I must be doing something wrong.
Let $V$ be a finite dimensional inner product space over $\mathbb{R}$ or $\mathbb{C}$, $v \in V$, $W \subseteq V$ a subspace of $V$, $w \in W$, and $u= \frac{w}{\|w\|} = \frac{w}{\sqrt{\langle w,w\rangle}}$.
I want to prove that $\frac{\langle v,w \rangle}{\langle w,w \rangle} = \langle v,u \rangle$.
We see that $w = \|w\|u$, so the left hand side becomes
$$ \begin{aligned} \frac{\langle v,w \rangle}{\langle w,w \rangle} &= \frac{\langle v,w \rangle}{\|w\|^2} \\ &= \frac{\langle v,\|w\|u \rangle}{\|w\|^2} = \frac{\overline{\langle \|w\|u,v \rangle}}{{\|w\|}^2} \\ &= \frac{\|w\|\overline{\langle u,v \rangle}}{\|w\|\|w\|} = \frac{\overline{\langle u,v \rangle}}{\|w\|} \\ &= \frac{\langle v,u \rangle}{\|w\|} \end{aligned} $$
But $\frac{\langle v,u \rangle}{\|w\|}$ is not always equal to $\langle v,u \rangle$!
Responding to ad2004's comment helped me see that I did indeed make a mistake.
I wrote "I want to prove that $\frac{\langle v,w \rangle}{\langle w,w \rangle} = \langle v,u \rangle$" and found that $\frac{\langle v,w \rangle}{\langle w,w \rangle} = \frac{\langle v,u \rangle}{\|w\|}$, so what I was trying to prove is not always true.
My mistake was forgetting the full projection formula: $\left(\frac{\langle v,w \rangle}{\langle w,w \rangle}\right)w = v^{\|} = \langle v,u \rangle u$. So what I actually wanted to prove is $\left(\frac{\langle v,w \rangle}{\langle w,w \rangle}\right)w = \langle v,u \rangle u$.
Since $\frac{\langle v,w \rangle}{\langle w,w \rangle} = \frac{\langle v,u \rangle}{\|w\|}$, the left hand side becomes $\frac{\langle v,u \rangle}{\|w\|}w = \langle v,u \rangle u$ as desired.