Relative position of planes in $\mathbb{R}^{5} $

Question

The exercise asks to describe a plane in $\mathbb{R}^{5} $ as the solution of a linear system of equations. So I've considered the following:

Considering hyperplanes of 4 dimensions embbeded in $\mathbb{R}^{5} $

Hyperplanes of 4 dimensions embbeded in 5 dimensions have the equation of the form $ax+by+cz+dw+eu=f$ (just as like planes of 2 dimensions embeded in 3 dimensions are of the form $ax+by+cz=d$).

So we can describe 2-d planes in $\mathbb{R}^{5} $ as the solution for the intersection of three 4-d hyperplanes.

$$ \left\{ \begin{array}{c} a_1x+b_1y+c_1z+d_1w+e_1u=f_1 \\ a_2x+b_2y+c_2z+d_2w+e_2u=f_2 \\ a_3x+b_3y+c_3z+d_3w+e_3u=f_3 \end{array} \right. $$

As the RHS is different of zero, we can say they are affine hyperplanes.

If my tought is correct how can I define the relative position of two such planes in $\mathbb{R}^{5} $? I'm strugling even to define when two such planes are paralel, since that if happens iff the cross product of their normal vector is equal to zero and their free coefficient are not equal each other, but cross product is not defined in 5 dimensions.

Is there any generalization to the relative position of affine subspaces? Apologize if something was not clear or if I'm misunderstanding some concept. Any external source or example would also be extremly helpful. Thanks.

Answer 1

You can describe a $2D$ plane in $\mathbb{R}^5$ vectorially as follows. Let $r \in \mathbb{R}^5$ be the coordinate vector of a point on this $2D$ plane then we can express $r$ as follows

$ r = r_1 + t_1 v_1 + t_2 v_2 $

where $r_1, v_1, v_2 \in \mathbb{R}^5 $ and $t_1, t_2 \in \mathbb{R} $.

Let another plane be given by

$ r = r_2 + t_3 v_3 + t_4 v_4 $

where $r_2, v_3, v_4 \in \mathbb{R}^5 $ and $t_3, t_4 \in \mathbb{R} $

These two planes will be parallel if the following two conditions are both satisfied:

1. $r_2$ does not lie in the first plane (or equivalently, $r_1$ does not lie in the second plane ).

2. $\text{Span}(v_1, v_2) = \text{Span}(v_3, v_4)$. That is, both $v_3$ and $v_4$ are linear combinations of $v_1$ and $v_2$

Another possibility for the two planes is being perpendicular to each other, and that will happen if

1. The two planes intersect.

2. $\text{Span}(v_1, v_2) \perp \text{Span}(v_3, v_4) $. That is, both $v_3$ and $v_4$ are perpendicular to $v_1$ as well as $v_2$.

If the two planes are not parallel, then we can seek their intersection by setting

$ r_1 + t_1 v_1 + t_2 v_2 = r_2 + t_3 v_3 + t_4 v_4 $

Then the solution of this equation is obtained by finding $t_1, t_2, t_3, t_4$ such that

$ [v_1, v_2, - v_3 , -v_4] [t_1, t_2, t_3, t_4]^T = r_2 - r_1 $

This is an overdetermined system of $5$ equations in $4$ unknows. It is possible that it has no solution, and this happens if

$[v_1, v_2, -v_3, -v_4, (r_2 - r_1)]$ has rank $5$.

If, however,

$\text{rank}([v_1, v_2, -v_3, -v_4]) =\text{rank}([v_1, v_2, -v_3, -v_4, (r_2 - r_1)]) = 4 $ then there will be a unique solution, i.e. the two planes intersect at a single point.

Moreover, if

$\text{rank}([v_1, v_2, -v_3, -v_4]) =\text{rank}([v_1, v_2, -v_3, -v_4, (r_2 - r_1)]) = 3 $, then there will be infinite solutions that lie on a the line $r_0 + t V_0 $. This follows from Gaussian elimination of the matrix of vectors $[v_1, v_2, -v_3, -v_4]$.

And by the same argument, if

$\text{rank}([v_1, v_2, -v_3, -v_4]) =\text{rank}([v_1, v_2, -v_3, -v_4, (r_2 - r_1)]) = 2 $, then there will be infinite solutions that lie on a the $2D$ plane $P_0 + t V + s W$. This also follows from Gaussian elimination of the matrix of vectors $[v_1, v_2, -v_3, -v_4]$.

Note that this last case means that the two given planes are identical, because we assumed that $v_1, v_2$ are linearly independent, so this menas that $v_3, v_4$ are linear combinations of $v_1, v_2$. And in addition we have $(r_2 - r_1)$ also a linear combination of $v_1, v_2$.