I have a question about Do Carmo's treatment of ruled surfaces in his classic Differential geometry of curves and surfaces.
At pages 193–194 (I have Dover's edition) he defines the line of striction of a ruled surface. To such end, he considers the ruled parametrization
\begin{align}
x \,\colon I \times \mathbb{R} &\to \mathbb{R}^{3}\\
(t,v) &\mapsto \alpha(t) + v w(t),
\end{align}
where $\alpha$ is a smooth curve in $\mathbb{R}^{3}$ and $w$ a smooth unit vector field along $\alpha$ such that $w'(t)\neq 0$ for all $t\in I$.
He thus searches for a "parametrized curve $\beta(t)$ such that $\langle \beta'(t),w'(t)\rangle =0$, $t\in I$, and $\beta(t)$ lies on the trace of $x$" (such curve is called the line of striction); in particular, he notes that $\beta$ lying on the trace of $x$ is equivalent to the existence of a function $u = u(t)$ such that
\begin{equation}\tag{1}\label{eq1}
\beta(t)= \alpha(t)+ u(t)w(t).
\end{equation}
All good so far. However, as soon as he has found such a curve, he goes on to prove that "the curve $\beta$ does not depend on the choice of the directrix $\alpha$ for the ruled surface." [emphasis added]
Question: Why do we need to prove that $\beta$ does not depend on $\alpha$?
EDIT: In reply to Ted Shifrin's comment, and to make the question self-contained, I am adding the proof of existence.
By definition of line of striction, we need to find a curve in the image of $x$ that is everywhere orthogonal to $w'$. As already mentioned, a curve $\beta$ is in the image of $x$ if and only if equation \eqref{eq1} holds for some function $u=u(t)$.
Differentiating \eqref{eq1} and dotting with $w'$, we obtain
\begin{align}
\beta'&= \alpha'+ u'w+uw',\\
\langle \beta',w' \rangle &= \langle \alpha',w'\rangle + u \langle w',w' \rangle.
\end{align}
From the last equation, we observe that the curve $\beta$ satisfies $\langle \beta'(t),w'(t)\rangle =0$ for all $t \in I$ exactly when
\begin{equation}
u=-\frac{\langle\alpha',w'\rangle}{\langle w',w' \rangle}.
\end{equation}
Hence, we conclude that the desired curve is given by
\begin{equation}
\beta=\alpha-\frac{\langle\alpha',w'\rangle}{\langle w',w' \rangle} w.
\end{equation}
EDIT 2: If I had to prove independence on $\alpha$, then I would do it as follows, without using the final expression of $\beta$.
Suppose that, for another directrix, the line of striction is different from $\beta$. Then there are two lines of striction, say $\beta_{1}$ and $\beta_{2}$. It follows that there exist two functions $u_{1}$, $u_{2}$ of $t$ such that $\beta_{i}=\alpha+u_{i} w$, with $i=1,2$. However, both of them are orthogonal to the same $w'$; as explained below, this fact implies $u_{1}=u_{2}$.
Differentiating $\beta_{i}= \alpha + u_{i} w$, we obtain $\beta_{i}'= \alpha' + u_{i}' w+u_{i}w'$, which in turn implies $\langle \beta_{i}',w'\rangle= \langle\alpha',w'\rangle + u_{i} \langle w',w'\rangle$. Since $\langle \beta_{1}',w'\rangle = \langle \beta_{2}',w'\rangle=0$, it follows that $\langle\alpha',w'\rangle + u_{1} \langle w',w'\rangle = \langle\alpha',w'\rangle + u_{2} \langle w',w'\rangle$, from which one concludes that $(u_{1}-u_{2}) \langle w',w'\rangle=0$. But $\langle w'(t),w'(t)\rangle \neq0$ for all $t$ by assumption, and so $u_{1}=u_{2}$.
Best Answer
After some thought, I believe I am able to answer my own question. Thanks for the helpful comments.
The answer of the question in the title is yes. We do need to prove that the line of striction is independent on the directrix, because otherwise it could be ill-defined. However, its well-definedness can already be observed from the fact that the solution of the problem (when using $\alpha$ as directrix) is unique. In other words, we do not need Do Carmo's strong proof, which uses the expression of the solution.
In EDIT 2 I have shown that, if the line of striction depended on the directrix, then the solution of the problem in $u=u(t)$ defined by \begin{equation} \begin{cases} \beta=\alpha+u w,\\ \langle \beta',w'\rangle=0, \end{cases} \end{equation} would not be unique. This proves that uniqueness implies independence, as desired.