Why is the derivative of the binormal vector parallel to the normal vector? In my notes, it says to consider an arc length parameterization r(s). Then we can show that B' points in the direction of N such that we can write B'=-N. I understand why ||B'||= but am unsure how to show why B' is parallel to N? Also, is the minus sign just for convention or is there a reason it is there? 
Any help would be greatly appreciated!
 A: $\def\vt{\mathbf{T}}
\def\vn{\mathbf{N}}
\def\vb{\mathbf{B}}
\def\vr{\mathbf{r}}$In what follows differentiation with respect to arc length is indicated with a prime. 
We have 
\begin{align*}
\vt &= \vr' \\
\vn &= \vt'/\|\vt'\| \\
\vb &= \vt\times\vn.
\end{align*}
Then 
\begin{align*}
\vb' &= \vt'\times\vn+\vt\times\vn' \\
&= \|\vt'\|\underbrace{\vn\times\vn}_{\mathbf{0}} + \vt\times\vn' \\
&= \vt\times\vn'.
\end{align*}
Note that $\vb$ is orthogonal to $\vt$ by definition of the cross product. 
Since $\|\vb\|=1$, $\vb'$ is also orthogonal to $\vb$. 
(Proof: $(\vb\cdot\vb)' = 2\vb\cdot\vb' = 0$.) 
Since $\vt,\vn,\vb$ form an orthogonal basis, it must be the case that $\vb'$ is proportional to $\vn$. 
We take 
$\vb' = -\tau\vn$. 
The minus sign is a convention. 
It is a version of the right-hand rule. 
Note that $\tau>0$ indicates a twisting of the coordinate frame about the $\vt$ axis in a right-handed manner. 
(Take your right hand and point your thumb in the $\vt$ direction. 
Twist your hand about this axis in the direction in which your fingers are pointing. 
This twist corresponds to $\tau>0$.) 
