A particle of unit mass with position vector $\vec {r(t)}$ at time $t$ is moving in space under the actions of certain forces. If $\vec r \times v = \vec c$, where $\vec c$ is a constant vector, prove that the motion takes place in a plane. Consider both $\vec c = \vec 0$ and $\vec c \not = \vec 0$.
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"
As in the linked answer, for $\vec c\neq \vec 0$, we get $0=\vec r\cdot (\vec r\times \vec v) = \vec r\cdot \vec c$, so the particle moves in the plane $0=\vec r\cdot \vec c$. (The box product is the determinant of the matrix with the 3 vectors as rows. If two rows repeat, the determinant is zero.)
For $\vec c=0$, this means that the velocity and the particle position are parallel. Thus, if there is some $t_0$ such that $\vec r(t_0)\neq \vec 0$, then $\vec r(t) = \vec r(t_0) \lambda (t)$ for some scalar valued function $\lambda (t)\in\mathbb R$. In particular....