# Tangent, Normal and Binormal vectors

Let $C$ be the graph of vector function $r(s)$ parametrized by arc length with $r'(0)= ({2\over3},{1\over3},{2\over3})$ and $r''(0)=(-3,12,-3)$.

I need to find tangent vector $T(0)$, normal vector $N(0)$ and binormal vector $B(0)$.

I found definitions of $T,N$ and $B$ in literature but I don't get how to plug in given values.

Edit: Thanks to anon's explanation I got (revison 2):

$T(0) = {r'(0)\over||r'(0)||} = {r'(0)\over1} = ({2\over3},{1\over3},{2\over3})$

$N(0) = {T'(0)\over||T'(0)||} = {r''(0)\over||r''(0)||} = {{r''(0)}\over{9\sqrt2}} = (-{1\over3\sqrt2},{4\over3\sqrt2},-{1\over3\sqrt2})$

$B(0) = T(0) \times N(0) = (-\sqrt2,0,\sqrt2)$

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What definitions did you find? You might want to include them in your question. – J. M. Oct 22 '11 at 10:34
@10smi: Your calculation of $\mathbf{N}$ is wrong, and subsequently so is your calculation of $\mathbf{B}$. There is a difference between the first and second derivatives of $\mathbf{r}$. You never answered J.M. or I's request to add in what facts you do know so we can help you better. – anon Oct 22 '11 at 11:37
I have added formulas from literature. I hope it is correct now. – 1osmi Oct 22 '11 at 12:42

You should know that $\mathbf{T}(s)=\dot{\mathbf{r}}(s)$ (because $\|\dot{\mathbf{r}}\|=1$) hence $\mathbf{N}(s)=\mathbf{T}_s/\|\mathbf{T}_s\|=\ddot{\mathbf{r}}(s)/\|\ddot{\mathbf{r}}(s)\|$ and of course the binormal can be computed with the other two using $\mathbf{B}(s)=\mathbf{T}(s)\times\mathbf{N}(s)$. Do you see how to plug in what information you have now for the answers?
Okay, so tangent vector is $T(0)=r'(0)$? – 1osmi Oct 22 '11 at 11:08