Well, I know that orthogonal projection is a linear transformation.
Let $t$, $n$ be two orthogonal (with $\|n\| = 1 =\|t\|$) non zero vectors that span a plane $\pi$. Then $T: \mathbb{R}^3 \to \mathbb{R}^3$ given by $ T(u) = {\rm Proj}_{\pi}(u) = \langle u,n\rangle n + \langle u,t\rangle t$ is a linear transformation.
I'm trying to show that for every $x,y \in \mathbb{R}^3$, $\langle x,y\rangle = \langle T(x),T(y)\rangle$.
What I've got: $$\langle T(x),T(y)\rangle = \langle (\langle x,n\rangle n + \langle x,t\rangle t),(\langle y,n\rangle n + \langle y,t\rangle t)\rangle =\langle x,n\rangle\langle y,n\rangle +\langle x,t\rangle \langle y,t\rangle,$$ since $u$ and $v$ are a orthonormal basis. I got stuck here.
Any help would be nice. Thank you.
The orthogonal projection is not, in general, an orthogonal map. Take for instance ${\bf n} = (1,0,0)$ and ${\bf t} = (0,1,0)$. Then $T(x,y,z)= (x,y,0)$, and although $\langle (1,0,1), (0,0,1)\rangle = 1$, we have $$\langle T(1,0,1), T(0,0,1)\rangle = \langle (1,0,0), (0,0,0)\rangle = 0 \neq 1.$$
Also, note that in general you need $\langle {\bf n},{\bf t}\rangle = 0$ for the relation ${\rm proj}_\pi({\bf u}) = \langle {\bf u},{\bf n}\rangle {\bf n} + \langle {\bf u},{\bf t} \rangle {\bf t}$ to hold.