Answer
Both $\mathbf{u}$ and $\mathbf{v}$ are always unit vectors.
Work Step by Step
All we need to do here is find the magnitude of both $\mathbf{u}$ and $\mathbf{v}$.
Magnitude of $\mathbf{u}$:
$\|\mathbf{u}\|=\sqrt{u_x^2+u_y^2}$
$=\sqrt{\cos^2{\theta}+(-\sin{\theta})^2}$
$=\sqrt{\cos^2{\theta}+\sin^2\theta}=\sqrt{1}=1$
So no matter what $\theta$ is, $\|\mathbf{u}\|=1$ and therefore $\mathbf{u}$ is always a unit vector.
Similarly
$\|\mathbf{v}\|=\sqrt{v_x^2+v_y^2}$
$=\sqrt{\cos^2\theta+\sin^2\theta}=\sqrt{1}=1$
By the same reasoning as above, $\mathbf{v}$ is always a unit vector.
