Linear velocity is cross product of angular velocity and position 
Why is linear velocity is cross product of angular velocity and position?
 A: If you take $\bf d\theta$ to be the (infinitesimal) angle swept about the rotation axis (about axial vector $\bf\omega$ in diagram) after a small displacement $d\bf r_j'$ then looking at the diagram, you can form the cross product $$\tag 1d\bf r_j'=d\theta\times r_j'$$
If this displacement occurs in a time interval given by $dt$ and you divide both sides of equation (1) by $dt$ you get $$\bf\frac{d\bf r_j'}{dt}=\frac{d\theta}{dt}\times r_j'$$ or $$\bf v=\omega\times r$$
A: 
2D case
the components of the position vector to  the point P, given in inertial system are
\begin{align*}
  &\vec{R}=\begin{bmatrix}
           x \\
           y \\
           z \\
         \end{bmatrix}=
\underbrace{\begin{bmatrix}
  \cos(\phi(t)) & -\sin(\phi(t)) & 0 \\
  \sin(\phi(t)) & \cos(\phi(t)) & 0 \\
  0 & 0 &1 \\
  \end{bmatrix}}_{\mathbf S}\,\begin{bmatrix}
           u_x \\
           u_y \\
           0 \\
         \end{bmatrix}
\end{align*}
thus the velocity
\begin{align*}
  &\vec{v}=\frac{d}{dt}\,\vec{R}=\frac{d}{dt}\,\phi(t)
  \begin{bmatrix}
  -\sin(\phi(t)) & -\cos(\phi(t)) & 0 \\
  \cos(\phi(t)) & -\sin(\phi(t)) & 0 \\
  0 & 0 &0 \\
  \end{bmatrix}\,\begin{bmatrix}
           u_x \\
           u_y \\
           0 \\      \end{bmatrix}=\vec{\omega}\times\vec{R}\quad,\text{where}\\
&  \vec{\omega}=  \begin{bmatrix}
           0 \\
           0 \\
           \frac{d}{dt}\,\phi(t) \\
         \end{bmatrix}     
\end{align*}
3D case
\begin{align*}
  &  \vec{R}=\mathbf{S}\left[~\phi_x~,\phi_y,~\phi_z~\right]\,\vec{u}
  \quad\Rightarrow\\
  &\vec v=\vec{\dot{R}}=\underbrace{\left[\frac{\partial\,\mathbf{S} }{\partial \phi_x}\,\dot{\phi}_x+
  \frac{\partial\,\mathbf{S} }{\partial \phi_y}\,\dot{\phi}_y+
  \frac{\partial \,\mathbf{S}}{\partial \phi_z}\,\dot{\phi}_z\right]}_{\mathbf{\dot{S}}}\,\vec u
  \overset{!}{=}\vec{\omega}\,\times\vec{R}\\
  &\text{where}\quad
  \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}}
\\ \omega_{{z}}&0&-\omega_{{x}}\\  
-\omega_{{y}}&\omega_{{x}}&0\end {array} \right]=\mathbf{\dot{S}}\mathbf{S}^T
\end{align*}

*

*$\mathbf S~$ is the transformation matrix between body-system and inertial-system

A: It's true by definition, there's really nothing more to it than that.
Angular velocity does not have to be represented by a vector, though it can be in 3 dimensions. We choose to represent it by the vector
$$\vec{w} = \vec{r} \times \vec{v}\tag1$$
And therefore following the cyclic properties of any cross-product, it must be true that
$$\vec{v} = \vec{w} \times \vec{r}\tag2$$
A: Angular velocity is a vector quantity and has both a magnitude and a direction. The direction is the same as the the angular displacement direction from which we defined the angular velocity. Where r is position vector and linear velocity is also vector quantity.
The resultent of vector product of any two vectors is a vector quantity.
