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For a unit speed curve

Show that $ \langle T^{'},B^{'}\rangle=- \kappa \tau. $

$T^{'} = \kappa N$ and $B^{'}=-\tau N $

(I'm just getting this from the $3×3$ matrix for $(T,N,B)'$ can use it definitionally from my text/lecture as well.)

I have $⟨\kappa N,\,-\tau N⟩ $ the dot product can be re written as $-\kappa \tau ⟨N,\, N⟩$ which is $1$ so. $ ⟨T^{'},B^{'}⟩ =-\kappa \tau$

I think you can pull a factor out of a dot product as long as you pull it out of every single term in the vector but I don't know if not this definitely isn't right.

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    This is fine. Remember that $\langle ax+by, cz\rangle = ac\langle x,z\rangle + bc\langle y,z\rangle$. At some point my free differential geometry text might be of use to you. You can download it by clicking in my profile. – 2017-01-13
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    Thanks I will check it out :) – 2017-01-13

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