Given a regular curve $a(s) \in \mathbb{R}^3$, we can attach a moving frame given by three orthogonal unit vectors, $T(s), N(s), B(s)$.
Now, my question is does $B'(s)$ (the derivative of the $B(s)$ unit vector, which we obtained from taking $T(s) \times N(s)$ only depend on the it's projection with respect to $N(s)$?
That is to say, $B'(s)= \tau N$, where $\tau$ is the torsion of the curve. I follow my books argument, and it seems to be technical rather than intuitive. To me, it seems like the rate of change of $B(s)$ should also depend on it's projection against $T$ also. It'd be possible for the rotation of the $B(s)$ vector to be completely in the $BT$ plane (the rectifying plane?), and if this is true, $B'(s) \neq 0$ but none of this change in position of $B(s)$ that it's derative represents will be in the direction of the $N$ vector. Am I making myself clear on why I'm confused?
In order for two unit vectors to stay perpendicular, the second must turn away from the first at precisely the same rate as the first turns toward the second. (Mathematically, this is saying $B'\cdot T = -T'\cdot B$.) You can prove this by the product rule, but it's also quite clear if you just rotate your hand, keeping the thumb and forefinger perpendicular.
Since by definition $T$ turns only toward $N$ (i.e., it rotates in the osculating plane, instantaneously), it doesn't turn at all toward $B$ and so $B$ can't turn at all toward $T$.