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Say I have a Line $P$ that cuts across 2 points $(0, 1, -1)$ and $(1,-1,0)$ in a space of $\mathbb{R}^3$.

If those were 2 vectors, I could say they span a plane, 3 vectors then the entire space. Since they are points and not vectors, what does a line across these points in $\mathbb{R}^3$ mean? I cannot say that it spans a line in the row/column space because this is a line.

I tried to find out the equations to this line by doing this: $$\begin{bmatrix} 0 & 1 & -1 & 1\\ 1 & -1 & 0 & 1 \end{bmatrix}\begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$$ $$\begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix}= t\begin{bmatrix} 2\\ 1\\ 0\\ -1 \end{bmatrix} + k\begin{bmatrix} -1\\ 0\\ 1\\ 1 \end{bmatrix}, \; \; k,t \in \mathbb{R}$$ Then I get 2 equations from the above: $$2x+y=1$$ $$-x+z=-1$$

Do these 2 equations represent the line $P$ that cut across the points $(0, 1, -1)$ and $(1,-1,0)$? But how do they represent because each of the 2 equations themselves is a line on their own.

I could continue to derive from the 2 equations to find out the solution set to $x$, $y$ and $z$: $$\begin{bmatrix} 2 & 1 & 0\\ -1 &0 & 1 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 1\\ -1 \end{bmatrix}$$ $$\begin{bmatrix} x\\ y\\ z \end{bmatrix} =\begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix} + c\begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix}$$

Again, at this point, I am also very confuse what this set of solution is representing. I know that this set of solution is for the 2 equations above but does it mean that if in the column space, by moving in any amount of $ $$\begin{bmatrix} x\\ y\\ z \end{bmatrix} =\begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix} + c\begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix}$ would let me reach the line $P$?

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In three dimensions, a line is uniquely determined by specifying any two planes whose intersection is the line in question... –  J. M. Sep 10 '11 at 16:25
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I think you are overlooking the definitions. First detail is that every vector has an implicitly defined starting point that is the origin. Otherwise vectors cannot have a direction right? Hence, it is a shortcut to define only the end point to name the vector.

Second, it is much better to phrase "...the linear combination of 2 vectors span a plane, and the linear combination of 3 vectors then the entire space...." So by themselves they don't span anything.

Regarding the the line, you should think of the line either as a collection of points or vectors such that their end points share something in common, namely being on the line.

Start by translating the line back to the origin and write it as $y = cx$

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Is it a must to go through the implicit form of the points to find out the equations before getting the explicit form of each of the $x$, $y$ and $z$? Can I get from the points to the explicit form without having to find out the implicit form first? –  xenon Sep 10 '11 at 18:08
    
Wait, wait @xEnOn, what exactly do you mean by "explicit" and "implicit" here. Given two points, it's easy to derive the parametric equations of the line passing through them... –  J. M. Sep 10 '11 at 18:31
    
By implicit form, I mean like expressing the line in terms of one or more equations for instance: $\left \{ \begin{bmatrix} x\\ y\\ z \end{bmatrix} \mid x+y+z=0 \; and \; x-y+2z=1\right \}$ and explicit I mean expressing the the line by having the variables $x,y,z$ as the subject like $\left \{ \begin{bmatrix} \frac{1}{2}-\frac{3}{2}t\\ -\frac{1}{2}+\frac{1}{2}t\\ t \end{bmatrix} \mid t \in \mathbb{R} \right \}$. So I was thinking if I could get to the explicit form directly without going through the implicit form first? –  xenon Sep 11 '11 at 3:45
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