3D Extension of a Fun Geometric Series Puzzle After being inspired by this question, and in particular Semiclassical's excellent response and generalisation, I thought of another generalisation to a 3-dimensional plane.:

Suppose you start at at some point in space, and then you move a
  distance $a$ meters from that point. Then, you rotate at an angle of
  elevation $\phi$ and azimuthal angle $\theta$ relative to your current
  position, and move forward $ar$ meters from that point, for $|r|<1$.
  You then rotate and move $ar^2$, then $ar^3$, and so forth. As you
  keep doing this ad infinitum, how far are you from your starting
  point?

Here's my idea for it. Suppose that we start at the point $(0,0,0)$ and our first movement takes us to $(0,0,a)$. The path to the next point is along the vector $(ar\sin\theta\sin\phi,ar\sin\theta\cos\phi,ar\cos\theta)$, the next one is along the vector $(ar^2\sin2\theta\sin2\phi,ar^2\sin2\theta\cos2\phi,ar^2\cos2\theta)$, and so forth. Hence, the distance from the origin is the magnitude
$$a\left\|\sum_{k=0}^\infty\pmatrix{r^k\sin k\theta\sin k\phi\\r^k\sin k\theta\cos k\phi\\r^ k\cos k\theta}\right\|$$
which unfortunately doesn't turn out very nice. Is there a simpler way to do this? Based on Semiclassical's response above, I imagine one could do it with quaternions, but I have no expericence with those. Is there an elegant solution to this problem?
 A: Introduction
You may find the solution elegant; the results, probably not.
A few results are needed in order to perform the summation of the vectors. Trigonometric identities we will need are:
$\sin{k\theta}\ \sin{k\phi} = \dfrac{\cos(k\theta-k\phi) - \cos(k\theta+k\phi)}{2} \tag{1}$
$\sin{k\theta}\ \cos{k\phi} = \dfrac{\sin(k\theta-k\phi) + \sin(k\theta+k\phi)}{2} \tag{2}$
Euler's formula for imaginary exponentials:
$e^{ix} = \cos{x} + i\ \sin{x}, \cos{x} = \operatorname{Re}(e^{ix}), \sin{x} = \operatorname{Im}(e^{ix})\ \ (\forall x \in \mathbb{R}) \tag{3}$
Also, for integer values of $k$ (basically de Moivre's formula)
$(e^{ix})^k = e^{ikx} = e^{i(kx)} = \cos{kx} + i\ \sin{kx}\ \ (\forall k \in \mathbb{Z}) \tag{4}$
Finally, 
$\Large\sum\limits_{k=0}^{\infty}\normalsize{z^k} = \dfrac{1}{1-z}\ \text{ for } |z|<1, z \in \mathbb{C} \tag{5}$
Summation of Components
Easiest First ($v_3$)
By (3), (4), (5) provided $|r|<1$ we have
$v_3 = \Large\sum\limits_{k=0}^{\infty}\normalsize{r^k\cos k\theta} = \Large\sum\limits_{k=0}^{\infty}\normalsize{r^k \operatorname{Re}(e^{ik\theta})} = \operatorname{Re}\Bigg(\Large\sum\limits_{k=0}^{\infty}\normalsize{(re^{i\theta})^k}\Bigg) = \operatorname{Re}\Bigg(\normalsize{\dfrac{1}{1-re^{i\theta}}}\Bigg)$ so
$v_3 = \operatorname{Re}\Bigg(\normalsize{\dfrac{1}{(1-r\cos\theta)-ir\sin\theta}}\Bigg) = \operatorname{Re}\Bigg(\normalsize{\dfrac{(1-r\cos\theta)+ir\sin\theta}{(1-r\cos\theta)^2+r^2\sin^2\theta}}\Bigg)$ 
Hence
$v_3 = \dfrac{1 - r\cos\theta}{1+r^2-2r\cos\theta} \tag{6}$
Then $v_1$
By (1) we have
$v_1 = \Large\sum\limits_{k=0}^{\infty}\normalsize{r^k\sin k\theta\sin k\phi} = \dfrac{1}{2}\Large\sum\limits_{k=0}^{\infty}\normalsize{r^k\cos(k(\theta-\phi))} + \dfrac{1}{2}\Large\sum\limits_{k=0}^{\infty}\normalsize{r^k\cos(k(\theta+\phi))}$
which consists of two terms of the same form as $v_3$, so can be expressed via (6) as
$v_1 = \dfrac{1}{2}\cdot\dfrac{1 - r\cos(\theta-\phi)}{1+r^2-2r\cos(\theta-\phi)} - \dfrac{1}{2}\cdot\dfrac{1 - r\cos(\theta+\phi)}{1+r^2-2r\cos(\theta+\phi)} \tag{7}$
This may or may not reduce nicely. That may depend on your standard for "nicely".
And a Bit More Work for $v_2$
First we establish an identity for the sum of sines (call this $s$).
Similar to before we have
$s = \Large\sum\limits_{k=0}^{\infty}\normalsize{r^k\sin k\theta} = \Large\sum\limits_{k=0}^{\infty}\normalsize{r^k \operatorname{Im}(e^{ik\theta})} = \dots = \operatorname{Im}\Bigg(\normalsize{\dfrac{(1-r\cos\theta)+ir\sin\theta}{(1-r\cos\theta)^2+r^2\sin^2\theta}}\Bigg)$ 
Hence
$s = \dfrac{r\sin\theta}{1+r^2-2r\cos\theta} \tag{8}$

By (2) we have
$v_2 = \Large\sum\limits_{k=0}^{\infty}\normalsize{r^k\sin k\theta\cos k\phi} = \dfrac{1}{2}\Large\sum\limits_{k=0}^{\infty}\normalsize{r^k\sin(k(\theta-\phi))} + \dfrac{1}{2}\Large\sum\limits_{k=0}^{\infty}\normalsize{r^k\sin(k(\theta+\phi))}$
Now applying (8) to each of the terms:
$v_2 = \dfrac{1}{2}\cdot\dfrac{r\sin(\theta-\phi)}{1+r^2-2r\cos(\theta-\phi)} + \dfrac{1}{2}\cdot\dfrac{r\sin(\theta+\phi)}{1+r^2-2r\cos(\theta+\phi)} \tag{9}$
