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If the osculating planes of a curve pass through a fixed point, the curve is a plane curve.

How to prove it?

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Parametrize the curve by arclength, as usual, by $\alpha(s)$. Say the fixed point is the origin. Then there are functions $a(s)$ and $b(s)$ so that $$\alpha(s) = a(s)\mathbf T(s) + b(s)\mathbf N(s).$$ Now differentiate and apply Frenet.

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  • $\begingroup$ Can we say the fixed point is the origin without loss of generality? I don't see how... $\endgroup$
    – Pthrp97
    Mar 20 '21 at 3:35
  • $\begingroup$ @Pthrp97 If all the planes pass through $\mathbf a$, just subtract that vector from $\alpha(s)$ and proceed. $\endgroup$ Mar 20 '21 at 3:38
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yes upon derivation, you should tao go to 0. Note that there is no B term. so the N' term will have a $\tau$ $\vec{B}$ term but this will go to zero.

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