Why is $\mathbf r(t) \times \mathbf r'(t)$ constant, with $\mathbf r(t): \mathbb R\to\mathbb R^3$?

Question

Let $\mathbf r(t): \mathbb R\to\mathbb R^3$ be a funtion of time representing a point subject to a radial acceleration, i.e. $\mathbf r''(t)$ is parallel to $\mathbf r(t)$. Prove that $\mathbf r(t) \times \mathbf r'(t) = \mathbf h$ (where $\times$ is the cross product), with $\mathbf h$ constant and that the trajectory is in a plane perpendicular to $\mathbf h$.

How should I prove this? (There is no initial position or velocity).

I'm posting this problem on math stack exchange because it is on my math book and i'm not even supposed to take a course in physics

Answer 1

Since $\mathbf r $ and $\mathbf r'' $ are parallel, we have (by properties of the cross product) that:

$$\mathbf r \times \mathbf r'' = \mathbf 0$$

Also, trivially

$$\mathbf r' \times \mathbf r' = \mathbf 0$$

Hence

$$ \begin{align} \mathbf 0 &= \mathbf r' \times \mathbf r' + \mathbf r \times \mathbf r'' \\ &= \frac{d}{dt}(\mathbf r \times \mathbf r') \end{align}$$

Which implies that

$$\mathbf r \times \mathbf r'= \mathbf h$$

where $\mathbf h$ is a constant vector. Take the dot product with $\mathbf r$ to give

$$ \mathbf r \cdot \mathbf h = \mathbf r \cdot(\mathbf r \times \mathbf r') = \mathbf 0 $$

and $\mathbf r \cdot \mathbf h = \mathbf 0$ is the equation of a plane through the origin, perpendicular to $\mathbf h$ .

Answer 2

Define $h(t) = r(t)\times r'(t)$. Differentiate $h'(t) = r'(t)\times r'(t) +r(t)\times r''(t) = 0 + 0$ to argue that $h(t)$ is constant. Now, argue that $r(t)\times h$ is always zero.