I have a unit vector defined by double projection angles, with $\alpha$ being the angle towards the positive $y$-axis from the positive $x$-axis, and $\beta$ being the angle towards the positive $z$-axis from the $xy$-plane.

Now I want to offset the coordinate system (change of basis) using intrinsic $z-y^\prime-x^{\prime\prime}$ Euler angles ($\psi$, $\theta$, $\phi$), as shown below (note: the new $X$-axis is in green, but the new $Y$ and $Z$ axes are not shown):

If the original vector remains the same, how can it be described using projection angles $A$ and $B$, relative to the new $XYZ$ coordinate system?


Original vector defined as a matrix in $xyz$ by the projection angles $\alpha$ and $\beta$:

$${{\vec v}_{xyz}} = \left[ {\matrix{ {\cos \beta \cos \alpha } \cr {\cos \beta \sin \alpha } \cr {\sin \beta } \cr } } \right]$$

Rotation matrix for the change in basis using yaw, pitch, and roll angles:

$${R_{xyz \to XYZ}} = \left[ {\matrix{ {\cos \psi \cos \theta } & {\cos \psi \sin \theta \sin \phi - \sin \psi \cos \phi } & {\cos \psi \sin \theta \cos \phi - \sin \psi \sin \phi } \cr {\sin \psi \cos \theta } & {\sin \psi \sin \theta \sin \phi - \cos \psi \cos \phi } & {\sin \psi \sin \theta \cos \phi - \cos \psi \sin \phi } \cr { - \sin \theta } & {\cos \theta \sin \phi } & {\cos \theta \cos \phi } \cr } } \right]$$

Application of transformation:

$$\eqalign{ & {{\vec V}_{XYZ}} = R\vec v = \cr & \left[ {\matrix{ {\cos \psi \cos \theta \cos \beta \cos \alpha + \cos \psi \sin \theta \sin \phi \cos \beta \sin \alpha - \sin \psi \cos \phi \cos \beta \sin \alpha + \cos \psi \sin \theta \cos \phi \sin \beta - \sin \psi \sin \phi \sin \beta } \cr {\sin \psi \cos \theta \cos \beta \cos \alpha + \sin \psi \sin \theta \sin \phi \cos \beta \sin \alpha - \cos \psi \cos \phi \cos \beta \sin \alpha + \sin \psi \sin \theta \cos \phi \sin \beta - \cos \psi \sin \phi \sin \beta } \cr { - \sin \theta \cos \beta \cos \alpha + \cos \theta \sin \phi \cos \beta \sin \alpha + \cos \theta \cos \phi \sin \beta } \cr } } \right] \cr} $$

Extracting the new double projection angles $A$ and $B$:

$$\eqalign{ & {{\vec V}_Z} = \sin B \cr & B = \arcsin \left( { - \sin \theta \cos \beta \cos \alpha + \cos \theta \sin \phi \cos \beta \sin \alpha + \cos \theta \cos \phi \sin \beta } \right) \cr & {{\vec V}_X} = \cos B\cos A \cr & A = \arccos \left( {{{\cos \psi \cos \theta \cos \beta \cos \alpha + \cos \psi \sin \theta \sin \phi \cos \beta \sin \alpha - \sin \psi \cos \phi \cos \beta \sin \alpha + \cos \psi \sin \theta \cos \phi \sin \beta - \sin \psi \sin \phi \sin \beta } \over {\cos B}}} \right) \cr} $$

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