I have a problem where i don't have the solution to since i like to practice odd problems.

I am not sure if i have done this correctly.

Determine whether the lines

L1: $\displaystyle 1 + t, y = 2 + 3t, z = 3 + t$

L2: $\displaystyle 1 + t, y = 3 + 4t, z = 4 + 2t$

are parallel, intersecting, or skew. If they intersect find the point of intersection.

Attempt:Take L2 and re-write it using t=s to make it simpler

L2: $\displaystyle 1 + s, y = 3 + 4s, z = 4 + 2s$

Make them parametric:

(1) $\displaystyle 1+t = 1+ s$

(2) $\displaystyle 2+3t = 3+4s$

(3)$\displaystyle 3+t = 4+2s$

Solve (1) for t

$\displaystyle 1+t = 1+ s$

$\displaystyle t = s$

Solve(2) for s

$\displaystyle 2+3t = 3+4t$

$\displaystyle s = -1$

so, $\displaystyle t = -1$ and $\displaystyle s = -1$

Solve (3) by subbing in t and s

$\displaystyle 3+(-1) = 4+2(-1)$

$\displaystyle 2 = 2$

Therefore, two lines intersect. Not parallel.

To find the point of intersection:

Solve L1 for $\displaystyle t = -1$

L1: $\displaystyle 1 + (-1), y = 2 + 3(-1), z = 3 + (-1)$

point: (0, -1, 2)

Is this correct? Or did i do this completely wrong?

Thank you