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($\times$ denotes the usual cross product)

Let $\alpha: I \to R^3$ be a smooth regular curve with non-zero curvature, parametrized by arc length. Given there exists $W: I \to R^3$ such that $W\times X = X'$ for every $X\in \{T, N, B\}$, where that last set is the Frenet-Serret frame of $\alpha$, we are asked to determine W. (probably in terms of the basis {T, N, B}).

I was able to conclude, using Frenet-Serret formulas that the $N$ component is $0$ since $W\times T = T' = kN$ therefore since $k\neq 0$, $N$ is orthogonal to $W$. So $W = aT + bB$ for some functions $a,b$. Now i'm stuck. Any hint?

Carla is my name
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  • https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas – Max Jun 15 '21 at 06:27
  • @Max Yes i am aware of those formulas and used them, but i don't know how to apply them here to proceed further. – Carla is my name Jun 15 '21 at 06:33
  • https://math.stackexchange.com/a/2248459/2633 aka https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Cross_product – Max Jun 15 '21 at 06:33
  • Now compare the two (in the N, T, B basis). – Max Jun 15 '21 at 06:34
  • You used $W \times T = T' = \kappa N$, but didn't finish saying how a,b were related to $\kappa$. After that use $W \times B = B' = -\tau N$. – AHusain Jun 15 '21 at 06:55

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