Say we have a ruled surface that is given by $x(u,v) = \alpha(u) + v\beta(u)$ with $\alpha'$ not equal to $0$ and $\| β \| = 1$.
If $\alpha'(u), \beta(u)$, and $\beta'(u)$ are linearly dependent for every $u$. I know that below (a, b, or c has to be true) one has to be true
a.) $\beta$ must be constant
b.) a function $\lambda$ exists so that $\alpha(u) + \lambda(u)\beta(u)$ is a nonzero multiple of $\beta(u)$ for all $u$.
c.) a function $\lambda$ exists so that $(\alpha+\lambda\beta)'(u) = 0$ for every $u$.
$d_1,d_2,d_3$ can be considered functions of $u$ that together will never be $0$,
so,
$d_1(u)α'(u) + d_2(u)β(u) + d_3(u)β'(u) = 0$ for every $u$.
I have been struggling with this proof for quite some time. I do not think that choice a would be possible and have narrowed it down to b and c, but am having trouble proving that one or the other is the correct choice for the ruled surface. Any hints or suggestions on walking through this problem would be appreciated!
HINT: Suppose $d_1=d_2=0$, $d_3\ne 0$. Then case A). Otherwise, start with what relation must hold in case C), and if it doesn't, show you're in case B).