Show that the intersection of a hyperplane and a plane is a line given the following conditions.

Question

I have been trying to prove that the abovementioned intersection is a line but I do not get to the conclusion. The problem gives the following restrictions. The dimension of the affine space A must be equal or greater than 3. There are no more restrictions. I already know that the intersection of a plane and a (n-1) affine variety can be a line, the whole plane or empty. However, I do not know how to conclude that the intersection is a line. Hope you can help me!

Answer 1

If I understand your question correctly, you ask whether the intersection of a plane with a hyperplane can be a line. The answer is yes. Consider, for example, in an affine space $\mathcal A^4$, referred to coordinates $x_i \;\;(i=1,\ldots, 4)$, the plane $$ x_1=0, \;\;x_2=0 $$ and the hyperplane $$x_3=0. $$ Their intersection is the line $$ x_1=0,\;\;x_2=0,\;\;x_3=0. $$

After all, this is intuitive in the ordinary case $\mathcal A^3$, where a hyperplane is itself a plane, and the intersection of two planes is a line, if they are not parallel.

Answer 2

A hyperplane in $\mathbb{R}^n$ is given by

$ a^T x = b \hspace{15pt}(1) $

where $ a, x \in \mathbb{R}^n $ and $ b \in \mathbb{R} $.

whereas a plane in $\mathbb{R}^n$ is the span of two vectors $v_1, v_2 \in \mathbb{R}^n $, and passes through $p_0$, i.e.

$ x = p_0 + V u \hspace{15pt}(2) $

where $ V = [v_1, v_2] $ and $ u = [u_1, u_2]^T $. Note that $p_0 \in \mathbb{R}^n, \ V \in \mathbb{R}^{n \times 2}$ and $u \in \mathbb{R}^2 $.

Substituting $(2)$ into $(1)$:

$ a^T (p_0 + V u) = b \hspace{15pt} (3) $

so that,

$ a^T V u = b - a^T p_0 \hspace{15pt}(4)$

Now $a^T V \in \mathbb{R}^{1 \times 2} $ while $b - a^T p_0 \in \mathbb{R} $.

So equation $(4)$ has the following solution in $u$

$ u = u_0 + t u_1 \hspace{15pt}(5)$

where $u_0 , u_1 \in \mathbb{R}^2, t \in \mathbb{R}$

Plug this into $(2)$, you get

$ x = p_0 + V (u_0 + t u_1) = w_0 + t w_1 \hspace{15pt} (5)$

where $ w_0 = p_0 + V u_0 , \ w_1 = V u_1 $

Equation $(5)$ is an equation of a line passing through $w_0$ and with a direction vector $w_1$.

However, this is all going to work provided the system

$ a^T (p_0 + V u) = b $

has a solution, which may not be the case, if $a^T V = 0 $ and $ b - a^T p_0 \ne 0 $

In this case, there is no intersection (the plane is parallel to the hyperplane). Further if $a^T V = 0 $ and $b - a^T p_0 = 0 $ , then the intersection is the whole plane given by $(2)$ (the plane is embedded in the hyperplane).