Showing a curve which lays on a sphere of radius 1 is plane

Question

Assuming $\alpha$ is a unit speed curve, I'm trying to prove that $\alpha$ is plane. By hypothesis, I know its curvature is such that $$\kappa=1$$ I'm trying to use the torsion's formula: $$\tau=\frac{\langle \alpha'\times\alpha'',\alpha'''\rangle}{\|\alpha'\times\alpha''\|^{2}}$$ to show it's equal to zero, but I have no idea how to use it.

Answer 1

If $\alpha$ is a unit speed curve that lies on the sphere of radius 1 we have $\langle \alpha, \alpha\rangle = 1 = \langle \alpha^\prime, \alpha^\prime\rangle$. From this, it follows $2\langle \alpha', \alpha\rangle =0$. Then differentiating once more, we get $0=\langle \alpha'',\alpha\rangle + \langle \alpha',\alpha'\rangle = \langle \alpha'',\alpha\rangle + 1$ which implies $\alpha'' = -\alpha$, since $1=k=\lVert \alpha''\rVert$, and $\lVert\alpha\rVert = 1$. Finally we have $\alpha '''= -\alpha'\implies \alpha'''\parallel\alpha '\implies \langle \alpha'\times \alpha'', \alpha'''\rangle =0\implies \tau = 0$