(Assume a unit speed parametrization)
Prove that an admissible curve $c:(a,b) \to \mathbb R^3$ (hence $c^{\prime} \times c^{\prime\prime}$ is never zero) with a zero torsion is contained in a plane.
I understand why this should be true, if there is no torsion element to the curve then $c$ can't twist into the third dimension.
$\mathbf T = c^{\prime}$
$\mathbf N = c^{\prime\prime}$
$\mathbf B = \mathbf T \times \mathbf N$
$\mathbf T^{\prime} = \kappa \mathbf N$
$\mathbf N^{\prime} = -\kappa \mathbf T + \tau\mathbf B$
$\mathbf B^{\prime} = -\tau\mathbf N$
So if the torsion $\tau$ is zero, $\mathbf B^{\prime} = \mathbf 0$, therefore $\mathbf B$ is a constant not equal to zero. I'm not sure where to go from here. If the normal vector of $\mathbf T$ and $\mathbf N$ is a constant, does that mean no matter what point you are at on the curve $\mathbf B$ points in the same direction?