= Solution
The <Lorentz boost> from A to B gives
$$
dt'=\gamma_v(dt-v\,dx/c^2),\quad dx'=\gamma_v(dx-v\,dt),\quad dy'=dy,\quad dz'=dz.
$$
Divide each spatial displacement by $dt'=\gamma_vdt(1-vu_x/c^2)$. The <velocity-addition formula> is
$$
\boxed{w_x=\frac{u_x-v}{1-vu_x/c^2},\qquad
w_y=\frac{u_y}{\gamma_v(1-vu_x/c^2)},\qquad
w_z=\frac{u_z}{\gamma_v(1-vu_x/c^2)}.}
$$
The same result follows by transforming the <four-velocity> and using $w_i=cU'^i/U'^0$. The transverse <velocities> change because the transformed time increment depends on $dx$, even though the transverse spatial coordinates are unchanged.
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