Outward conormal of a surface boundary ($uds=d\mathbf{r}\times n$)

On the smooth boundary of an orientable surface, let $n$ be the chosen unit normal vector and $t$ the positively oriented unit tangent vector used by Stokes theorem. The outward conormal is $u=t\times n$. It is a unit vector, tangent to the surface and orthogonal to its boundary. This convention turns $(n\times v)\cdot d\mathbf{r}$ into $u\cdot vds$ by cyclic invariance of the scalar triple product.