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This is not a question on my homework, just one from the book I'm trying to figure out. They want me to find the intersection of these two lines: \begin{align} L_1:x=4t+2,y=3,z=-t+1,\\ L_2:x=2s+2,y=2s+3,z=s+1. \end{align} But they do not provide any examples. Flipping to the back it tells me that they do intersect and at the point $(2,3,1).$ How did they arrive at this answer? I'm just hoping to understand because I cannot derive any answer.

Thank you for your time,

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    $\begingroup$ set $4t+2 = 2s+2,$ $3 = 2s+3,$ $-t+1=s+1$ and find both $s$ and $t$ and then check that it all worked correctly. $\endgroup$ – Will Jagy Jan 31 '15 at 20:04
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They intersect each other when all their coordinates are the same.

If we call $L_1=\langle x_1,y_1,z_1\rangle$ and $L_2=\langle x_2,y_2,z_2\rangle$ then you have to solve the system: $$x_1=x_2\Longrightarrow4t+2=2s+2,$$ $$y_1=y_2\Longrightarrow3=2s+3,$$ $$y_1=y_2\Longrightarrow1-t=s+1.$$

Note that the system is solved for $t=0=s$. When you plug $t=0$ in $L_1$ you get $\langle 2,3,1\rangle$. The same happens when you plug $s=0$ in $L_2$.

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  • $\begingroup$ Okay, so I have two unknowns, and three equations. So for the first one I find the relation that $2s=4t\implies s=2t$. How does this then allow me to find anything? $\endgroup$ – jm324354 Jan 31 '15 at 20:11
  • $\begingroup$ @bd1251252 take a look at the second equation. This gives you the answer straightaway! $\endgroup$ – Vladimir Vargas Jan 31 '15 at 20:17
  • $\begingroup$ Doesn't this just tell me $s=0$? $\endgroup$ – jm324354 Jan 31 '15 at 20:19
  • $\begingroup$ The only thing I see is that if the end numbers on $s$, i.e. if $s=0$, are (2,3,1) just like the answer. But I don't see how this gives me a point of intersection. $\endgroup$ – jm324354 Jan 31 '15 at 20:21
  • $\begingroup$ @bd1251252 The two lines intersect when they have the same values. That's why we need to check the values for $t$ and $s$ at which $x_1=x_2,y_1=y_2,z_1=z_2$. $\endgroup$ – Vladimir Vargas Jan 31 '15 at 20:23

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