What do solutions of a system of linear equations mean?

Question

I'm currently studying linear algebra mainly vectors and analytic geometry. I'm having trouble seeing "what goes where".

For example, in 2D coordinate system, if we are given the lines

$y1 = 2x + 1$

and

$y2 = 3x + 5$

and we must find their intersection, we put $y1 = y2$. Why? Because that is the point where they intersect, that is, that is the point where they are equal. Hence the above equation.

However, now that I'm working in 3D space, I'm having trouble with solving linear equations.

For example, I have two lines:

$p1... x = 1 + t, y = -1 + 2t, z = t$

and

$p2... x = 1 - t, y = 3 - t, z = t$

If I want to check if they intersect, I get that they don't, because if I put $x = x$, that is

$1 + t = 1 - t$

I get that

$t = 0$

and if I put that $y = y$, I get that

$- 1 + 2t = 3 - t$

which says that

$t = \frac 34$

and somehow from that I should conclude that by $t$ from $x = x$ not being equal to $t$ from $y = y$ the lines do not intersect. I fail to see that connection not only in this example, but in many others.

Can someone clarify that? What do solutions of a linear equations mean?

Thanks in advance!

Answer 1

"However, now that I'm working in 3D space, I'm having trouble with solving linear equations. For example, I have two lines: p1...x=1+t,y=−1+2t,z=t and p2...x=1−t,y=3−t,z= t"

It's a bad idea to use the same letter, t, for both lines. Instead write the second line as x= 1- s, y= 3- s, z= s

"If I want to check if they intersect, I get that they don't, because if I put x=x, that is 1+t=1−t I get that t=0"

No, that point does not necessarily correspond to the same value of the parameter. Instead x= x gives 1+ t= 1- s from which t= -s

and if I put that y=y, I get that −1+2t=3−t"

No, you get that 1- 2t= 3-s and, from that, s= 2t+ 2

"which says that t=34 and somehow from that I should conclude that by t from x=xnot being equal to t from y=y the lines do not intersect. I fail to see that connection not only in this example, but in many others." No, you have only used two equations, for x and y and two lines in two dimensions always intersect or are parallel. You should have, from x=x, t= -s and, from y= y, s= 2t+ 2. We can solve those two equations for s and t: Since t= -s, t= -2t+ 2 so that 3t= 2, t= 2/3. And then s= 2(2/3)+ 2= 4/3+ 2= 10/3.

Now, from "z= z", t= s which is not true.

"Can someone clarify that? What do solutions of a linear equations mean?"

If there had been values of s and t such that x= x, y= y, z= z for the two sets of equations then that one point would lie on both lies- it would be the point of intersection. Here's perhaps a better example, where the two lines do intersect- line 1 given by x= 2t-1, y= t+2, z= 1 and line 2 given by x= 3t+ 1, y= t+ 3, z= t+ 1 where I have written both lines using the same parameter, t, since lines are often given that way. But the first thing I would do is write one of the lines with a different parameter, say the second line as x= 3s+ 1, y= s+ 3, z= s+ 1. Then "x= x" gives 2t- 1= 3s+ 1, "y= y" gives t+ 2= s+ 3, and "z= z" gives 1= s+ 1. That last equation immediately gives s= 0. Putting that into the equation for "y= y", t+ 2= 0+ 3 so t= 1. Finally, setting s= 0 and t= 1 in the "x= x" equation, 2(1)- 1= 3(0)+ 1 or 1= 1. Since that equation is true, the two lines intersect at the point where t= 1 in the first set of equations, s= 1 in the second set. That is, the two lines intersect at x= 2(1)- 1= 3(0)+ 1= 1, y= 1+ 2= 0+ 3= 3, z= 1= 0+ 1= 1. That is, the two lines intersect at (1, 3, 1).

Answer 2

Denote the two lines by $\ell_1$ and $\ell_2$. So $$ \ell_1 = \{ (1+t,-1+2t,t) \mid t \in \mathbb R \} $$ and $$ \ell_2 = \{ (1 - s, 3 - s, s ) \mid s \in \mathbb R \}. $$ Does $\ell_1$ intersect $\ell_2$? In other words, is there a point that belongs to both $\ell_1$ and $\ell_2$? In other words, is it possible to find numbers $t$ and $s$ such that $$ \tag{$\spadesuit$}(1 + t, -1 + 2t, t) = (1 - s,3 - s, s). $$ If it were possible to find such numbers $s$ and $t$, it would follow from equation ($\spadesuit$) that $$ \tag{$\heartsuit$} t = s$$ and also \begin{align} & 1 + t = 1 - s \\ \tag{$\diamondsuit$}\implies & t = -s . \end{align} Equations ($\heartsuit$) and ($\diamondsuit$) together imply that $t = s = 0$. But then it is not true that $-1 + 2t = 3 - s$, because $-1 \neq 3$. This is a contradiction. Therefore, it is not possible to find a point that belongs to both $\ell_1$ and $\ell_2$. The two lines do not intersect.

Answer 3

To start with, if you want to plot those two lines $p_1,p_2$ in the same coordinate system and study their intersection, then you should not use the same parameter $t$ for both $p_1$ and $p_2$. You should instead choose two independent parameters, so one gets something like this: $$p_1: \,\,\,\, x=1+s, \,\,\,\, y=-1+2s, \,\,\,\, z=s $$ $$p_2: \,\,\,\, x=1-t, \,\,\,\, y=3-t, \,\,\,\, z=t $$

The second point is, these equations for $p_1$, $p_2$ are not a system of linear equations. They are instead parametric equations for the lines $p_1,p_2$, where the $s$ variable is the parameter for the line $p_1$ and the $t$ variable is the parameter for the line $p_2$ (formally, $p_1$ is the image of the function $f(s)=(1+s,-1+2s,s)$ and similarly for $p_2$).

If you want to check whether the lines $p_1,p_2$ intersect, your job is not to solve the system of equations. Instead, your job is to solve for values of the parameters $s,t$ so that $(1+s,-1+2s,s)=(1-t,3-t,t)$. And that translates into a system of three equations in two unknowns $$\begin{align*} 1+s &= 1-t \\ -1+2s &= 3-t \\ s &= t \end{align*} $$ Of course, this is an overdetermined system, so it is possible that this system has no solution. In geometric terms, this system has a solution if and only if the two lines do indeed intersect. And with little trouble I am sure you can find that this system has no solution hence these lines do not intersect.