – Jonas Jan 14 '19 at 17:15
  • Does this generalise to more than 3 dimensions? – user1438310 Dec 08 '20 at 19:35
  • @user1438310: yes, it does, as the relationships hold for cubes/spheres of any dimension. – Jonas Dec 20 '20 at 13:27