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L1 = <3,4,1> + t <2,-1,3> L2 = <1,3,4> + s <4,-2,5>

I'm trying to see if these lines are parallel, skew, or intersect. I've already discovered that they are not parallel. I was thinking about setting them equal to each other at the point t=s and seeing if there is a solution, which would prove that they intersect, and if their isn't one that would prove that they skew.

So, <3,4,1> + t <2,-1,3> = <1,3,4> + t <4,-2,5>

How do I solve this?

ShaneBird
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  • Do you have any other information? For example, if the question says that these two lines are in the plane, or at least in the same plane in space, they must either be parallel or intersect. Solving the equations simultaneously should give you your answer. – 123 Sep 01 '14 at 23:09

2 Answers2

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That is not exactly how you should proceed. The point is that if the two lines have a point in common, then the point will have its associated value of $t$ on the first line and its value of $s$ on the second line ; there is no reason for these values to be the same. So what you want to see is if there exists realy numbers $t,s$ such that $$ (3,4,1) + t (2,-1,3) = (1,3,4) + s(4,-2,5), $$ i.e. find $s$ and $t$ such that the point can be on both lines. You get linear equations : $$ 3+2t = 1+4s, \quad 4-t = 3-2s, \quad 1+3t = 4+5s. $$ You can put all the $t$'s and $s$'s on the left sides and all the constants on the right sides to get a linear system of equations $$ \begin{bmatrix} 2 & -4 \\ -1 & 2 \\ 3 & -5 \\ \end{bmatrix} \begin{bmatrix} t \\ s \end{bmatrix} = \begin{bmatrix}-2 \\ -1 \\ 3 \end{bmatrix}. $$ Now your two lines intersect if and only if this linear system has a solution. I leave it up to you to figure it out ; feel free to ask for help if you need to.

Added : If you don't know linear algebra, we can try solving the system "manually". Adding the first equation with twice the second, we get $$ 11 = (3+2t)+2(4-t) = (1+4s) + 2(3-2s) = 7 $$ which shows the system has no solution, so your pair of lines are skew.

Hope that helps,

Patrick Da Silva
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  • I don't understand the matrices completely, I remember being exposed to them in precalc but that was a long time ago. I think I'll have to study them a bit because it seems to be a common way to solve problems. – ShaneBird Sep 01 '14 at 23:20
  • @ShaneBird : It is a very common way, linear algebra is very powerful ; you should definitely learn it if you wish to pursue in any area of science. Concerning this answer, you can go around without linear algebra. I'll edit my answer to comment on that. – Patrick Da Silva Sep 02 '14 at 11:57
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It isn't a good idea to assume $t=s,$ as that adds another condition that hasn't already been given, and makes a solution less likely. There is a reason that two distinct parameters were given.

Rather, assume that $$\langle 3,4,1\rangle+t\langle 2,-1,3\rangle=\langle 1,3,4\rangle+s\langle 4,-2,5\rangle\\\langle 3,4,1\rangle+\langle 2t,-t,3t\rangle=\langle 1,3,4\rangle+\langle 4s,-2s,5s\rangle\\\langle 3+2t,4-t,1+3t\rangle=\langle 1+4s,3-2s,4+5s\rangle$$

So, we can instead consider this as a linear system with $3$ equations in two variables:

$$\begin{cases}3+2t=1+4s\\4-t=3-2s\\1+3t=4+5s\end{cases}$$

Can you take it from there? (Also, do you see what the assumption that $t=s$ is a bad idea?)

Cameron Buie
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  • I do see why t = s is a bad idea now, the point of intersection is likely at a point where t is not equal to s. – ShaneBird Sep 01 '14 at 23:14
  • Could I try using elimination to find a value for t and then try to solve for s from there? – ShaneBird Sep 01 '14 at 23:17
  • Absolutely. You should see whether or not a solution can exist in short order. – Cameron Buie Sep 01 '14 at 23:18
  • So I got the value s = 6 and t = 11 using the method of elimination but plugging that into some of the equations gave me invalid answers. Is that enough evidence to prove that the lines do not intersect? – ShaneBird Sep 01 '14 at 23:29
  • Yes, indeed! Alternately, note that adding twice the second equation to the first yields $11=7,$ so at that point we're done. – Cameron Buie Sep 02 '14 at 04:05