# Determining whether the lines are parallel, intersect, or skew

#### icelated

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

Last edited:

#### chiro

MHF Helper
Hey icelated.

Confirming you answers with Octave, we get:

>> >> A = [1, -1, 0; 4, -3, -1; 2, -1, -1]
A =

1 -1 0
4 -3 -1
2 -1 -1

>> rref(A)
ans =

1 0 -1
0 1 -1
0 0 0

This confirms your results as correct.

#### icelated

Thank you so much for responding.