Let $x$ and $y$ be two vectors in $\mathbb{R}^d$ with $\mid x \mid = \mid y \mid$. Find a unit vector $u$ such that $P_u x = y$ Let $u \in \mathbb{R}^d = V$ be a unit vector and set $W = \text{span}(u) ^{\bot}$ (with respect to the dot product). The reflector across W is $P_u = I_d - 2uu^T$.
Let $x$ and $y$ be two vectors in $\mathbb{R}^d$ with $\mid x \mid = \mid y \mid$. Find a unit vector $u$ such that $P_u x = y$.
I know that $P_u$ is orthogonal, $v \in V \Rightarrow v = cu + w$ unique $c \in \mathbb{R}$ and $w\in W$, and $Pv = -cu + w$.
If $P_u x = -cu + w = y$, then $u = \frac{-1}{c} (y - w)$. I would need to show that this is a unit vector, but this is where I am stuck. I'm not sure if this is the route I want to take or if there is something else that I am missing. 

If $x=y$ choose any non-zero vector perpendicular to $x$.  Otherwise $u:=(x-y)/\|x-y\|$. In this case
$$x\mapsto x-2\frac{\langle x,x-y\rangle}{\|x-y\|^2}(x-y)=\frac{\|x\|^2x-2\langle x,y\rangle x+\|y\|^2x-(2\|x\|^2x-2\|x\|^2y-2\langle x,y\rangle x+2\langle x,y\rangle y)}{\|x\|^2-2\langle x,y\rangle+\|y\|^2}.$$
As $\|x\|=\|y\|$ this dramatically simplifies to
$$\frac{2\|x\|^2y-2\langle x,y\rangle y}{2\|x\|^2-2\langle x,y\rangle}=y.$$ 
