A vector $\vec{r}$ in space has to be expressed using basis $\hat{e}$ as a reference and coordinate $\vec{v}$, such that,

$\displaystyle \vec{r} = \hat{e} \cdot \vec{v} = \sum_{i} \hat{e}_i v_i$

A transform of a vector $R(\vec{r})$ is done by

$\vec{r'} = R(\vec{r}) = \hat{e} \cdot R \cdot \vec{v}$

This transform can be view as

$(\hat{e}\cdot R) \cdot \vec{v} = \hat{e} \cdot ( R \cdot \vec{v})$

The first one is change the basis and keep the coordinate, and the later is change the coordinate and keep the basis.

The Euler angle, defined as,

$\vec{r'} = \hat{e} \cdot R_{z}(\alpha) \cdot R_{y}(\beta) \cdot R_{z}(\gamma) \cdot \vec{v}$

Since the rotational matrix is fixed and not based on any basis, how to understand this transform can be viewed differently.

The implementation of the rotation usually like this:

$\vec{v'} = R_{z}(\alpha) \cdot R_{y}(\beta) \cdot R_{z}(\gamma) \cdot \vec{v}$

And the rotation order is $\alpha \to \beta \to \gamma$. In this order, the rotation is using body (or current, intrinsic) axis.

If the rotation order is $\gamma \to \beta \to \alpha$, the rotation is using fixed (or global, external, Laboratory) axis.

To understand this, we have to add back the basis $\hat{e}$. When using the body axis, the left most matrix actually act on the basis (operationally)

$\hat{e}\cdot R_{z}(\alpha) \cdot R_{y}(\beta) \cdot \vec{v} \to \hat{e'} \cdot R_y(\beta) \cdot \vec{v}$

When using the fixed axis,

$\hat{e}\cdot R_{z}(\alpha) \cdot R_{y}(\beta) \cdot \vec{v} \to \hat{e} \cdot R_z(\alpha) \cdot \vec{v'} \to \hat{e} \cdot \vec{v''}$

Thus, if we using Euler angle $\alpha \to \beta \to \gamma$ in fixed frame, the correct matrix is

$R_{z}(\gamma) R_y(\beta) R_z(\alpha)$

In the following, we use fixed frame, unless specified.

Since using Euler angle $\alpha \to \beta \to \gamma$ in changing frame, the rotation $\gamma$ is fixed on the new axis $\hat{n}(\theta = \beta , \phi = \alpha)$. However, the matrix

$R(\hat{n}, \omega) = R_z(\phi) \cdot R_y(\theta)\cdot R_z(\omega)$

is NOT a rotation on the axis with the angle. As we can see that, for a zero rotation $\omega = 0$, any vector will still be transformed by $R_z(\phi) \cdot R_y(\theta)$ .

Now, suppose we have a rotation axis $\hat{n}$ and an rotation angle $\omega$, the rotation matrix can be constructed as follow.

Notice that the rotational matrix $R$ is a unitary transform, its eigen-vectors are $P$,

$\displaystyle R\cdot P = P \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & a + i b & 0 \\ 0 & 0 & a-i b \end{pmatrix}$

$a^2 + b^2 = 1$

$|\textrm{Arg}(a + i b)| = \omega$

Using $\omega$, solve $a = \cos(\omega) , b = \sin(\omega)$.

Since the eigen vector of $R_z(\alpha)$ are

$\displaystyle p_0 = \hat{e_z}, p_{\pm} = \frac{1}{2}(\hat{e_x}\pm i \hat{e_y})$

Since $p_0 \to \hat{n}(\theta, \phi)$, thus,

$\displaystyle p_{\pm} \to \frac{1}{2}( \hat{e_x'} \pm i \hat{e_y'} )$

such that

$\hat{e_i'} = R_z(\phi)\cdot R_y(\theta) \cdot \hat{e_i}$

Notice that the corresponding eigen vector for the eigen value $a \pm i b$ is $p_{\mp}$.

Then the rotational matrix is

$R = P\cdot D\cdot P^{-1}$

Another to look at this problem is that, an operator, can be “decomposed” into its eigen vectors and eigen-values, using Dirac notation,

$\displaystyle R(\omega) = \sum_{i} |v_i \rangle \omega_i \langle v_i |$

The rotation round the vector can be transformed use $R_z(\omega)$ from $(0,0) \to (\theta, \phi)$, Thus, the vector  has to be transform, so the operator

$\displaystyle R(\hat{n}, \omega) = \sum_{i} R_z(\phi) R_y(\theta) |v_i \rangle \omega_i \langle v_i | R_y(-\theta) R_z(-\phi)$

$R(\hat{n}, \omega) = R_z(\phi) \cdot R_y(\theta) \cdot R_z(\omega) \cdot R_y(-\theta) \cdot R_z(-\phi)$

This is a relation between fixed frame rotation to a body rotation! $\hat{n}$ can be the body z-axis, y-axis, or x-axis in particular.

For example, Let $\hat{n}$ be the rotated y-axis, such that

$\displaystyle \hat{n} = R_z(\alpha) \cdot \hat{y} , (\theta, \phi) = (\frac{\pi}{2}, \alpha+\frac{\pi}{2})$

$\displaystyle R(\hat{y'}, \omega) = R_z(\alpha+\frac{\pi}{2}) \cdot R_y(\frac{\pi}{2}) \cdot R_z(\omega) \cdot R_y(-\frac{\pi}{2}) \cdot R_z( -\alpha - \frac{\pi}{2})$

Notice that

$R_y(\frac{\pi}{2}) \cdot R_z(\omega) \cdot R_y(-\frac{\pi}{2}) = R_x(\omega)$

$R_z(\frac{\pi}{2}) \cdot R_x(\omega) \cdot R_z(-\frac{\pi}{2}) = R_y(\omega)$

The above formula are not difficult to imagine, decompose the middle operator into the eigen-system and use Dirac notation. For example, the rotate of the z-axis around y-axis with 90 degree is the x-axis. Thus,

$\displaystyle R(\hat{y'}, \omega) =R_{yB}(\omega) = R_z(\alpha) \cdot R_y(\omega) \cdot R_z(-\alpha)$

We recovered the relation of the Lab frame to body frame for y-axis!

Also, given a vector $\vec{r}$ rotated by $\omega$ to $\vec{r'}$ around unit vector $\hat{n}$ is

$\vec{r'} = \vec{r} \cos(\omega) + \hat{n} (\hat{n}\cdot \vec{r}) (1-\cos(\omega)) + \hat{n} \times \vec{r} \sin(\omega)$