Identify $\Bbb C \cong \Bbb R^2$ in the usual way. The oriented angle formed by the pair $(z,w)$, that is, the angle from $z$ to $w$, counterclockwise, is the number $\arg(w\overline{z})$. This number is defined modulo $2\pi$.
Ok, you want the angle from $(1,0)$ to $(0,1)$. Write $z = 1$ and $w = i$ there. So: $$\arg(w\overline{z}) = \arg(i) = 90º.$$
So $v_1$ is obtained from $v_2$ after a rotation of $90º$ counterclockwise. If you let these guys swap roles, meaning, if you want the angle from $(0,1)$ to $(1,0)$, then call $z = i$ and $w = 1$. So: $$\arg(w\overline{z}) = \arg(-i) = 270º.$$
Hence $v_2$ is obtained from $v_1$ after a rotation of $270º$ counterclockwise (or equivalently, $90º$ clockwise).
I'm assuming a uniform continuous distribution, assuming the probability distributions of the two points are independent, and assuming you mean "close to orthogonal" rather than strictly orthogonal.
It is true that "all locations on the sphere are equally probable candidates for both points." Assuming independent distributions,
all locations are equally probable for the second point even given a known position for the first point.
The point is that the part of the sphere close to the "equator" that you randomly selected when you placed the first point
is much larger than the part that is close to that point or to the antipodal point.
If you go to a high enough dimension, most of the sphere will be "close" to the "equator," for whatever your definition of "close" is.
The remarks above, of course, amount merely to a restatement of the claim in slightly different terms.
Better would be an actual proof.
Suppose we have a hypersphere of $n$ dimensions.
To simplify the calculations, let the radius of the sphere be $1.$
The total volume inside this sphere is therefore
$$
V_n(1) = \frac{\pi^{n/2}}{\Gamma\left(\frac n2 + 1\right)}.
$$
We assume a uniform distribution over the volume inside the sphere. That is, when we pick a random point according to this distribution, the probability that the point lands inside a particular region of hyperspace inside the sphere is proportional to the volume of that region.
Now arbitrarily pick a number $q$ with $0 < q \leq 1$ and consider the probability that the distance $r$ between the center of the hypersphere and the random point will be less than $q.$
This probability is the volume inside the hypersphere of radius $q$ divided by the total volume inside the hypersphere, namely,
$$
P(r \leq q) = \frac{V_n(q)}{V_n(1)}
= \frac{\frac{\pi^{n/2}}{\Gamma\left(\frac n2 + 1\right)} q^n}
{\frac{\pi^{n/2}}{\Gamma\left(\frac n2 + 1\right)}}
= q^n.
$$
Now suppose $n$ is very large, for example, $n = 10000,$ and let's see what is the probability that the random point lands within a distance $q = 0.999$ from the center of the hypersphere. This is
$$
P(r \leq 0.999) = 0.999^{10000} \approx 0.000045173.
$$
So the random point will land in this region less than $0.005\%$ of the time.
The other $99.995\%$ of the time, the point lands in the thin hyperspherical shell within just $0.001$ units from the surface.
This doesn't happen because the point was somehow attracted to the surface of the hypersphere or because we skewed the distribution to something non-uniform to make it close to the surface. This is a distribution that is strictly uniform by volume, with no other influences. But it almost always lands very close to the surface simply because that's where almost all the volume is.
High dimensions are weird!
Now let's randomly choose two points and measure the angle between them.
In order to measure the angle, after the points have already been chosen
(so that we can no longer unduly influence the random placement of the points), we choose one of the points as the pole of a set of spherical coordinates.
We do this only so that we can use the "latitude" in that coordinate system to measure the angle between the two points.
The chance that the angle will be between $\phi_1$ and $\phi_2$
(where these are both angles between $0$ and $180$ degrees)
is then proportional to how much of the sphere lies between the lines of latitude at angles $\phi_1$ and $\phi_2$ from the pole.
A general formula for this probability in high dimensions is rather ugly, I think. It is related to the question
what is the surface area of a cap on a hypersphere?
The probability density, however, is dealt with in the answer to
Distribution of an angle between a random and fixed unit-length $n$-vectors.
If the angle $\phi$ is measure in radians, the density over the range
$0 \leq \phi\leq \pi$ is
$$
f(\phi)=\frac{\sin^{n-2}\phi}{\int_0^\pi\sin^{n-2}\theta\,\mathrm d\theta}.
$$
Now let's take our $10000$-dimensional hypersphere again and let's see what the probability is that the angle is within $1/50$ radian (a little more than one degree) from a right angle. This is
$$
P(\frac\pi2 - \frac1{50} \leq \phi \leq \frac\pi2 + \frac1{50})
= \frac{\int_{\pi/2 - 1/{50}}^{\pi/2 + 1/{50}}\sin^{9998}\phi\,\mathrm d\phi}{\int_0^\pi\sin^{9998}\theta\,\mathrm d\theta}
\approx 0.95449.
$$
So there is a better than $95\%$ chance that the angle will be between
$88.85$ and $91.15$ degrees.
Best Answer
What is “heading”? To me it means you establish a local coordinate system at point $A_1$, and giving the azimuth (and perhaps elevation) of $A_2$ with respect to this coordinate system. So what directions do you have in this coordinate system?
(Getting all signs correct means taking the cross product in the right order. There are many possible conventions. Instead of discussing this at length, just try out one generic example and if the signs are wrong, swap the order in the cross product.)
So now you have three basis vectors of an orthonormal basis, and you can compute the coordinates of your points with respect to this basis simply by computing the corresponding dot products. If I call these coordinates $u,e,n$ (in direction up, east resp. north), then the azimuth would be
$$\tan\theta=\frac en\qquad\theta=\operatorname{arctan2}(e,n)$$
with zero meaning north and positive angles going east from there. If you point a laser in the given direction at $A_1$ and adjust the elevation correctly, you get it to point at $A_2$ unless something (like the earth) gets in the way.