My physics professor (at a T10 university) was unable to explain this. He mentioned something about a bivector, but fundamentally did not say why $w_{z} = \frac{d\theta}{dt}$, and that is what my question is. I would also like more intuition on why the angular velocity is in the $z$-direction. He said that the accurate representation of a plane is a bivector, but that is complex, so the next best thing is a vector perpendicular to the plane, which makes sense. What does not make sense is that $\theta(t)$ is a measurement in the $xy$-plane, yet the derivative $w_{z} = \frac{d\theta}{dt}$ is a magnitude in the $z$-direction.
Please do not tell me about the right hand rule unless you're going to explain why it works. I have background in lower-division math courses only, single and multi variable calculus, linear algebra, and differential equations, so please direct your explanation at that. I am in the physics series for engineers and am not a physics major, but I am curious why this works.
My question is fundamentally why the magnitude of the vector we are using to characterize our plane, which rotation occurs about, is $d\theta/dt$. Should the derivative of $\theta$ not be tangent to it, rather than perpendicular to the plane in which $\theta$ occurs?
Consider an object circulating the origin, with its position given by
$ \vec{r} = (\cos \theta , \sin \theta , 0) $
Then the velocity is by definition given by $\dfrac{d \vec{r}}{dt}$
$ \vec{v} = \dfrac{d \vec{r}}{dt} = (- \sin \theta , \cos \theta, 0 ) \dfrac{d \theta}{dt} $
Note that $\vec{v}$ is perpendicular to $\vec{r}$. We need to relate the vector $\vec{v}$ and the vector $\vec{r}$. This can be done if we define the angular velocity vector to be
$ \vec{\omega} = (0, 0, \dfrac{ d \theta }{dt} ) $
Then one can see that
$ \vec{v} = \vec{\omega} \times \vec{r} $
And this is the motivation the vector angular velocity is defined as such.