Chapter 2: Space Curves
Section 2.6: Binormal and Torsion
Prove that B′s is necessarily along N.
Since Bs is a unit vector, B·B′=0, so that B′ is orthogonal to B. That puts B′ somewhere in the osculating plane, that is, the plane formed by T and N.
Now B is orthogonal to T, so B·T=0. Differentiating gives B′·T+B·T′=0. Rearranging gives
= −B·κ N
where T′=κ N is the basis for the definition of the curvature κ.
Thus, B′, already in the osculating plane, is now orthogonal to T. Hence, it must be along N, which was to be established. Incidentally, since B′ is along N, it must be that it is a multiple of N, and that multiple is taken as a measure of the torsion by writing B′= −τ N.
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