So we have been given the following:
$$\frac{dr}{dt} = (-3,2,0)$$
$$\frac{d^2r}{dt^2} = (0,3,-3)$$
$$\frac{d^3r}{dt^3} = (0,0,1)$$
With the information above, I have found the unit tangent vector by doing: $T = \frac{\frac{dr}{dt}}{|\frac{dr}{dt}|} = \frac{(-3,2,0)}{\sqrt{3^2+2^2+0^2}} = (\frac{-3}{\sqrt{13}},\frac{2}{\sqrt{13}},0)$
I have found the curve $k = \frac{|\frac{dr}{dt} \times \frac{d^2r}{dt^2}|}{|\frac{dr}{dt}|^3} = \frac{3\cdot \sqrt{22}}{13\cdot \sqrt{13}}$
Lastly i have found the torsion to be $\tau =\frac{(\frac{dr}{dt} \times \frac{dr^2}{dt^2})\cdot\frac{dr^3}{dt^3}}{|\frac{dr}{dt} \times \frac{dr^2}{dt^2}|^2} = -\frac{1}{22}$
Now i am having difficulties with finding the unit normal vector, N.
Best Answer
Since the acceleration is $a=(0,3,-3)$ and the unit tangent vector is $T=\left(-\frac3{\sqrt{13}},\frac2{\sqrt{13}},0\right)$, the normal acceleration is$$a_N=a-\langle a,T\rangle T=\frac1{13}(18,27,39).$$Since $N$ is a unit vector such that $a_N$ is a multiple of $N$, $N=\pm\frac1{\sqrt{286}}(6,9,-13)$. Which one? That can be determined using the fact that$$\langle a,N\rangle=\frac{\|v\times a\|}{\|v\|}\geqslant0.$$Since $\left\langle(0,3,-3),(6,9,-13)\right\rangle>0$,$$N=\frac1{\sqrt{286}}(6,9,-13).$$