I am told to prove that the derivative of a vector with constant module is perpendicular to the vector. Here is my approach, though I'm not sure about if it's correct or not:
Let $A(t)$ be a constant vector with $|A(t)|=C$, where C is a constant. Then,
$|A(t)|=\sqrt{x(t)^2+y(t)^2+z(t)^2}=C$, and therefore, $|A(t)|'=(\sqrt{x(t)^2+y(t)^2+z(t)^2})'=0$, thus: $(x(t)^2+y(t)^2+z(t)^2)'=0$. Since all terms are $>=0$, we can deduce $x'(t)=y'(t)=z'(t)=0$ (I think here's what I'm doing wrong).
Now we can say $A(t)·A'(t)=x(t)x'(t)+y(t)y'(t)+z(t)z'(t)=0$, and therefore they are perpendicular.
I think this is wrong because of $x'(t)=y'(t)=z'(t)=0$, which would imply the derivative is the vector 0, which has no sense. But I haven't got a better approach.