Vector Approximation on The Unit Sphere I am trying to solve the following approximation problem on the general unit sphere $S^{n-1} \subseteq \mathbb{R}^n$. First, let $\theta \in (0, \pi)$ be a fixed angle, and $r \in S^{n-1}$ a fixed unit vector. Then you can define a subset $R \subseteq S^{n-1}$ containing all the vectors that are at most an angle of $\theta$ away from $r$.
\begin{equation}
R = \{x \in S^{n-1} : \langle x, r \rangle \ge \cos(\theta)\}
\end{equation}
The approximation problem I am trying to solve is now as follows.

Given an arbitrary $v \in S^{n-1}$, find a vector $w \in R$ that is closest to $v$. In other words, solve the following minimization problem:
\begin{equation} w = \text{argmin } \|v - x\|_2 \quad\text{for all } x \in R \end{equation}

I have tried finding the solution intuitively, by considering what happens in $\mathbb{R}^3$, and basically hoping my solution there generalizes to $\mathbb{R}^n$. The proposed solution I came up with is as follows.

First, if $v \in R$, then the solution is trivially given by $w = v$. So now assume $v \notin R$, which means $\langle v, r \rangle < \cos(\theta)$. Then define $x = v - \langle v, r \rangle r$, consequently the solution is given by
  \begin{equation} w = \frac{\sin(\theta)}{\|x\|} \cdot x + \cos(\theta) \cdot r \ \end{equation}

The idea is, if $v \notin R$, you assume the solution $w$ will be both in the plane spanned by $v$ and $r$, and also on the "border" of the set $R$, in other words $\langle w, r \rangle = \cos(\theta)$. So you consider the affine hyperplane which passes through the point $\cos(\theta) \cdot r$ and is perpendicular to the vector $r$. This is basically the hyperplane that "slices" your unit sphere between $R$ and $R^c$ and thus contains the aforementioned border. If you intersect this hyperplane with the plane spanned by $v$ and $r$, eventually you should get the solution I came up with, assuming my calculations were correct.
Now I have been able to prove that my solution is both contained in $S^{n-1}$, and $R$, simply by working out the inproducts $\langle w, w \rangle = 1$ and $\langle w, r \rangle = \cos(\theta)$. However I cannot seem to prove the original statement, that this $w$ indeed minimizes the norm $\|v - x\|_2$. To cut down potential work done by people trying to help, I'm going to put some of these results down here as well. Below I will define $\langle v, r \rangle = \cos(\phi) < \cos(\theta)$, and also introduce a vector $u \in R$ such that $\|u - v\| \le \|w - v\|$ or equivalently, $\langle u, v \rangle \ge \langle w, v \rangle$. If all my work is correct, the following statements should be true.
\begin{align*}
\langle v, w \rangle &= \sin(\theta)\sin(\phi) + \cos(\theta)\cos(\phi) \\
\langle u, w \rangle &= \frac{\sin(\theta)}{\sin(\phi)}\langle u, v \rangle - \frac{\sin(\theta)}{\tan(\phi)} \langle u, r \rangle + \cos(\theta)\cos(\phi) \\
&\ge 1 + \cos(\theta)(\cos(\phi) - \cos(\theta))
\end{align*}
Can anyone help me figure out how to prove my solution indeed works? Or alternatively, point out why it doesn't?
Footnote: In writing this I became aware of the pathological case where $v = -r$, which would mean the solution set contains the entire border of $R$ (right?). This leaves me wondering if this is the only case when multiple solutions exist or not. If it seems that multiple solutions can be found in non-pathological cases, it would be of great help if I were able to explicitly define the entire solution set, if at all possible.
 A: Okay, I think I might have figured it out. I would kindly request someone to look over my work to make sure I haven't made any mistakes. The trick was to work towards $u$'s unit length in the system of inequalities. Here goes...
Assume $u \in R$ is a better solution than $w$, i.e. $\langle u, v \rangle \ge \langle w, v \rangle$. Since inproducts are invariant under orthogonal transformations, we can perform such a transformation to allow $r = e_1$, and $v = (\cos(\phi), \sin(\phi), 0, \dots, 0)$ for some $\phi \in (0, \pi)$ to be in the xy-plane (the latter of which is trivially true in the case $n = 2$). From the definition of $w$ it immediately follows that $w = (\cos(\theta), \sin(\theta), 0 \dots 0)$. Since $u \in R$, it follows that $u_1 = \langle u, r \rangle \ge \cos(\theta)$. Secondly we know that
\begin{equation}
\langle v, u \rangle = \cos(\phi) u_1 + \sin(\phi) u_2
\end{equation}
on one hand, per definition of the inproduct. On the other hand, we also know that
\begin{equation}
\langle v, u \rangle \ge \langle v, w \rangle = \sin(\theta)\sin(\phi) + \cos(\theta)\cos(\phi)
\end{equation}
per the assumption on $u$. Since $\phi \in [0, \pi)$, this means $\sin(\phi) \ge 0$. Putting the two previous (in)equalities together we can deduce that
\begin{equation}
u_2 \ge \sin(\theta) + \frac{\cos(\theta) - u_1}{\tan(\phi)}
\end{equation}
However since we already know that $u_1 \ge \cos(\theta)$, it immediately follows that $u_2 \ge \sin(\theta)$. Since $u$ is supposed to be a unit vector, it can only possibly be $u = (\cos(\theta), \sin(\theta), 0 \dots 0) = w$.
