Instantaneous Axis Of Rotation
Let a Cartesian coordinate system be attached to a rigid body, say with its origin at the center of gravity of that body.
Then, because the body is rigid, any movement of the coordinate system will perserve the length of the base vectors and
the right angles between them. That is: any movement with respect to the bodies' origin is described by an orthogonal matrix
(with positive determinant, because volumes are preserved as well). Such movements are commonly called rotations.
Rotations in 3-D space are represented by:
$$
\vec{r_1}(t) = R(t)\,\vec{r} \qquad ; \qquad
\left[ \begin{array}{c} x_1 \\ y_1 \\ z_1 \end{array} \right](t) =
\left[ \begin{array}{ccc}
R_{11}(t) & R_{12}(t) & R_{13}(t) \\ R_{21}(t) & R_{22}(t) & R_{23}(t) \\ R_{31}(t) & R_{32}(t) & R_{33}(t)
\end{array} \right]
\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]
$$
Thus it is assumed that the vector $\vec{r}$ is a constant,
$R(t)$ is a real valued orthogonal $3\times 3$ matrix and its coefficients are functions
of time $t$ . Because $R$ is orthogonal, it is known that this matrix times
its transpose is equal to the identity matrix (transpose = inverse). Differentiation
according to the product rule and known rules for matrix transposition then gives:
$$
R(t)R^T(t) = I \quad \Longrightarrow \quad \dot{R}(t)R^T(t) + R(t)\dot{R}(t)^T = 0
\quad \Longrightarrow \quad \dot{R}(t)R^T(t) = - \left[\dot{R}(t)R^T(t)\right]^T
$$
Define:
$$
\Omega(t) = \dot{R}(t)R^T(t) \qquad \Longrightarrow \qquad \Omega(t) = -\Omega^T(t)
$$
It can be easily shown that the $\Omega$ matrix must have the following form:
$$
\left[ \begin{array}{ccc}
0 & - \omega_z & +\omega_y \\ +\omega_z & 0 & -\omega_x \\ -\omega_y & +\omega_x & 0
\end{array} \right]
$$
The vector $\;\vec{\omega}$ , not at all by coincidence, is defined in such a way that:
$$
\vec{\omega} = \left[ \begin{array}{c} \omega_x \\ \omega_y \\ \omega_z \end{array} \right]
\quad \Longrightarrow \quad \Omega\,\vec{r} =
\left[ \begin{array}{ccc}
0 & - \omega_z & +\omega_y \\ +\omega_z & 0 & -\omega_x \\ -\omega_y & +\omega_x & 0
\end{array} \right]
\left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =
\left[ \begin{array}{c}
\omega_y z - \omega_z y \\ \omega_z x - \omega_x z \\ \omega_x y - \omega_y x
\end{array} \right] = \vec{\omega} \times \vec{r}
$$
Where $\times$ denotes the usual cross product of two vectors. Consequently:
$$
\dot{\vec{r}}_1(t) = \dot{R}(t)\vec{r} = \dot{R}(t)R^T(t)R(t)\vec{r}=\vec{\omega}(t)\times\vec{r}_1(t)
$$
Thus the velocity of a vector rotating in time is a rotation of the same vector around an axis
$\vec{\omega}(t)$ that itself is moving in time. Especially the fact that in the general case everything is
"moving in time" makes it not very easy to comprehend conceptually.