I'm trying understand the converse of example $6.2$ of these notes, but I'm stuck when is stated that $x''(t)$ is parallel to $N$, where $N$ is the unit normal to a plane $\Pi$. The author argues that $x''(t)$ is parallel to $N$ because $x(t)$ is an unit speed geodesic. I know that
$$||x'(t)||^2 = 1 \Longrightarrow \langle x'(t), x'(t) \rangle = 1 \Longrightarrow \langle x''(t), x'(t) \rangle = 0 \Longrightarrow x''(t) \perp x'(t), $$
but what I do not understand is why I can state that $x''(t)$ is parallel to $N$. I think I need to argue using the fact that the planar curve has not torsion ($\tau \equiv 0$) and using the Frenet–Serret formulas, but I couldn't conclude that $x''(t)$ is parallel to $N$.
Thanks in advance!
$x''(t)$ is parallel to $N$ by definition of geodesic, because it is supposed that $x = x(t)$ is a unit speed geodesic.