Prove that $\vec r ~\times \vec a = \vec 0 $ implies $\vec r ~\times \vec v = \vec c $, where $\vec c$ is a constant vector.

Question

A particle of unit mass with position vector $\vec {r(t)}$ at time $t$ is moving in space under the actions of certain forces. Prove that $\vec r ~\times \vec a = \vec 0 $ implies $\vec r ~\times \vec v = \vec c $, where $\vec c$ is a constant vector.

I know that this question has already been posted, but the answer only considered the case given $\vec c = \vec 0$. I am not looking for a full on answer but rather a hint...

Can I also use the triple product to prove it for $\vec c \not = \vec 0$ ? Though is use of the triple product for the case $\vec c = \vec 0$ even valid? Wouldn't it give me a normal vector to the plane equal to $\vec 0$ ?

This section of the textbook deals with "The unit tangent, the principal normal, and the osculating plane of a curve"

Answer 1

Hint: compute the time derivative of $\vec r\times \vec v$.

Hint 2: If $f'(t)=0$ for all $t$, what can we say about $f$?