Show that any curve is a geodesic on the surface generated by its binormals.
Normal property says that, any curve $u=u(s),v=v(s)$ on a surface $\textbf{r}=\textbf{r}(u,v)$ is a geodesic iff the principal normal at every point on the curve is normal to the surface. We have the identities : $$U(s)=\frac{d}{ds}\bigg(\frac{\partial T}{\partial u'}\bigg)-\frac{\partial T}{\partial u}=\textbf{r}_1.\textbf{r}'',V(s)=\frac{d}{ds}\bigg(\frac{\partial T}{\partial v'}\bigg)-\frac{\partial T}{\partial v}=\textbf{r}_2.\textbf{r}''$$ where $\displaystyle\textbf{r}_1=\frac{\partial\textbf{r}}{\partial u}$ and $\displaystyle\textbf{r}_2=\frac{\partial\textbf{r}}{\partial v}$ and $\displaystyle T=\frac{1}{2}(Eu'^2+2Fu'v'+Gv'^2)$. Now we also have $$\mathbf{r}''=\frac{d^2\textbf{r}}{ds^2}=\frac{d\textbf{t}}{ds}=\kappa\textbf{n}$$ by first Frenet-Serret equation. So clearly $\textbf{r}''$ is the principal normal. Now to conclude the result using normal property we are left to show that $$\boxed{\textbf{r}_1.\textbf{r}''=\textbf{r}_2.\textbf{r}''=0}$$ so that $\textbf{r}''$ is also the surface normal to $\textbf{r}$. But how to show this using that surface is generated by binormals I have no idea. Any help is appreciated.
