geodesic line on intersecting planes I have a problem and I don't understand how to solve it, please help!
Prove that if two surfaces in $R^3$ intersect along a curve that is geodesic on both surfaces, and the tangent planes to the surfaces at any point of the curve do not coincide (in this situation, the surfaces are said to intersect transversely), then this curve is straight.
 A: Let $\gamma:(-a,a)\rightarrow \mathbb{R}^3$ be a local arclength parametriziation of the curve in question.
This means $|\gamma^\prime(t)|^2 =1$ for every $t$ and, consequently, $\langle \gamma^{\prime\prime},\gamma^\prime \rangle = 0$. The claim is equivalent to showing $ \gamma^{\prime\prime} = 0$ (why?).
Now since the curve is a geodesic in the two surfaces -- assuming you know the geometric consequences of this statement --  you know that the tangential component of $\gamma^{\prime\prime}$ in $M_1$ vanishes. Let me write $(\gamma^{\prime\prime})^{M_1;T}$ for that component.
I.e. if $e_1(t) $ is a normal vector to $\gamma$ in $M_1$ of length $1$,
$$(\gamma^{\prime\prime})^{M_1;T} = 0 =\langle \gamma^\prime,\gamma^{\prime\prime}\rangle \gamma^\prime+ \langle e_1(t),\gamma^{\prime\prime}\rangle e_1(t)   $$
Since (see the introductory remark) $\langle \gamma^\prime,\gamma^{\prime\prime}\rangle=0 $, also 
$$ \langle e_1(t),\gamma^{\prime\prime}\rangle = 0$$
The same is true for a vector field $e_2$ of length $1$ along $\gamma$ which is normal to $\gamma$ in the second surface, $M_2$, say.
If the surfaces meet transversally, $e_1$ and $e_2$ are linearly independent, and together with $\gamma^\prime$ they span $T_{\gamma(t)}\mathbb{R}^3$. The claim then follows by elementary linear algebra.
