Determining whether the lines are parallel, intersect, or skew

Aug 2011
117
0
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
Sep 2012
6,608
1,263
Australia
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.