Distance between projections

Question

Let $x,y,z \in \mathbb R^2$ such that $||x|| = ||y||= ||z|| = 1$. Project $z$ onto the lines defined by $x$ and $y$ as follows: \begin{equation} z_x = (z^\text{T}x) x, \ \ z_y = (z^\text{T}y) y, \end{equation}

Is it necessarily true that $||z_x - z_y|| \leq ||z|| = 1$? If so, how can we prove it?

(If the answer can be generalized to $\mathbb R^n$, that would be cool too...)

Answer 1

$\mathbb{R}^2$

Fix $z$, let $x$ run around the unit circle, and forget $y$ for now.

The locus of projections of $z$ on $x$ is the circle with diameter 1 containing the origin and $z$.

The same locus for $y$.

Since the diameter is $||z||=1$, any two points are no farther away from each other than 1. QED.

$\mathbb{R}^n$

The projection $z_{xy}$ of $z$ on the $xy$ plane is no longer than $z$ and the statement follows from the above for $z_{xy}$.

PS. The details are left for the reader...