How to show something is a straight line

I'm trying to show that if alpha(s) is a straight line if and only if all its tangent lines are parallel.

Pf/ I know that I will need the Frenet Serret Theorem and my stab at it is:

Assume all the tangent lines of a(s) are parallel. So the tangent vector T is the same for all points xo on the curve a(s) and the values of T(s) of any two points on the curve are parallel. Thus T(s) is constant, and T'(s)=0 which implies that the curvature is zero, and thus a(s) must be a straight line.

-

I'm not so sure you want to use the Frenet Serret equations in their full form with curvature and torsion. These quantities involve the normal which is ambiguous for lines in three dimensions. The usual equation for the normal suffers division by zero for a line.

That said, I don't think you need the Frenet Serret formulas.

Suppose a curve is a line. It is clear that all the tangent lines have the same direction vector and are hence paralell.

Conversely, suppose all the tangent lines to a curve $\alpha(s)$ are parallel. It follows that $\alpha'(s) = v_o$ for a particular direction vector $v_o$ and all $s \in dom(\alpha)$. Now, integrate and suppose $\alpha(s_o)=r_o$, it follows $\alpha(s) = r_o+v_o(s-s_o)$. Hence our curve is a line.

-
Thanks James; just wondering, what was missing in my proof? –  mary Sep 9 '12 at 6:17
@mary I suppose you should mention the forward direction of the iff is clear. Geometrically I think your argument is valid. The question is how do you know zero curvature implies a straight line? (I believe it, but why?) What is the definition of curvature, etc... I wanted to avoid these questions in my answer. I suppose the larger point to make here is that we don't need the Frenet formulas because the case considered is easy enough that the basic calculus of space curves is adequate to show the claim. I do like your argument. –  James S. Cook Sep 9 '12 at 17:24
Thanks for explaining the ambiguity –  mary Sep 9 '12 at 23:06