Let $c: I \rightarrow \mathbb{R}^3$ be a regular curve, $V: I \rightarrow \mathbb{S}^2$ a vector field and $a < b$. Then we call $$ f: (a,b)\times I \rightarrow \mathbb{R}^3, \quad f(s,t):= c(t) +sV(t) $$ a ruled surface.
Show that $f$ has gaussian curvature $K(s, t) \leq 0$.
For the first fundamental form, I obtained $$ G = \begin{pmatrix} 1 & \langle V, c'\rangle \\ \langle V, c'\rangle & \lvert c' + sV' \rvert^2 \end{pmatrix} $$
and for the second fundamental form $$ B = \frac{1}{\lvert V \times ( c' + sV') \rvert} \begin{pmatrix} 0 & \langle V', V \times (c' + sV') \rangle \\ \langle V', V \times (c'+sV') \rangle & * \end{pmatrix} $$ and thus (with $V \perp V'$) that $$ K = \frac{\det B}{\det G} = \frac{-2 \langle V', V\times c' \rangle}{\lvert c' + sV' \rvert^2 -2\langle V, c' \rangle} \cdot \frac{1}{\lvert V \times (c' + sV') \rvert}. $$
I know that this question was already answered here: Gaussian and Mean Curvatures for a Ruled Surface.
However, there are additional assumptions made such as $c' \perp V'$ and $\lvert V' \rvert = 1$. I don't know how to apply that as my case is a bit more general.
Two comments:
First, you can see on geometric grounds that any ruled surface has $K\le 0$, as the rulings are asymptotic curves, and one cannot have asymptotic curves when $K>0$.
Second, you know that $\det G>0$ (always) and it's clear (from the $0$ in the $11$-entry) that $\det B\le 0$. So $\det B/\det G \le 0$.