Intersections and Proximities of p-planes.

 In $D$-dimensions, $p_1$-plane and $p_2$-plane has intersection of $(D-p_1-p_2)$-plane at least or with more dimensions. If $D-p_1-p_2<0$, they can have 0 to $\min \{p_1,p_2\}$ dimension proximity object or intersections. (i.e. no intersection guaranteed). This is because parameter of each dimension of plane gives constraint to the $D$-dimensional coordinate equation.

The two objects may be written by

$$\overrightarrow{x}(a)={\overrightarrow{v}}_0+\sum _{k=1}^{p_1}{a}_k{\overrightarrow{v}}_k,$$

$$\overrightarrow{y}(b)={\overrightarrow{w}}_0+\sum _{k=1}^{p_2}{b}_k{\overrightarrow{w}}_k.$$

The proximity or intersection maybe given by

$$\min |\overrightarrow{x}-\overrightarrow{y}|,$$

which may be represented by this extrema equation:

$$\{\begin{array}{l}0=\frac{\partial }{\partial a_k}{|\overrightarrow{x}-\overrightarrow{y}|}^2=2(\overrightarrow{x}-\overrightarrow{y})\cdot \frac{\partial }{\partial a_k}(\overrightarrow{x}-\overrightarrow{y})=2(\overrightarrow{x}-\overrightarrow{y})\cdot \frac{\partial \overrightarrow{x}}{\partial a_k}\\ 0\frac{\partial }{\partial b_k}{|\overrightarrow{x}-\overrightarrow{y}|}^2=2(\overrightarrow{x}-\overrightarrow{y})\cdot \frac{\partial }{\partial b_k}(\overrightarrow{x}-\overrightarrow{y})=-2(\overrightarrow{x}-\overrightarrow{y})\cdot \frac{\partial \overrightarrow{y}}{\partial b_k}\end{array}$$

$$\Rightarrow \{\begin{array}{l}0=(\overrightarrow{x}-\overrightarrow{y})\cdot \frac{\partial \overrightarrow{x}}{\partial a_k}={\overrightarrow{v}}_k\cdot (\overrightarrow{x}-\overrightarrow{y})\\ 0=(\overrightarrow{x}-\overrightarrow{y})\cdot \frac{\partial \overrightarrow{y}}{\partial b_k}={\overrightarrow{w}}_k\cdot (\overrightarrow{x}-\overrightarrow{y})\end{array}$$

$$\Rightarrow \{\begin{array}{l}0={\overrightarrow{v}}_k\cdot (\overrightarrow{x}-\overrightarrow{y})={\overrightarrow{v}}_k\cdot ({\overrightarrow{v}}_0-{\overrightarrow{w}}_0)+\sum _{l=1}^{p_1}{\overrightarrow{v}}_k\cdot {\overrightarrow{v}}_l{a}_l-\sum _{l=1}^{p_1}{\overrightarrow{v}}_k\cdot {\overrightarrow{w}}_l{b}_l\\ 0=-{\overrightarrow{w}}_k\cdot (\overrightarrow{x}-\overrightarrow{y})=-{\overrightarrow{w}}_k\cdot ({\overrightarrow{v}}_0-{\overrightarrow{w}}_0)-\sum _{l=1}^{p_2}{\overrightarrow{w}}_k\cdot {\overrightarrow{v}}_l{a}_l+\sum _{l=1}^{p_2}{\overrightarrow{w}}_k\cdot {\overrightarrow{w}}_l{b}_l\end{array}$$

The equations may be re-written by

$$\left(\begin{array}{ccc}\ast & \ast & \ast \\ {\overrightarrow{v}}_k\cdot ({\overrightarrow{v}}_0-{\overrightarrow{w}}_0)& {\overrightarrow{v}}_k\cdot {\overrightarrow{v}}_l& -{\overrightarrow{v}}_k\cdot {\overrightarrow{w}}_l\\ -{\overrightarrow{w}}_k\cdot ({\overrightarrow{v}}_0-{\overrightarrow{w}}_0)& -{\overrightarrow{w}}_k\cdot {\overrightarrow{v}}_l& {\overrightarrow{w}}_k\cdot {\overrightarrow{w}}_l\end{array}\right)\left(\begin{array}{c}1\\ a_l\\ b_l\end{array}\right)=\left(\begin{array}{c}\ast \\ 0\\ 0\end{array}\right)$$

or

$$\left(\begin{array}{cc}{\overrightarrow{v}}_k\cdot {\overrightarrow{v}}_l& -{\overrightarrow{v}}_k\cdot {\overrightarrow{w}}_l\\ -{\overrightarrow{w}}_k\cdot {\overrightarrow{v}}_l& {\overrightarrow{w}}_k\cdot {\overrightarrow{w}}_l\end{array}\right)\left(\begin{array}{c}{a}_l\\ b_l\end{array}\right)=\left(\begin{array}{c}-{\overrightarrow{v}}_k\cdot ({\overrightarrow{v}}_0-{\overrightarrow{w}}_0)\\ {\overrightarrow{w}}_k\cdot ({\overrightarrow{v}}_0-{\overrightarrow{w}}_0)\end{array}\right)$$

With convention of column vector $\overrightarrow{v}$'s, and 

$${{\mathbf{m}}_0}_{(D\times 1)}:={\overrightarrow{v}}_0-{\overrightarrow{w}}_0,{\mathbf{M}}_{(D\times (p_1+p_2))}:=\left(\begin{array}{cc}{\overrightarrow{v}}_k& -{\overrightarrow{w}}_k\end{array}\right),{\mathbf{a}}_{((p_1+p_2)\times 1)}=\left(\begin{array}{c}{a}_k\\ b_k\end{array}\right),$$

the equations will be re-written by

$$({\mathbf{M}}^T\mathbf{M})\mathbf{a}=-{\mathbf{M}}^T{\mathbf{m}}_0.$$

So, the solution of parameters is

$$\mathbf{a}=-({\mathbf{M}}^T\mathbf{M}{)}^{-1}{\mathbf{M}}^T{\mathbf{m}}_0$$

unless ${\mathbf{M}}^T\mathbf{M}$ is singular.

If singular, to separate singular (freed) and non-singular (constraint) components, we need to diagonalize it. We may try SVD to avoid catastrophic result.

$$({\mathbf{M}}^T\mathbf{M})=:\mathbf{U}\mathbf{\Sigma }{\mathbf{V}}^T$$

where $U$ and $V$ are orthogonal (Hermitian) and $\mathrm{\Sigma }$ is a nonnegative diagonal matrix.

The parameters will be rotated by the orthogonal matrices to diagonalize the matrix to see the free variable.

$$\mathbf{U}\mathbf{\Sigma }{\mathbf{V}}^T\mathbf{a}=-{\mathbf{M}}^T{\mathbf{m}}_0\Rightarrow \mathbf{\Sigma }{\mathbf{V}}^T\mathbf{a}=-{\mathbf{U}}^T{\mathbf{M}}^T{\mathbf{m}}_0=:\left(\begin{array}{c}{m}_r\\ 0\end{array}\right)$$

where in the rows with the zero values in $\mathrm{\Sigma }$, the vector component is also zeros, so the solution for that row is indefinite. We know indefinite because it is zero divide by zero.

$${\mathbf{V}}^T\mathbf{a}={\mathbf{\Sigma }}^{-1}\left(\begin{array}{c}{m}_r\\ z\end{array}\right)=\left(\begin{array}{c}{\mathrm{\Sigma }}_r^{-1}{m}_r\\ 0^{-1}\cdot 0\end{array}\right)$$.

Now, substitute the indefinite numbers to free parameters $z_i$.

$${\mathbf{V}}^T\mathbf{a}=\left(\begin{array}{c}{\mathrm{\Sigma }}_r^{-1}{m}_r\\ z_i\end{array}\right)\Rightarrow \mathbf{a}=\mathbf{V}\left(\begin{array}{c}{\mathrm{\Sigma }}_r^{-1}{m}_r\\ z_i\end{array}\right).$$

Also, we can find the shortest distance using the equation and the solution. The squared distance is

$${|\overrightarrow{x}-\overrightarrow{y}|}^2={|{\mathbf{m}}_0+\mathbf{M}\mathbf{a}|}^2={\mathbf{m}}_0^T{\mathbf{m}}_0+2{\mathbf{a}}^T{\mathbf{M}}^T{\mathbf{m}}_0+{\mathbf{a}}^T{\mathbf{M}}^T\mathbf{M}\mathbf{a}$$

$$={\mathbf{m}}_0^T{\mathbf{m}}_0+{\mathbf{a}}^T{\mathbf{M}}^T{\mathbf{m}}_0={\mathbf{m}}_0^T(\mathbf{I}-\mathbf{M}({\mathbf{M}}^T\mathbf{M}{)}^{-1}{\mathbf{M}}^T){\mathbf{m}}_0$$