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I have proved that a planar curve of zero curvature is a straight line. It follows from the Frenet equations. But now I need to prove that if $\varkappa=0$, then the space curve $\mathbf{r}(t)$ is planar. From the condition and the Frenet equations it follows that $$ \left\{ \begin{aligned} \frac{d}{ds}\mathbf{v}&=k(s)\mathbf{n}(s),\\ \frac{d}{ds}\mathbf{n}&=-k(s)\mathbf{v}(s),\\ \frac{d}{ds}\mathbf{b}&=0.\\ \end{aligned} \right. $$

But how can be technically deduced from these equations that the curve is planar?

Update: from a related question planar curve if and only if torsion I have realized that I need to show that $(\mathbf{r}(t)-\mathbf{r}(t_0))\cdot\mathbf{b}(t)=0$ for any $t$ and some $t_0$. The question now is how to do that. I appreciate any help.

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Hint: So $\mathbf{b}(s)$ is a constant vector $\mathbf{b}(0).$ Now, consider the plane passing through $\mathbf{r}(0)$ with the normal vector $\mathbf{b}(0).$ Does the plane contain $\mathbf{r}(s)?$

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The plane contains $\mathbf{r}(s)$ if $(\mathbf{r}(s)-\mathbf{r}(0))\cdot\mathbf{b}(0)=0$. I need to show that. –  Sergey Filkin Apr 4 '12 at 7:44
    
@SergeyFilkin: Yes, indeed. Work with $<\frac{d}{ds}(\mathbf{r}(s)-\mathbf{r}(0)), \mathbf{b}(s)>$ where $<.,.>$ is the usual inner product in $\mathbb{R}^3.$ –  Ehsan M. Kermani Apr 4 '12 at 7:49
    
did you mean $\frac{d}{ds}<(\mathbf{r}(s)-\mathbf{r}(0)), \mathbf{b}(s)>$ ? –  Sergey Filkin Apr 4 '12 at 7:52
    
Both are the same, since $\mathbf{b}(s)$ is constant. –  Ehsan M. Kermani Apr 4 '12 at 7:56
1  
I get it, thanks for support! –  Sergey Filkin Apr 4 '12 at 8:15

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