How to determine if two line intersects each other?!

Determine if line L1 intersects line L2.

L1: [x,y,z]= [4, -3, 2] + t[1, 8, -3]
L2: [x,y,z]= [2, -19, 8] + s[4, -5, -9]

Please provide a theoretical solution on how to figure it out, I know how to find it graphically. Thanks for your time!

if the lines intersect, there's a value of t and a value of s for which they both produce the same values of x, y, and z. we solve for s and t.

4 + t = x = 2 + 4s → 4 + t = 2 + 4s → 4s - t = 2
-3 + 8t = y = -19 - 5s → -3 + 8t = -19 - 5s → 5s + 8t = -16

5s + 8t = -16
32s - 8t = 16
-------------------
37s ∙ ∙ ∙ = 0
∙ ∙ ∙ ∙ s = 0 ∙ ∙ ∙ ∙ ∙ and so t = -2

check it:
[x,y,z] = [2,-19,8] + 0[4,-5,-9] = [2,-19,8] and
[x,y,z] = [4,-3,2] - 2[1,8,-3] = [2,-19,8]

so yes, the lines intersect.

note: David ignored the possibility of skew lines which are not parallel yet don't intersect.

If these two lines intersect, you will have three equations in two unknowns. Pick two equations and solve for t and s. Then see if your solution also holds for the third equation. If it does the lines intersect.

4 + t = 2 + 4s

-3 + 8t = -19 -5s

s = 0 and t = -2

Now go the the third equation

2 - 3t = 8 - 9s

Notice that s = 0 and t = -2 works in the third equation.

The point of intersection is [2,-19,8] + 0[4,-5,-9] = [2,-19,8]

The lines intersect if and only if they are not parallel.

The lines are parallel if and only if their direction vectors are parallel.

Since [1, 8, -3] is not a multiple of [4, -5, -9] (by multiple I mean multiplied by any real number), we determine that the direction vectors are not parallel.

So the lines intersect.