I'm trying to comprehend the solution to the following problem.
Let $S=\left\{(x, y, z) \in \mathbb{R}^{3}: x^{2}+2 y^{2}+3 z^{2}=4\right\} \subset \mathbb{R}^{3}$ be an ellipsoid. Show that the intersection $S \cap\left\{(x, y, z) \in \mathbb{R}^{3}: z=0\right\}$ is a simple closed geodesic in $S$.
Here is the official solution: The intersection $C=S \cap\left\{(x, y, z) \in \mathbb{R}^{3}: z=0\right\}$ is an ellipse in the plane $\{z=0\}$. The unit normal to $S$ is proportional to $\nabla f$, which is contained in the plane $\{z=0\}$ for all $(x, y, z) \in S \cap\{z=0\} .$ Hence the curvature vector of $C$ is proportional to the unit normal of $S$ along $C$, from which it follows the geodesic curvature of $C$ is zero and $C$ is a geodesic.
The only part that I'm confused about is why the curvature vector of $C$ is proportional to the unit normal of $S$ along $C$. Could anyone help me understand this?