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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.

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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 $\lambda$.

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  • $\begingroup$ 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? $\endgroup$ – proofromthebook Apr 26 '16 at 8:58
  • $\begingroup$ Show it the same way you show two sets are equal in general. $\endgroup$ – Ted Shifrin Apr 26 '16 at 13:57

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