# Orthogonality of ruling and directrix of a ruled surface

Let $M$ be a ruled surface of $\mathbb{R}^3$ with a regular parametrization given by:

$$x(u,v)= \alpha(u) + v\beta(u)$$

where $\alpha' \neq 0$ and $||\beta || = 1$.

I want to show that $<\alpha'(u), \beta(u)>=0$ (inner product)

I am trying to show that we can choose $\alpha$ to be orthogonal to $\beta$ and still get the same surface, or at least upto an isometry.

I have therefore substituted $\alpha(u)$ by $\alpha(u) - <\alpha(u), \beta(u)>\beta(u)$ and if this can be done without altering the original surface then the desired result follows.

But I can't justify my argument. I believe it is true merely out of the vague notion that we can choose the ruling and the directrix to always be orthogonal, though I don't know if this is actually true, or if it is, how do I prove it.

Well, you say you want to show that $\langle\alpha'(u),\beta(u)\rangle = 0$. But, as you go on to say, you really want to choose a different directrix curve $\tilde\alpha(u)$ so that you'll have $$\langle\tilde\alpha'(u),\beta(u)\rangle = 0\tag{\star}.$$ So you want to choose a scalar function $\lambda(u)$ so that, setting $\tilde\alpha(u) = \alpha(u)+\lambda(u)\beta(u)$, ($\star$) will hold. Differentiate and figure out how to choose $\lambda$.
• I did the computations and got $\lambda(u) = - \int_I <\alpha'(u),\beta(u)> \,du$. However, how can I be sure that this new directrix give sthe same ruled surface? – proofromthebook Apr 26 '16 at 8:58