Geodesics which are lines of curvature to surfaces in Euclidean space How to show that if a curve C in a surface is both a line of curvature and a geodesic, then C is a plane curve.
Thanks
 A: As Ryan Budney mentioned in the comments, a curve (with non-vanishing curvature) is planar if and only if its torsion is zero.  Therefore, we will show that the torsion is zero.
Notation: Let $\{t, n, b\}$ denote the Frenet frame, and let $N$ denote the normal vector to the surface.  Let $\kappa_n$ denote the normal curvature, and $\kappa$ the (extrinsic) curvature.
Edit: Hm, I'm thinking that it might be better to write the solution as a series of hints.


*

*Show that being a geodesic implies that $\kappa = \kappa_n$ and $n = \pm N$.

*Show that being a line of curvature implies that $$\frac{dN}{ds} = -\kappa_nt.$$

*Conclude that the torsion is zero by using $\tau = \frac{dn}{ds} \cdot b$.
Note: It may be helpful to assume that $\kappa \neq 0$ and $\kappa_n \neq 0$.  This is an okay assumption because if either $\kappa = 0$ or $\kappa_n = 0$, then the curve reduces to a line (why?) and the result is trivial.
A: Here is another way showing explicitely the plane containing $C$. Let me call $S$ the surface whose line of curvature $C$ is geodesic.
Let $p \in C$ and let $v_p$ a non-zero tangent vector to $S$ perpendicular to $C$ at $p \in S$. By using the intrisic parallel transport of $S$ extend $v_p$ to a parallel vector field $v$ along $C$. Since $C$ is geodesic $v$ is always perpendicular to $C$ hence $v$ is always an eigenvector of the shape operator of $S$.
We claim that $v$ regarded as vector field in $\mathbb{R}^3$ (along $C$) is constant. 
That is to say, $D_{C'} v$ the derivative of $v$ along $C$ in $\mathbb{R}^3$ is identically zero. Indeed, the tangent component (to the surface $S$) of $D_{C'}v$ is zero due to the construction of $v$ i.e. such tangent component is the intrinsic covariant derivative of $S$. The normal component of $D_{C'} v$ is the second fundamental form of $S$ w.r.t. $v$ and $C'$ which is also zero since both $C'$ and $v$ are eigenvectors of the shape operator of $S$.  Thus $C$ is contained in the plane through $p$ normal to the vector $v_p$.
An even simpler proof :  consider the cross $C' \times N$. The derivative of the cross along $C$ is zero hence the cross is constant along $C$. Then $C$ is contained in the plane through any of its points perpendicular to the cross.
