What does $\vec{\omega}\times\vec{r}$ equals to in circular motion? I know that $\vec{v}=wr\hat{\theta}$ in uniform circular motion. This equation looks like a result of a cross product.
Yesterday, I started to learn Basic Dynamics of Rigid Bodies. My teacher wrote $\vec{v}=\vec{\omega}\times\vec{r}$ in the lecture. But I have never seen this. This equation also equals to $\vec{v}={\omega}{r}\sin(\vec{\omega},\vec{r})\hat{?}$ If $\sin(\vec{\omega},\vec{r})$ equals to $\sin(\pi/2)$ then the equation equals to $\vec{v}={\omega}{r}\hat{?}$
I wonder that how the unit vector $\hat{?}$ is equal to $\hat{\theta}$. When I try to figure it out, I can't. Can you explain it, please?
 A: Your first equation works when you have already isolated the plane of rotation.  It treats $\omega$ as a scalar.  Sometimes we can't isolate it into a 2d case like this, such as if there are angular accelerations or other considerations.  To handle the full 3d case, we define rotation using a vector, $\vec{\omega}$.  This vector is defined to have a magnitude equal to the $\omega$ from the first equation, and a direction which is is at right angles to the rotation.
Now, if you're a purist, that cross product might bug you.  In these cases, we're not actually using the cross product of two vectors, but the product of a bivector, which is easier to trace down to why it is the correct tool to use.  It just so happens that in 3 dimensions, the math for cross products and bivectors is identical, and historically we've found teaching cross products easier than introducing bivectors.
A: you can write any  vector with his magnitude and unity direction
$$\vec{v}=|\vec{v}|\,\vec{n}\tag 1$$ 
with
$|\vec{v}|=|\vec{\omega}|\,|\vec{r}|\,\sin(\theta) $
where $\theta$ is the angle between $\vec{\omega}$ and $\vec{r}$
thus equation (1) 
$$\vec{v}=|\vec{\omega}|\,|\vec{r}|\,\sin(\theta)\,\vec{n}$$
where the vector n is perpendicular to the vector omega  and r 
$\vec{n}\perp\vec{\omega}\quad ,\vec{n}\perp\vec{r}$ with $||\vec{n}||=1$
