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B is a binormal vector where B = T x N. T is the unit tangent and N is the principal normal vector. Basically, the derivative of the binormal vector is perpendicular to both the unit tangent and the binormal vector.

I expanded everything out using B = T x N and the cross product differentiation rule but I ended up with (N' x T) . (T x N).

iuppiter
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    By definition, $T$ is an unit vector and $N$ is another unit vector in the direction of $T'$. Notice $$|T| = 1 \implies T\cdot T = 1 \implies T \cdot T' = 0 \implies T \cdot N = 0 \ \implies |B| = |T \times N| = \sqrt{|T|^2 |N|^2 - (T\cdot N)^2} = 1$$ $B = T \times N$ is yet another unit vector and hence $B\cdot B' = 0$. – achille hui Apr 19 '13 at 03:10

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Yes, the most basic observation is that $B$ is always a unit vector, so its derivative must be a vector orthogonal to $B$. But from the formula you had, $N' \times T$ must be in the $N$ direction, hence orthogonal to $T\times N$.

Ted Shifrin
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