# Two-Dimensional Lorentz Velocity Transformation Problem

Two spaceships A and B are approaching along perpendicular directions, as seen from earth. If A is observed by a stationary Earth observer to have velocity $$𝑢_𝑦$$ = -0.90c and B to have velocity $$𝑢_𝑥$$ = +0.90c, determine the speed of ship A as measured by the pilot of ship B.

To solve the problem, I split it into two parts: first part for the $$x$$ component the velocity of A, and second part for the $$y$$ component. Let $$v$$ and $$v'$$ be the velocities of ship A in frame S and S' respectively. Similarly, let $$w$$ and $$w'$$ be the velocities for ship B. We set frame S to be attached to Earth while S' is attached to spaceship B. From Lorentz' velocity transformation formula, $$v_x'=\dfrac{v_x-w_x}{1-\dfrac{v_xw_x}{c^2}}=-w_x=-0.9c$$ since $$v_x=0$$. Similarly for $$v_y$$, $$v_y'=\dfrac{v_y-w_y}{1-\dfrac{v_yw_y}{c^2}}=v_y=-0.9c$$ since $$w_y=0$$. Then I used Pythagoras' Theorem to obtain the total velocity of ship A, and...well here's the result: $$v=\sqrt{v_x^2+v_y^2}\approx 1.273c$$

At this point, I think my problem clear. As we know, nothing can exceed the speed of light, so it is not possible for ship A to be travelling at a speed higher than $$c$$. What did I do wrong in my steps? Every example I found online was one-dimensional. I looked for answers on this website, and I found mixed answers more confusing than the problem itself. I'm suspecting that my mistake lies in the assumption that Pythagora's Theorem holds for relativistic velocities. Help, please?

$$u'_x=\frac{u_x-v}{1-\frac{vu_x}{c^2}}=\frac{0-v}{1-\frac{0}{c^2}}=-v=-0.9c$$
Remember that the velocity of $$B$$ shall be considered as $$v$$ in the above link. $$(u_x=v=0.9c)$$ Therefore, we can write:
$$u'_y=\frac{u_y\sqrt{1-\frac{v^2}{c^2}}}{1-\frac{vu_x}{c^2}}=u_y\sqrt{1-\frac{v^2}{c^2}}=-0.9c×\sqrt{1-0.9^2}=-0.392c$$
$$u'=\sqrt{u^{\prime 2}_x+u^{\prime 2}_y}=0.982c$$