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!

If $r(t)=f(t)ê (t)$ then
$$ \frac{d \vec r}{dt} = f'(t)ê (t) + f(t)\frac{dê (t)}{dt} $$ 
$$ \vec r \times \frac{d \vec r}{dt} = \vec r \times f'(t)ê (t) + \vec r \times f(t)\frac{dê (t)}{dt} = \vec r \times f(t)\frac{dê (t)}{dt} = 0$$
Thus $\frac{dê (t)}{dt} = 0$ or  $\vec r$ is parallel to $\frac{dê (t)}{dt}$
But $\vec r$ is perpendicular to $\frac{dê (t)}{dt}$ because of the property - "A vector of constant magnitude is perpendicular to it's derivative."
Therefore, $\frac{dê (t)}{dt} = 0$ i.e. there is no change in the direction vector.
