The system of linear equations $AX=B$ such as $A=\pmatrix{1 \;\;\;1 \;\;\;2 \\ 1 \;\;\;1 \;\;\;1 \\ 2\;\;\;2 \;\;\;2}, X=\pmatrix{x\\y\\z}, B=\pmatrix{3\\1\\2}$ has the following solutions (infinitely many): $\{x=t,y=-1-t,z=2\}, t \in \mathbb R.$
My question is about geometrical understanding/intuition. I understand that, since the vector $\pmatrix{t\\-1-t\\2}$ is a solution to $AX=B$, we can say that $t\pmatrix{1\\1\\2} + (-1-t)\pmatrix{1\\1\\2} + 2\pmatrix{2\\1\\2}$ which can be further written as $(-1)\pmatrix{1\\1\\2} + 2\pmatrix{2\\1\\2}$ (since $\pmatrix{-1\\-1\\-2} + \pmatrix{4\\2\\4} = \pmatrix{3\\1\\2}$). Therefore, we have a linear combination of 2 vectors of the matrix A.
Please, can you confirm or extend my reasoning:
1) This means that the vector $\pmatrix{3\\1\\2}$ is in the span of $\pmatrix{1\\1\\2}$ and $\pmatrix{2\\1\\2}$.
2)The vector $\pmatrix{3\\1\\2}$ is in the plane formed by $\pmatrix{1\\1\\2}$ and $\pmatrix{2\\1\\2}$; if it were not, then there would not be any solution.
3) 2 3D-vectors can only span a plane in 3D.
4) Now in terms of intersection of lines/hyperplanes, etc. I think we can say that the solution is a line (but why, really (in terms of geometrical interpretation I mean).. it is not the 'intersection of something' here?). Since the line's equation is y=mx+b (or in vector form $\vec{OQ} = \vec{OP}+\lambda \vec{d}$, with $\vec{d}$ a vector in the direction of the line and $\vec{OP}$ a point on this line), is it correct then to say that $\vec{OP}=\pmatrix{4\\2\\4}$ and $\vec{d}=\pmatrix{1\\1\\2}$ and $\lambda=t$ ? (I sometimes have trouble "seeing" the parameters of the line or hyperplane when there are infinitely many solutions; this example is a way of clarifying that. (Indeed, if we set $\lambda = -1$, then we get the vector $\pmatrix{3\\1\\2}$ out of the vectorial line equation... but I would like the "explanation" part of how/why?)
Please note that I plotted them in Matlab as
[X,Y]=meshgrid(-10:0.1:10);
surf(X,Y,1-X-Y);
hold on;
surf(X,Y,(3/2)-X/2-Y/2);
hold on;
plot3(4+t,2+t,4+2*t)
However, the line as plotted on the figure is not there where I would have thought... maybe I made a mistake?
If you want to obtain the solution set as the intersection of planes, you can look at what the system $Ax=b$ means. We have, $$ \begin{cases} 1 x + 1y + 2z = 3 \\ 1 x + 1 y + 1z = 1 \\ 2 x + 2 y + 2z = 2 \end{cases} $$ These are all planes in $\mathbb{R}^3$. In this case, the bottom two equation gives the same plane, which intersects the top plane as a line. This means there are infinitely many points (all on a line) which lie on all three planes.
Another way to think of the solution set as a line is using the solution you found. Write $u = (0,-1,2)^T$ and $v=(1,-1,0)^T$. Then, $$ \begin{bmatrix} t \\ -1 - t \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \\ 2 \end{bmatrix} + t \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = u + t v $$ Note that no matter what other representation we use for the line, the direction part, $v=(1,-1,0)^T$ cannot change.
What is special about this vector? It turns out that $Av=0$ and $Au = b$. Then, no matter what we pick as $t$, $$ A(u+tv) = Au + tAv = b + t\cdot 0 = b $$
So we see that once we have found a point which gives the solution (in this case $u$), that adding anything in the null space of $A$ to this point will still give a solution. In the case of this matrix, the null space is one dimensional, so the solution set is a line.
If the null space were two dimensional, then it would be a plane. That means once we find a solution (if a solution exists) we could add anything in the null space to it and still have a solution. In this case, the solution set would be a plane.