Orthogonal projection of p on hyperplane M is given by $p - \langle v,p\rangle v$

Question

I am trying to solve this exercise:

The hyper-plane $M$ contains the zero vector and has a normal vector v with $||v|| = 1$. Show that the orthogonal projection of $p$ on $M$ is given by $p - \langle v,p\rangle v$.

So, the normal vector stands orthogonal on the Hyperplane $M$. But I don't see how to get started.

Thanks for any help!

Answer 1

The orthogonal projection of $p$ onto $M$ will be some vector $p+t v$ such that $p+tv \in M$.

Compute $t$.

You know that $M$ has the form $M=\{ x | \langle v ,x \rangle = \alpha \}$ for some $\alpha$. Since $0 \in M$, you can figure out $\alpha$.