Distance between affine spaces using least squares method

Question

Let's represent an $n$-dimensional affine space $L$ as an affine frame, so an origin point $p$ and linear basis $\{ e_1, e_2, ..., e_n \}$. How to compute the minimum Euclidean distance between affine spaces $L_1$ and $L_2$ using the least squares method?

Answer 1

Here is how I would try: let the two spaces be ${\cal A}$ and ${\cal B}$ and let $A\in {\cal A}$ and $B \in {\cal B}$ be two points that minimize $d(A, B)^2$. A necessary and sufficient condition is that the line $AB$ is orthogonal to the affine manifolds ${\cal A}$ and ${\cal B}$. Let the coordinates of $A$ and $B$ in the frame $(p, e_1,\ldots,e_n)$ be

$$A(a_1,\ldots, a_n),\quad B(b_1,\ldots, b_n)$$

Let ${\cal A}$ be defined by affine conditions $$\varphi_i(a_1,\ldots, a_n) = \alpha_i$$ Let ${\cal B}$ be defined by affine conditions $$\psi_j(b_1,\ldots, b_n) = \beta_j$$ Assuming $e_1,\ldots,e_n$ is an orthonormal basis, the orthogonality condition leads to the linear system $$ \left\{ \begin{array}\\ \varphi_i(a_1,\ldots, a_n) = \alpha_i\\ \psi_j(b_1,\ldots, b_n) = \beta_j\\ B - A = \sum_i {\lambda_i\varphi_i}\\ B - A = \sum_j \mu_j\psi_j \end{array} \right. $$ Solve this in the unknowns $a_k$, $b_k$, $\lambda_k$, $\mu_k$. The distance between the manifolds is then $d(A, B)$.

Numerical example: take the two straight lines in ${\mathbb{R}}^{3}$ defined respectively by the equations

$$\left\{\begin{array}{rcl}x+y+z&=&1\\ 2 x-y+3 z&=&2 \end{array}\right.$$

and

$$\left\{\begin{array}{rcl}x-2 y-z&=&{-1}\\ 3 x-4 y+z&=&1 \end{array}\right.$$

The above procedure leads to the system

$$\left\{\begin{array}{rcllllllllll}1&=&{a}_{1}&{+{a}_{2}}&{+{a}_{3}}&&&&&&&\\ 2&=&2 {a}_{1}&{-{a}_{2}}&{+3} {a}_{3}&&&&&&&\\ {-1}&=&&&&{b}_{1}&{-2} {b}_{2}&{-{b}_{3}}&&&&\\ 1&=&&&&3 {b}_{1}&{-4} {b}_{2}&{+{b}_{3}}&&&&\\ 0&=&{a}_{1}&&&{-{b}_{1}}&&&{-{{\lambda}}_{1}}&{-2} {{\lambda}}_{2}&&\\ 0&=&&{a}_{2}&&&{-{b}_{2}}&&{-{{\lambda}}_{1}}&{+{{\lambda}}_{2}}&&\\ 0&=&&&{a}_{3}&&&{-{b}_{3}}&{-{{\lambda}}_{1}}&{-3} {{\lambda}}_{2}&&\\ 0&=&{a}_{1}&&&{-{b}_{1}}&&&&&{-{{\mu}}_{1}}&{-3} {{\mu}}_{2}\\ 0&=&&{a}_{2}&&&{-{b}_{2}}&&&&{+2} {{\mu}}_{1}&{+4} {{\mu}}_{2}\\ 0&=&&&{a}_{3}&&&{-{b}_{3}}&&&{+{{\mu}}_{1}}&{-{{\mu}}_{2}} \end{array}\right.$$

The unique solution is

$$A = \left(\frac{11}{195} , \frac{46}{195} , \frac{46}{65}\right) \qquad B = \left(\frac{1}{5} , \frac{2}{15} , \frac{14}{15}\right)$$

$$\left({{\lambda}}_{1} , {{\lambda}}_{2}\right) = \left(\frac{4}{195} , \frac{{-16}}{195}\right) \qquad \left({{\mu}}_{1} , {{\mu}}_{2}\right) = \left(\frac{4}{30} , \frac{{-6}}{65}\right)$$

The distance between the two lines is

$$d \left(A , B\right) = \sqrt{\frac{16}{195}}$$