Vector calculus - Vector Path Properties

Question

I've got a question about cross product properties and vectors.

So, $\mathbf T, \mathbf N $ and $\mathbf B $ are the unit tangent, normal and binormal vectors to a $C^3$ path. And I need to prove that $\frac{\mathrm d \mathbf B}{\mathrm dt} \cdot \mathbf B = 0 $

so far, I've done the following:

$$ \frac{\mathrm d\vec {\mathbf B}}{\mathrm dt} = \frac{\mathrm d}{\mathrm d t} \Bigl[\mathbf T (t) \times \mathbf N(t)\Bigl] $$ so $$ \frac{\mathrm d\vec {\mathbf B}}{\mathrm dt} \cdot \mathbf B = \Bigl(\frac{\mathrm d}{\mathrm d t} \Bigl[\mathbf T (t) \times \mathbf N(t)\Bigl]\Bigl) \cdot \Bigl(\mathbf T (t) \times \mathbf N(t)\Bigl)$$

$$ = \Biggl(\frac{\mathrm d \mathbf T}{\mathrm d t} \times \mathbf N + \frac{\mathrm d \mathbf N}{\mathrm dt} \times \mathbf T\Biggl) \cdot \Bigl(\mathbf T \times \mathbf N \Bigl)$$

$$ = \Biggl(\frac{\mathrm d \mathbf T}{\mathrm d t} \times\frac{\frac{\mathrm d \mathbf T}{\mathrm d t}}{\big\lvert\frac{\mathrm d \mathbf c}{\mathrm d t}\big\rvert} + \frac{\mathrm d \mathbf N}{\mathrm dt} \times \mathbf T\Biggl) \cdot \Bigl(\mathbf T \times \mathbf N \Bigl) $$

$$ = \require{cancel}\Biggl(\biggl(\frac{1}{\big\lvert\frac{\mathrm d \mathbf c}{\mathrm d t}\big\rvert}\biggl)\biggl(\cancel{\frac{\mathrm d \mathbf T}{\mathrm d t} \times \frac{\mathrm d \mathbf T}{\mathrm d t}}\biggl)\Bigg) + \frac{\mathrm d \mathbf N}{\mathrm dt} \times \mathbf T\Biggl) \cdot \Bigl(\mathbf T \times \mathbf N \Bigl)$$

$$ = \Bigg( \frac{\mathrm d \mathbf N}{\mathrm dt} \times \mathbf T\Biggl) \cdot \Bigl(\mathbf T \times \mathbf N \Bigl)$$

From here though I'm not sure how to proceed. I know that $$\mathbf N (t) = \frac{\frac{\mathrm d \mathbf T}{\mathrm d t}}{\big\lvert\frac{\mathrm d \mathbf T}{\mathrm d t}\big\rvert}$$ and $$ \mathbf T (t) = \frac{\frac{\mathrm d \mathbf T}{\mathrm d t}}{\big\lvert\frac{\mathrm d \mathbf c}{\mathrm d t}\big\rvert}$$ but otherwise I'm not sure how to proceed. Any help would be fantastic, cheers :)

Answer 1

Hint: Prove $||\mathbf B||=\text{Constant}$, then $$\dfrac{d}{dt}||\mathbf B||^2=2\frac{\mathrm d \mathbf B}{\mathrm dt} \cdot \mathbf B = 0$$