|z|^2 = |x|^2 + |y|^2 - 2|x||y|cos(theta)
Start from the following facts:
(since the vectors make a triangle)
2) Show that ax+by+cz+dt = (0,0,0,0) if and only if a=b=c=d=0. This is done by showing that the system of simultaneous equations has only this solution. You will discover this if you solve the system in the normal way.