My question is regarding two similar exercises from Apostol Calc. Vol. I : Given two fixed nonzero vectors $\vec{A}$ and $\vec{B}$ making an angle $\theta$ where 0<$\theta$<$\pi$. A motion with position vector $\vec{r(t)}$ satisfies the differential equation $\vec{r'(t)}=A \times \vec{r(t)}$ and the initial condition $\vec{r(0)} = B $. Prove the following:
- The acceleration vector $\vec{a(t)}$ is orthogonal to $\vec{A}$.
- Compute the speed, which is constant.
- Compute the curvature, which is constant.
Also, while not part of the exercise, I am wondering if this differential equation can be solved with only the given values, or would more information be needed.
My attempt at a solution seems to involve an excessive amount of cross-products. Firstly, the velocity vector $\vec{v(t)}$ is orthogonal to the position $\vec{r(t)}$ because of $\vec{r'(t)}=\vec{A} \times \vec{r(t)}$ so we have a fixed length position vector, equal to the initial condition $\vec{r(0)} = B$. This must indicate the motion is on a sphere. The magnitude of the velocity is then $|A||B|\sin{\theta}$. The acceleration vector $\vec{a(t)}$ is given by $A \times\vec{v'(t)}=A \times (A \times \vec{r(t)})$ so is orthogonal to A? Then we have a unit tangent vector $\vec{T}$ equals $\frac {\vec{A} \times \vec{r(t)}}{|A||B|\sin{\theta}}$ so the normal acceleration T' is $\frac {(A \times (A \times \vec{r(t)}))}{|A||B|\sin{\theta}}$ leading to curvature $\frac {|\vec{T'}|}{|\vec{v(t)}|}$ which equals $\frac {|A||A||B|\sin{\frac {\pi}{2}}\sin{\theta}}{|A||B|\sin{\theta}|A||B|\sin{\theta}}$. Assuming the vector A is unit length ,this matches the books answer $\frac {1}{|B|\sin{\theta}}$ but shouldn't curvature equal the inverse of the radius of the sphere, $\frac {1}{|B|}$ ? I doubt my method as it feels quite messy. Any help is appreciated. (This is the first time I posted here so hopefully my description is clear).