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I was reading this question about Bertrand curves and I got a bit confused about some steps of the solution given by the author.

In section a) of the problem it's stated that if $\alpha$ and $\beta$ are Bertrand mates then $$ \beta = \alpha + r N $$ for some scalar function $r$. From here the author arrives at the following equation $$ r' = \beta' \cdot N - \alpha' \cdot N = \beta' \cdot N =0 $$ I understand that $\alpha' \cdot N= 0$ since we define the normal vector to $\alpha$ to be orthogonal to the tangent vector, but I don't follow why $\beta' \cdot N =0$. I thought that it could maybe be the case that $\alpha' \parallel \beta'$, but on section b) it's shown that this is not the case, and in fact, there's even an explicit formula for the constant angle between the tangent vectors. I've searched many places for similar answers but every place I see seems to obviate this step, but I just can't seem to see why.

I'm sure there's a simple piece of reasoning I'm not seeing here. Can someone tell me what it is that I'm missing? Thank you!

Robert Lee
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  • Because by definition at the top of the page Bertrand mates share the normal line at the corresponding points, so $\beta'\perp N$ by definition. In 3D two lines being perpendicular to a third does not imply that they are parallel to each other, so $\alpha' \parallel \beta'$ does not follow from both being perpendicular to $N$. – Conifold Nov 02 '20 at 10:34
  • Ohhhh, of course. I was implicitly still thinking of plane curves. So is it just that by hypothesis $N$ is normal to both curves? – Robert Lee Nov 02 '20 at 10:36
  • The "hypothesis" being the definition of Bertrand mates. – Conifold Nov 02 '20 at 10:39
  • Yes, that's what I meant. Thank you for the explanation. It was, indeed, something obvious. – Robert Lee Nov 02 '20 at 10:40

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