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Let us assume that $\alpha(s)$ is a unit speed curve with $\kappa > 0$. I'm trying to find the vector function $w(s)$ such that

$T' = w \times T,\quad N' = w \times N,\quad B' = w \times B.$

I see that this is an application of Frenet Serret with $T' = \kappa\times N, N' = \frac{T'}{|T|}$, and $B' = \langle-\tau, N\rangle$ and I am just not seeing how to group these guys to get what I want, i.e., vector $w$. Do I need to use the def of an osculating plane or the right hand rule? Thanks

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    I think $N'=T'/|T|$ should be wrong. Also other reported formulas seems to be wrong: $\kappa\times N$, but $\kappa$ is not a vector.2012-10-08

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If you do the procedure in my comment you find that $w=\kappa B -\tau T$ where $\kappa$ is the curvature and $\tau$ is the torsion of $\alpha$.

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From Frenet we have:

$\left[ \begin{array}{c} T' \\ N' \\ B' \end{array} \right] = \left[ \begin{array}{ccc} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0 \end{array} \right].\left[ \begin{array}{c} T \\ N \\ B \end{array} \right],$

where the curvature and torsion are respectively denoted $\kappa$ and $\tau.$

So from the first row from the Frenet we have, $T' = \kappa N$ and then crossing both sides by $T(s)$, we get $w(s) \times T = b N \times T + c B \times T = \kappa N,$ which only holds if $b = 0$ and $c = \kappa$.

From the second row of the Frenet frame we have, $N' = -\kappa T + \tau B$ and then crossing both sides by $N(s)$, we get, $w(s) \times N = a T \times N + c B \times N = -\kappa T + \tau B,$ which only holds if $a = \tau$ and $c = \kappa$.

Finally, the third row of the Frenet frame, $B' = -\tau N$ and then crossing both sides by $B(s)$, we get $w(s) \times B = a T \times B + b N \times B = - \tau N,$ which only holds if $a = \tau$ and $b=0.$

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I just happen to have done this problem and here is a link to my solution:

http://andrew-exercise.blogspot.com/2016/01/differential-geometry-and-its_11.html

The vector you are looking for a called the Darboux vector and is used to describe the rotation of a rigid body following the curve.

My conclusion seems to have an opposite sign with the answer from @PAD, but I am pretty sure I am correct as the problem on my book requested for the verification of a given solution.