Does projection operation preserve perpendicularity?

Question

Let $C$ be a nonempty convex closed subset of $R^n$, and $\pi_C$ be an orthogonal projection mapping from $R^n$ onto $C$. Suppose that $y \in R^n$ is perpendicular to $x \in R^n$. Is it possible to prove/disprove that $y$ is also perpendicular to $\pi_C (x)$ (i.e. the orthogonal projection of $x$ onto $C$)?

Answer 1

No, for example, take standard projections along last coordinate, that is $$P(x) = (x_{1},...,x_{n-1},0)$$ where $x=(x_{1},x_{2},...,x_{n})$.
If x and y are perpendicular then $\sum_{k=1}^{n} x_{k}y_{k} = 0$ and this does not imply that $\sum_{k=1}^{n-1} x_{k}y_{k} = 0$

Answer 2

No. Consider $x=(1,1)$ and $y=(-1,1)$ which has $x \cdot y = 0$ and take $\pi_c(x)$ the projection onto the first coordinate. We can now see that $\pi_c(x)=(1,0)$ so $\pi_c(x) \cdot y = -1 \neq 0$.