Solución
Tenemos
$$\begin{aligned} \|\bold{u}\times\bold{v}\| &= \|\bold{u}\|\cdot\|\bold{v}\|\cdot sen\theta\\ &= \sqrt{0^2+4^2+0^2}\cdot\sqrt{0^2+0^2+(-3)^2}\cdot sen\frac{\pi}{2}\\ &= 4(3)(1)=12 \end{aligned}$$