Why does this implication work regarding the normal and binormal vectors 
I'm reading my notes on geometry of curves and trying to understand the intuition behind the definition of torsion. Specifically, I don't understand why $\|b\|=1$ implies that $b\perp{}b'$ and hence implieds that $b'\parallel{}n$. I understand the proceeding line though.
Any help in ways of thinking about this would be much appreciated.
 A: If $\|b\|=1,$ then $b\cdot b=1.$ If you differentiate this, you get $2b\cdot b'=0,$ so $b\perp b'.$ Since we also know that $b\perp t,b\perp n$, this tells us that $b'$ must be parallel to either $t$ or $n$ (we're in $\mathbb{R}^3$, so non-zero elements can only be perpendicular to two directions). Since the first part of the proof showed that $b'\perp t$, we must have that $b'$ is parallel to $n$.
A: For any parametrized, differentiable unit vector $X(s)$ we have
$X(s) \cdot X(s) = \Vert X(s) \Vert^2 = 1, \tag 1$
and if we differentiate this equation with respect to $s$ we obtain, via the product rule for derivatives,
$2X^\prime(s) \cdot X(s) = X^\prime(s) \cdot X(s) + X(s) \cdot X^\prime(s) = 0, \tag 2$
which forces
$X^\prime(s) \cdot X(s) = 0, \tag 3$
that is, 
$X^\prime(s) \bot X(s). \tag 4$
Clearly this argument applies to $\mathbf b$, whence
$\mathbf b^\prime \bot \mathbf b. \tag 5$
Since, as our OP kam has shown,
$\mathbf b^\prime \bot \mathbf t, \tag 6$
the fact that we are working in $\Bbb R^3$ forces
$\mathbf b^\prime \parallel \mathbf n; \tag 7$
in greater detail, if we write
$\mathbf b^\prime = p \mathbf t + q \mathbf n + r \mathbf b, \tag 8$
generally possible since $\mathbf t$, $\mathbf n$, $\mathbf b$ form an orthonormal frame in $\Bbb R^3$, then it is clear from (5), (6) that 
$p = r = 0, \tag 9$
leaving us with
$\mathbf b^\prime = q \mathbf n, \tag{10}$
and now a simple change of notation
$q = -\tau \tag{11}$
allows us to write
$\mathbf b^\prime = -\tau \mathbf n. \tag{12}$
The perhaps more conventional derivation of this equation stems from expressing $\mathbf n^\prime$ in terms of $\mathbf t$, $\mathbf n$, and $\mathbf b$; indeed, we define $\mathbf n$ via
$\mathbf t^\prime = \kappa \mathbf n, \; \Vert \mathbf n \Vert = 1 \tag{13}$
when
$\kappa = \Vert \mathbf t^\prime \Vert > 0; \tag{14}$
since
$\Vert \mathbf t \Vert = 1, \tag{15}$
as in (1)-(4) we have
$\kappa \mathbf n \cdot \mathbf t = \mathbf t^\prime \cdot \mathbf t = 0, \tag{16}$
whence in light of (14), 
$\mathbf n \cdot \mathbf t = 0, \tag{17}$
from which we have, again in light of the product rule,
$\mathbf n^\prime \cdot \mathbf t + \mathbf n \cdot \mathbf t^\prime = 0, \tag{18}$
or
$\mathbf n^\prime \cdot \mathbf t = -\mathbf n \cdot \mathbf t^\prime = -\mathbf n \cdot \kappa \mathbf n = -\kappa; \tag{19}$
from (13), again as in (1)-(4), 
$\mathbf n^\prime \cdot \mathbf n = 0. \tag{20}$
We have thus obained the components of $\mathbf n^\prime$ along $\mathbf t$ and $\mathbf n$; but since we are working in $\Bbb R^3$ we may need a third component to fully express $\mathbf n^\prime$; we acccomodate this prospect by defining
$\mathbf b = \mathbf t \times \mathbf n, \tag{21}$
so that
$\mathbf b \cdot \mathbf n = \mathbf b \cdot \mathbf t = 0, \tag{22}$
and in light of (17), 
$\Vert \mathbf b \Vert = 1; \tag{23}$
thus $\mathbf t$, $\mathbf n$, and $\mathbf b$ form an orthonormal triad in $\Bbb R^3$ and so we may complete the description of $\mathbf n^\prime$ by taking
$\mathbf n^\prime \cdot \mathbf b = \tau, \tag{24}$
which defines the torsion of our (tacitly) given curve.  Then
$\mathbf n^\prime = -\kappa \mathbf t + \tau \mathbf b, \tag{25}$
and from (22) we find
$\mathbf b^\prime \cdot \mathbf n + \mathbf b \cdot \mathbf n^\prime = 0, \tag{26}$
which via (24) yields
$\mathbf b^\prime \cdot \mathbf n = -\mathbf b \cdot \mathbf n^\prime = -\tau. \tag{27}$
Since by (1)-(6)
$\mathbf b' \cdot \mathbf b = \mathbf b^\prime \cdot \mathbf t = 0, \tag{28}$
we conclude from (27) that
$\mathbf b^\prime = -\tau \mathbf n, \tag{29}$
which in turn gives us (7).
Note Added in Edit, Friday 20 December 2019 10:06 AM PST:  The arguments (1)-(4) that $X^\prime(s) \bot X(s)$, (17)-(19) that $\mathbf n^\prime \cdot \mathbf t = -\mathbf n \cdot \mathbf t^\prime$ and (24), (26)-(27) that $\mathbf b^\prime \cdot \mathbf n = -\mathbf b \cdot \mathbf n^\prime$ may in fact be succinctly subsumed under the observation that, for any two parametrized, differentiable vectors $X(s)$ and $Y(s)$ such that
$X(s) \cdot Y(s) = c, \; \text{a constant}, \tag{30}$
we have
$X^\prime(s) \cdot Y(s) = -X(s) \cdot Y^\prime(s), \tag{31}$
which follows from (30) upon differentiation using the product rule; then taking 
$Y(s) = X(s) \tag{32}$
yields
$X^\prime(s) \cdot X(s) = 0, \tag{33}$
i.e.,
$X^\prime(s) \bot X(s). \tag{34}$
End of Note.
