Prove that if $r(t)\times \frac{dr(t)}{dt}=0$, then $r(t)$ has a fixed direction. Let $r(t)$, where $t$ is a parameter $(t ∈R)$, be a position vector such that $r(t)\times \frac{dr(t)}{dt}=0$
I am asked to show that r(t) has a ﬁxed direction.
A hint says: Let $r(t) = f(t)\hat e(t)$ where $\hat e(t)$ is a unit vector). 
Could someone tell me how to use the hint? May I ask for a proof?
EDIT: I did some searching and found this question Show fixed direction of a position vector
But I cannot understand the answer. Could someone please use the hint to explicitly show that? Thanks in advance!

This is not true. A counterexample:
$$ r(t) = \begin{cases} (0,t^2) & \text{for } t\le 0 \\ (t^2,0) & \text{for } t> 0 \end{cases} $$
If you have the additional assumption that $r(t)\ne 0$, then your property holds, and something like Tim the Enchanter's proof should work.
