I'm learning differential geometry, specifically the theory of surfaces, and need help with the following exercise:
Consider the surface $S : \vec x = \vec x(u,v) \in C^3$ for which the vector $\vec n_u$ and $\vec n_v$ are linearly dependent. Show that the Gaussian curvature of the surface is identically zero.
Note: we use $\vec x(u, v)$ to denote the parametric representation of the surface with parameters $u$ and $v$. Also, we use $\vec n$ to denote the unit normal vector to the surface at a point $P$.
Unfortunately I wasn't able to do much with this one. I can state the obvious by saying that, if the Gaussian curvature is identically zero, then it must be that
$$\frac{LN - M^2}{EG - F^2} = 0 \tag{1}$$
where $E, F, G$ and $L, M, N$ are the coefficients of the first and second fundamentals forms, respectively. However, I don't think relation $(1)$ is going to be of any help to solve this exercise.
I clearly have to make use of the fact that $\vec n_u$ and $\vec n_v$ are linearly dependent but I don't know how I should start.
Differentiating $\vec{x}_u \cdot \vec{n} = 0$ as suggested by BigMathTimes, I have found the following:
$$L = -\langle {\vec x_u} , {\vec n_u} \rangle = \langle {\vec x_{uu}} , {\vec n} \rangle, \quad M = -\langle {\vec x_u} , {\vec n_v} \rangle = \langle {\vec x_{uv}} , {\vec n} \rangle, \quad N = -\langle {\vec x_v} , {\vec n_v} \rangle = \langle {\vec x_{vv}} , {\vec n} \rangle.$$
Use the fact that $\vec{x}_u \cdot \vec{n} = 0$, and differentiate with respect to $u$ and $v$. Do a similar computation with $v$ instead of $u$. See what this makes the second fundamental form. Can you now use the fact that $n_u$ and $n_v$ are dependent to show that the determinant vanishes?