Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 25H / b / Solution

If $\tau =0$ everywhere, the third of the Frenet-Serret formulas gives $B^{\prime }=0$, so $B$ is a constant vector. MoreoverThus $\alpha (s)\cdot B$ is constant, and the image of $\alpha$ lies in the plane perpendicular to $B$.