Do these two lines intersect? 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?
 A: 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,
A: 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?)
