Igor, I think that this article can be useful for you:

https://people.kth.se/~johanph/habc.pdf

Yes, this is useful, thank you.

I am just curious as to what the answer would be for 2D landscapes. Then we could extrapolate that to a 3D projection (or even higher). For the record, the distance is real and follows a principle for the smallest entanglement metric, which is 2D in a 3D wave function as a bi directional projection for a photon pair. So, the 3D versus the 2D is essentially a special dual solution.

At the first look at the problem, it seems correct to assume the distinguished collection points be distributed according to the Poisson rule. Then, there is no particle within the distance  R  from arbitrary fixed point with probability    exp[ - D 4 \pi R^3 / 3 ],  where   D  is the average density of the particles. This is also the probability, that the distance to the nearest particle is greater than  R.

Conclution: The cdf of the distance to the particle nearest to an arbitrary point  equals 1 - exp[ - D 4 \pi R3 / 3 ]. By the Markov property, this holds for arbitrary point of the distinguished collection.

@Joachim Yes, it definitely looks  like this. Now, the next problem. How much it would change  if the medium is stretched. Note, that due to Poisson ratio particles are in fact closer together though the density drops.   Decreased density would  moves them apart. 

Thus you are talking about non-homogeneous Poisson point process. Then, the problem is in evaluation of the volume of the corresponding ellipsoid. Regards

Thank you, and that would probably solve the problem of angle distribution as well.

Simingly, yes:) BUT the calculations must be performed very carefully!

Regards

Additional important remark: The same question related to the whole infinite set of the distinguished points has obvious answer  \delta_0 . Joachim