The Stokes-Einstein rotational diffusion relation tells us that we can write down a rotational diffusion coefficient for a sphere as:

$D_r \approx \frac{k_B T}{\zeta_f} \approx \frac{k_B T}{(8 \pi \eta)(r)^3}$

Where $k_B$ is Boltzmann's constant, $T$ is the temperature in Kelvin, $\zeta_f \approx (8 \pi \eta)(r)^3$ is the friction, $\eta$ is the viscosity of the medium (e.g. $\approx 1 \space cP$ in pure water), and $r$ is the radius of the sphere.

For an example calculation of $D_r \approx 0.2 \space rad^2/s$ (or $Hz$) of a $\approx 1 \space \mu m$ radius sphere in pure water at room temperature, please see here: http://www.microurl.info/YOtwp (this shortened URL is going to WolframAlpha).

My question is the following: Take a point on the sphere undergoing rotational diffusion in solution, and select a random point on its surface. Define a vector $v_1$ between this random point and the point at the center of the sphere. Let $v_2$ be another randomly oriented vector. What probability distribution do we have for the time until a minimum angle $\theta$ between $v_1$ and $v_2$ is achieved?

Alternatively, if we start with the vectors $v_1$ and $v_2$ being within a minimum angle $theta$, what time distribution do we have until the minimum angle becomes larger than $\theta$?