I will only hint at the solution, leaving it to you to make things explicit. This way you will have to chase through the definitions, which, I claim, is a good exercise.
For simplicity I assume this is about 2-dimensional surfaces in $\mathbb{R}^3$. I also assume you know the 2nd fundamental form of a hypersurface can be expressed as the (covariant) derivative of the normal to the hypersurface along that hypersurface (up to sign).
If two surfaces $M_1, M_2$ touch along a curve, they share the normal along the curve (up to sign) (why?). The curve is a line of curvature iff it's tangent is an eigenvector of the 2nd fundamental form. This means: if $\phi$ parametrizes the curve along which the surfaces intersect and if $\nu$ denotes the normal you need to show that along $\phi$:
$$ \nabla^{M_1}_{\phi'} \nu = \lambda \phi' \Rightarrow \nabla^{M_2}_{\phi'} \nu = \sigma \phi' $$
where $\lambda$, $\sigma$ are scalar functions along $\phi$ and $\nabla^M$ denotes the derivative along the surface (i.e. the orthogonal projection of the derivative in the ambient space onto the tangent space of the surface). If you know how to compute this, the answer to i) should be quite obvious.
As for ii): convince yourself that the line of intersection is a curvature line in the plane. Since the surface touches the plane it is located in one of the halfspaces the plane defines. Write the surface locally as a graph of a nonnegative function $u$ over the plane, use the fact that $u$ has a minimum along the line of intersection. Check the implications of this observation for the second derivative and, in consequence, for the second fundamental form of the surface.
are you enrolled in UofCalgary PMAT 423? I have the same question as you.
(=>) this is what I was thinking as well. Just remember that we're supposed to show that γ is perpendicular to the principal normal of γ, not to γ". Use γ" = kn (not N which is the standard unit normal of the surface).
(<=) here we have to use the idea that the principal normal of γ is perpendicular to γ at every point on γ. Since kn = t' = γ", where k is the curvature of γ (not the surface), this implies γ" is perpendicular to γ at every point which then implies γ" is perpendicular to γ'. A property of the cross product is that if A x B = C then C is perpendicular to A and to B so (N x γ') is perpendicular to γ'. Since Y' is perpendicular to γ" and γ" is parallel to N (because γ lies on the surface) then (N x γ') must be perpendicular to γ" as well. From this we get γ" • (N x γ') = 0 = kg. Since the geodesic curvature equals 0, then the surface must be geodesic.
I think this is the correct way to do this question but it both directions just seems too simple. If you have any ideas I'd love to here it.
Best Answer
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.