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The one-dimensional kernel of $d{K}_x$ is the tangent to its level curve; the metric gives two unit normals and $d{K}_x(e_x)>0$ selects one. Isometries preserve the metric, $K$, and this orientation choice, so they carry $e_x$ to $e_{\varphi (x)}$. Therefore $g(x)=d{K}_x(e_x)=|\nabla K(x)|$ is isometry invariant.

Orient the plane $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal $\mathbf{n}$. The signed distance of a point $\mathbf{p}$ from the plane isIts absolute value is the ordinary distance from a point to a plane. The closest point in the plane is $\mathbf{p}-s\mathbf{n}$, so the displacement from the point to its projection is $-s\mathbf{n}$. Reversing the normal and $d$ reverses $s$ without changing the geometric plane or the unsigned distance. A sphere of radius $q$ centred at $\mathbf{p}$ has sphere-plane intersection radius $\sqrt{q^2-s^2}$ when $|s|<q$.