 Solution

An orthogonal projector sends $x$ to the foot $Px$ of its perpendicular onto the image of a linear map $\mathrm{range}P$. The residual $r=x-Px$ is orthogonal to the entire range, not just to the single vector $Px$. Indeed, for any range vector $Pv$,In particular $\boxed{(Px{)}^T(x-Px)=0}$. The orthogonal projection diagram illustrates the resulting right triangle.