Characterization of a line segment

Question

I want to prove that a curve $\alpha: I \to \mathbb R ^2$ parameterized by arc length is a segment of a line iff the intersection of all its tangents is not empty.

To prove the interesting direction, I formalized the hypothesis as saying that $p_0 = \alpha (s) + \lambda (s) T_\alpha(s)$ for certain $\lambda(s)$.

But then I did two apparently innocuous steps and got a nonsensical result. Namely, I multiplied by the normal vector $N(s)$. Since the normal vector its orthogonal to the tangent vector, we get $p_0 N(s) = \alpha(s) N_\alpha (s)+ \lambda (s) T_\alpha(s)N_\alpha (s) = \alpha(s) N_\alpha (s)$.

But since the normal vector is unitary, I can multiply again by it to eliminate it from the equation, which gets me $p_0 = \alpha(s)$ which is obviously wrong.

What is going on?

Answer 1

HINT: In your equation, $p_0$ is a constant vector function. So differentiate your equation and use Frenet. This is the always the standard approach in basic differential geometry!

Answer 2

Remember that multiplication means inner product. Therefore, if you have $<p_0,N(S)>=<a(s),N_a(s)>$ you cannot deduce that $p_0=a(s)$. You are on the right track however.

Let me present a different approach. Using the implicit function theorem , locally you can assume your curve is a graph of some function. Therefore Rolle's theorem stands. Therefore for any two points $p_1,p_2$ (not on a local extremum) in the region there exists a tangent parallel to the line joining these points. But this tangent intersects the tangents at $p_1,p_2$.

If all the points lie in an extremum then we have nothing to prove because locally our curves is a line and since it is smooth also globaly.