How to generate a random plane that contains a point and its antipode on a sphere?

Question

Given a point and its antipode on a unit sphere, how can I generate a random plane containing them? My idea is to pick a random point on the unit circle from $\theta = 0$ to $\theta = \pi$ and use that as the third point to generate a plane. Of course, this strategy would not work if the random point I choose happens to have a z-coordinate of $0$, but otherwise is this approach valid?

Answer 1

This will give you the desired distribution if the unit circle is in a plane orthogonal to the axis through the two points $P$ and $P'$. Let $\mathbf n={P-P'\over\|P-P'\|}$ be a unit vector in the direction of this axis. If $\mathbf u$ and $\mathbf v$ are unit vectors that are orthogonal to each other and to $\mathbf n$, then a unit circle orthogonal to $\mathbf n$ can be given parametrically as $\cos\theta\,\mathbf u+\sin\theta\,\mathbf v$. Suitable vectors $\mathbf u$ and $\mathbf v$ are pretty easy to compute: take any unit vector in the orthogonal complement of $\operatorname{span}\{\mathbf n\}$ as $\mathbf u$ and $\mathbf n\times\mathbf u$ as $\mathbf v$. It’s also common to generate the first vector $\mathbf u$ by taking the cross product of $\mathbf n$ with one of the standard basis vectors and normalizing.

Answer 2

I initially said here your approach would work, but now I think I want to reconsider this. If the initial point is close to the top of the sphere, then it samples close to randomly; if your initial point $P$ has $z$ even close to $0$, then you will be biased, because choosing the third point close to either $P$ or $-P$ corresponds to a small fraction of the choices for $\theta$, but a large fraction of the possible directions that the plane could go in.

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Instead, if you choose a random point on the sphere to be the third point (e.g., by taking three samples from the standard normal distribution, and normalizing to get a point on the sphere) then by symmetry every plane will be equally likely.