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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.2012-04-04
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    @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.$2012-04-04
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    did you mean $\frac{d}{ds}<(\mathbf{r}(s)-\mathbf{r}(0)), \mathbf{b}(s)>$ ?2012-04-04
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    Both are the same, since $\mathbf{b}(s)$ is constant.2012-04-04
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    Ok, then $<\mathbf{v}(s),\mathbf{b}(s)> = 0 \Rightarrow \frac{d}{ds}<(\mathbf{r}(s)-\mathbf{r}(0)), \mathbf{b}(s)> = 0 \Rightarrow <(\mathbf{r}(s)-\mathbf{r}(0)), \mathbf{b}(s)> = Const$. We need to show that the $Const$ is $0$.2012-04-04
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    Yeap, set $s=0.$2012-04-04
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    I get it, thanks for support!2012-04-04
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    I'm wondering about the need of $b(s)$ to be constante. I could simply affirm and show $r$ is in the plane determined by $b(0)$ and by the point $r(0)$, is there something wrong with this approach?2014-04-22