How does the solution look like in $\mathbb R^4$?

Question

The problem came to me when I learned to calculate the solution of the equation.

here is the matrix $\mathit A$: $$ A= \begin{bmatrix} 1 & 2 & 2 & 2\\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \\ \end{bmatrix} $$ The task is solving $\mathit A \mathbf x= \mathbf b $ using rref form. The answer is:
let $\mathbf b$=$\begin{bmatrix} 1\\5\\6 \end{bmatrix}$，$\mathbf x_{complete}$ = $\begin{bmatrix} -2\\0\\3/2\\0 \end{bmatrix}$+$c_1\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}$+$c_2\begin{bmatrix}2\\0\\-2\\1\end{bmatrix}$, where $c_1$ and $c_2$ are real numbers.

The solution is correct, which I don't think it is necessary to show details here. Because the calculation procedure is a routine.
And what I don't have a clue is as follows:

The nullspace of $\mathit A$ is a two-dimensional subspace of $\mathbb R^4$, the solutions to the equation $\mathit A\mathbf x=\mathbf b$ form a plane parallel to that through $\mathit x_p$=$\begin{bmatrix}-2\\0\\3/2\\0\end{bmatrix}$.

For the particular $\mathbf b$, I can get infinite vector $\mathbf x_{complete}$ here, since we can go on with the different value of $\mathbf c_1$ and $\mathbf c_2$. And all these vector $\mathit x_{complete}$ formed a plane in $\mathbb R^4$. And the plane goes through $\mathit x_p$=$\begin{bmatrix}-2\\0\\3/2\\0\end{bmatrix}$.
But why the plane is parallel to the nullspace, I just stuck here. I think the plane and the nullspace and the $\mathit x_p$ should be intersected in the origin.

Here is another question popping up when I am writing this post. There may be another $\mathbf b$ from which I can get another $\mathit x_p$. Then what the result will be? Well, I think I just try to figure out myself first for this one.

Answer 1

The plane represents your solution set. If the plane and the nullspace intersect, then, a point in the nullspace is also a solution to your problem. That is not true. Therefore, we can not expect the plane and the nullspace to intersect.

Also, the plane is not a subspace, but it is an affine space. Thus, it does not necessarily contain the origin.

For any $b$, the plane will be an affine space (a shifted version of the null space of $A$).

Answer 2

The question arises, which plane through $x_p$? For a point does not determine a plane. I guess the answer is obvious: the reference is to the solution set of $Ax=b$. This solution set is indeed a plane through $x_p$.

So, the answer seems to be that the null space is parallel to that plane.

And that is indeed true. For the null space is $c_1(-2,1,0,0)^t+c_2(2,0,-2,1)^t, c_1,c_2\in \Bbb R$.

The solution space, on the other hand, is just the translation of this plane by $x_p$. (As such, it is an affine space, but not a vector space.)

As to your last question, the answer is that this is always the scenario: the general solution set $S$ is equal to the set of all sums of a particular solution and elements of the kernel. That is, $S=\{x_p+y| y\in\operatorname{ker}A\}$. Here $x_p$ can be any particular solution.