Showing $(v - \hat{v})\,\bot\,v$ $\fbox{Setting}$
Let $V$ be an inner-product space with $v \in V$.
Suppose that $\mathcal{O} = \{u_1, \ldots, u_n\}$ forms an orthonormal basis of $V$.
Let $\hat{v} = \left\langle u_1, v\right\rangle u_1 + \ldots + \left\langle u_n,v\right\rangle u_n$ denote the Fourier Expansion of $v$ with respect to $\mathcal{O}$.
$\fbox{Question}$
How do we show that $\left\langle v - \hat{v}, \hat{v} \right\rangle  = \left\langle v, \hat{v} \right\rangle - \left\langle \hat{v}, \hat{v} \right\rangle$ is $0$?
 A: Hint: Plug in the definition of $\tilde{v}$ and use that $\mathcal{O}$ is an orthonormal basis and that $\langle-,-\rangle$ is bilinear.
A: $$
\begin{align}
\langle v-\hat{v},\hat{v}\rangle
&=\left\langle\color{#C00000}{v}-\color{#00A000}{\sum_{k=1}^n\langle v,u_k\rangle u_k},\color{#0000FF}{\sum_{k=1}^n\langle v,u_k\rangle u_k}\right\rangle\\
&=\left\langle\color{#C00000}{v},\color{#0000FF}{\sum_{k=1}^n\langle v,u_k\rangle u_k}\right\rangle-\left\langle\color{#00A000}{\sum_{j=1}^n\langle v,u_j\rangle u_j},\color{#0000FF}{\sum_{k=1}^n\langle v,u_k\rangle u_k}\right\rangle\\
&=\color{#0000FF}{\sum_{k=1}^n\langle v,u_k\rangle}\langle\color{#C00000}{v},\color{#0000FF}{u_k}\rangle-\color{#00A000}{\sum_{j=1}^n\langle v,u_j\rangle}\color{#0000FF}{\sum_{k=1}^n\langle v,u_k\rangle}\langle\color{#00A000}{u_j},\color{#0000FF}{u_k}\rangle\\
&=\sum_{k=1}^n\langle v,u_k\rangle^2-\color{#00A000}{\sum_{j=1}^n\langle v,u_j\rangle}\color{#0000FF}{\langle v,u_j\rangle}\tag{$\ast$}\\
&=\sum_{k=1}^n\langle v,u_k\rangle^2-\sum_{j=1}^n\langle v,u_j\rangle^2\\[6pt]
&=0
\end{align}
$$
$(\ast)$ since $\langle u_j,u_k\rangle=1$ when $k=j$ and $0$ otherwise.
