Euclidean distance ($d_2(x,y)=\Vert x-y{\Vert }_2$)

The Euclidean distance between $x,y\in {\mathbb{R}}^d$ is

If a surface lies outside a sphere centered at the origin, its minimum distance to the sphere is its minimum radial distance minus the sphere radius. For each surface point, the triangle inequality gives that lower bound, and the radial point on the sphere attains it. Thus minimizing the squared norm on the surface suffices.

For a line $Ax+By+C=0$ with $A^2+B^2>0$, its unit normal is $(A,B)/\sqrt{A^2+B^2}$. Choose any point on the line and project the displacement to $(x_0,y_0)$ onto this normal. The absolute value of that projection gives the title formula. This is the shortest Euclidean distance, because any displacement to a different point on the line has the same normal component and an additional orthogonal tangential component.

For distinct points ${\mathbf{r}}_1,{\mathbf{r}}_2$, the cross product in the numerator has magnitude twice the area of the triangle with third vertex ${\mathbf{r}}_0$. The same double area is the base length $|{\mathbf{r}}_2-{\mathbf{r}}_1|$ times the perpendicular height from ${\mathbf{r}}_0$, proving the displayed Euclidean distance. Equivalently, let $\mathbf{u}={\mathbf{r}}_2-{\mathbf{r}}_1$ and project ${\mathbf{r}}_0-{\mathbf{r}}_1$ onto the line direction. The squared distance is $|{\mathbf{r}}_0-{\mathbf{r}}_1{|}^2-[\mathbf{u}\cdot ({\mathbf{r}}_0-{\mathbf{r}}_1){]}^2/|\mathbf{u}{|}^2$; Lagrange's identity gives the same cross-product formula.

For two planes $r\cdot b=\lambda$ and $r\cdot c=\mu$ with linearly independent normal vectors, put $B=b\cdot b$, $C=c\cdot c$, $D=b\cdot c$ and $\Delta =BC-D^2>0$. Their intersection is a line parallel to $b\times c$. Its closest point to the origin is orthogonal to that direction, hence has the form $a=\alpha b+\beta c$. The constraints give $B\alpha +D\beta =\lambda$, $D\alpha +C\beta =\mu$, soFor every other point $a+t(b\times c)$, orthogonality gives squared distance $|a{|}^2+t^2|b\times c{|}^2$, proving the global minimum and its uniqueness. The positive determinant $\Delta =|b\times c{|}^2$ also proves that an intersection exists for every real $\lambda ,\mu$. Parallel normals require a separate consistency check and do not define a unique intersection line.

With fixed positive scale estimates $s_j$, this is Euclidean distance in standardized coordinates. It removes marginal units but ignores correlations. Small estimated variances can greatly amplify measurement noise; a zero scale must be handled separately.