Let f(t) be a smooth curve such that $f'(t)\not=0$ for all t. Then we can define unit tangent vector T by $T(t)=\frac{f'(t)}{||f'(t)||}$. So $$T'(t)=\frac{f'(t)\times(f''(t)\times f'(t))}{||f'(t)||^3}$$
Now, assume that f '(t) and f ''(t) are not parallel. Then $T'(t)\not=0$, so we can define unit principal normal vector N by $$N(t)=\frac{T '(t)}{||T '(t)||}$$.
Now, how to show that $$N(t)=\frac{f'(t)\times(f''(t)\times f'(t))}{||f'(t)||*||f''(t)||\times ||f'(t)||}$$
My attempt: $$N(t)=\frac{f'(t)\times(f''(t)\times f'(t))}{||f'(t)||^3*||T'(t)||}$$
I don't understand how to compute $$||T'(t)||=\frac{||f'(t)\times (f''(t)\times f'(t))||}{||\left(||f'(t)||^3\right)||}$$
If any member knows the correct answer to this question, may reply with correct answer.