How is angular velocity defined in this case (and why)? Suppose we have two particles with trajectories
$$ \vec{r}_{1}(t) = (\cos ct, \sin ct, 1) \quad\text{ and }\quad \vec{r}_{2}(t) = (-\cos ct, -\sin ct, -1). $$
On one hand, we could say the angular velocity of the two particle system is $\vec{\omega} = c \,\vec{e}_{z}$, because both particles orbit around the $z$-axis at angular frequency $c$.
On the other hand, $\vec{r}_{1}\times \dot{\vec{r}}_{1} = (-c\cos ct, -c\sin ct, c)$ and $\vec{r}_{2}\times \dot{\vec{r}}_{2} = (-c\cos ct, -c\sin ct, c)$. Both of these vectors point away from the $z$-axis and are continuously changing direction. It would be a mistake to add these vectors, but nonetheless this makes it seem like $\vec{\omega} \ne c \,\vec{e}_{z}$.
Which of these two conclusions is correct? More to the point, I'm not sure I understand angular velocity as well as I thought, so what exactly is the correct, consistent way to define angular velocity in the abstract? It's not clear in the general case.
 A: Let's talk about angular momentum first, because that's what $\vec{r}\times\dot{\vec{r}}$ is (up to a factor of the mass). If you think about what is the angular momentum of a particle moving in a straight line, it should be clear that the outcome depends on where your origin is. It's the same with circular motion. In general, angular momentum is defined with respect to some origin.
(Angular momentum conservation arises when the environment is isotropic with respect to the chosen origin. If particles are moving in the circles you describe, this is clearly not the case. Still, that motion could arise in a environment that is invariant with respect to rotations about the $z$ axis only, e.g. the particles could be orbiting a $z$-oriented line of charge, or they could be moving in a $z$-oriented magnetic field. Then the $z$ component of the angular momentum would be conserved, and you find this to be the case.)
The same idea applies to angular velocity. You can think of the angular velocity about a particular origin (i.e. orbital angular velocity), which is just your $(\vec r\times \dot{\vec r})/r^2$. As you show, for a particle moving in a circle about the $z$ axis, the angular velocity about an origin point that is not the circle's center is not simply a constant pointing along $\vec e_z$.
Alternatively, you can also think of angular velocity about an axis, which is often relevant for rigid bodies. The Wikipedia article on angular velocity has some discussion of the distinction.
A: It is sometimes worth drawing a diagram(s) to visualise the situation with both undergoing anticlockwise rotations about the z-axis.

The first to note is that $\dot z =  0$ and nor $c$.
Hopefully from the diagrams and doing the appropriate sums you can show that $\omega_{\rm z}$ is the same for both motions.
A: We could write $\vec r_2= - \vec r_1$.
Particle 1 is rotating anticlockwise around z axis in a plane parallel with $(x,y)$ at $z=1$.
Particle 2 is rotating anticlockwise around z axis in the same plane but is diametrically opossed to particle 1.
Both particles 1 and 2 have the same angular speed $\vec \omega= c*\vec e_3$.
(The angular speed is perpendicular to the plane of rotation when the particle rotates with constant speed)
A: with:
$$ \vec R_1=\left[ \begin {array}{c} \cos \left( ct \right) \\ 
\sin \left( ct \right) \\ 1\end {array} \right]\quad,
\vec v_1=\left[ \begin {array}{c} -\sin \left( ct \right) c
\\  \cos \left( ct \right) c\\  0
\end {array} \right] \\
\vec R_2=\left[ \begin {array}{c} -\cos \left( ct \right) 
\\  -\sin \left( ct \right) \\  -1
\end {array} \right] \quad,
\vec v_2=\left[ \begin {array}{c} \sin \left( ct \right) c
\\  -\cos \left( ct \right) c\\  0
\end {array} \right] 
$$
hence
$$\vec\omega_1=\frac{\vec R_1\times\vec v_1}{r_1^2} =\frac 12 c \left[ \begin {array}{c} \,-\cos \left( ct \right)  
\\   \,-\sin \left( ct \right)  
\\ 1\end {array} \right] \\
\vec\omega_2=\frac{\vec R_2\times\vec v_2}{r_2^2}=\vec\omega_1
$$
where $~r_1=r_2=\sqrt{2}$
now  the angular velocity is:
$$\vec \omega_1 =\omega_1\,\hat\omega_1=\frac 12 c\sqrt{2}\,
\left[ \begin {array}{c} -\frac 12\,\sqrt {2}\cos \left( ct \right) 
\\  -\frac 12\,\sqrt {2}\sin \left( ct \right) 
\\  +\frac 12\,\sqrt {2}\end {array} \right] \ne {c\,\vec e_z}
$$
so your first equation for the angular momentum is wrong
but you obtain equal results with this equation
$$\omega_1=\frac{|\vec v_1|}{r_1}=\frac{c}{\sqrt{2}}=\frac 12\,\sqrt{2}\,c$$
