This is a vector algebra question I'm struggling with--- it's low dimensional, but I still can't visualize 4D in the same way I can lower dimensions. It's also a toy problem for a harder version, but I figure I should try to understand a smaller version first.
I have the following line:
$$\ell(t) = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 1\end{bmatrix} + t\begin{bmatrix} 1 \\ -1 \\ 1 \\ -1\end{bmatrix}$$
in $\mathbb{R}^4$, and two hyperplanes:
$$H_1 = \left\{ x : \sum_{i=1}^4 x_i = 2\right\} \text{ and } H_2 = \{ x : x_1 + x_2 = 1\}$$
The line is a $1$-dimensonal affine subspace, and lies on both $H_1$ and $H_2$. Moreover, $H_2 \neq H_1$, so $H_1 \cap H_2$ forms a $2$-dimensional affine subspace. What I seek is some $H_3$, preferably in "standard form" $\{x : \langle n, x \rangle =1 \}$ so that $\ell(\mathbb{R}) = H_1 \cap H_2 \cap H_3$.
Can someone give a hint on how to solve this problem?
EDIT:
Nevermind. I found it: $\{x : x_2 + x_3 = 0\}$.
The given line is of the form $p = v_0 + t v_1$. Hyperplanes on which this line exists are of the form $a^T (p - v_0) = 0 $, so you want to solve
$ v_1^T x = 0 $
This results in the orthogonal subspace to $v_1$. Thus the vector $a$ is given by
$a = r u_1 + s u_2 + t u_3 $
Direct solution of $[1, -1, 1, -1] x = 0$ leads to
$u_1 = [1, 1, 0, 0], u_2 = [-1, 0, 1, 0], u_3 = [1, 0, 0, 1]$
$H_1$ corresponds to $u_1 + u_2 + u_3$, while $H_2$ corresponds to $u_1$. You can construct $H_3$ by setting different (not all zero) combinations of $r,s,t$.