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Relative angular velocities add as vectors: ${\mathbf{\omega }}_{B/C}={\mathbf{\omega }}_{B/A}+{\mathbf{\omega }}_{A/C}$. Geographic pole latitudes and longitudes must first be converted into Cartesian vector directions.

Let $\hat{x}$ point to the equatorial meridian at zero longitude, $\hat{y}$ to the equator at $90^{\circ }$ E, and $\hat{z}$ to the north pole. Use a right-hand positive Euler angular velocity vector. Relative rotations add as vectors, not as geographic coordinates:ConsequentlyHere $\lambda ={\tan }^{-1}(12/5)$ is the longitude of the positive Euler pole, and its latitude $\varphi$ is zero. The antipodal pole with the opposite angular-velocity sign is an equivalent convention; mixing a pole reversal with an unchanged sign would give the wrong motion.

At latitude $\varphi$ and longitude $\lambda$, $\mathbf{r}=R(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$. Motion is ${\mathbf{v}}_{B/C}={\mathbf{\omega }}_{B/C}\times \mathbf{r}$. A meridian boundary has horizontal normal ${\hat{e}}_{\lambda }=(-\sin \lambda ,\cos \lambda ,0)$, so its across-boundary velocity isFor a transform fault this must vanish along the whole meridian, not only at an isolated point. Thus $5\cos \lambda +12\sin \lambda =0$. Select the solution in the western hemisphere:The other solution is the antipodal meridian in the eastern hemisphere. Geometrically, the transform meridian is in a great-circle plane perpendicular to the Euler angular velocity vector. Its outgoing tangent direction at either pole, expressed in the fixed $x,y$ axes, is ${\hat{t}}_F=(12,-5)/13$.