Show that the lines L1 and L2 are skew lines, with parametric...

show that the lines L1 and L2 with parametric equation:
L1: x = 1 + t , y = -2 + 3t , z = 4 - t
L2: x = 2s , y = 3 + s , z = -3 + 4s
are skew lines

firstly verify that these lines are not parallel:

coefs: L1: 1,3,-1
L2: 2,1,4

as you can see one line is not the multiple of the other

secondly verify if there is and intersection point

1+t=2s
-2+3t=3+s

solve for t and s
t=11/5 and s=8/5

use in third remaining equations:
does
4-t=-3+4s
9/5 is not 17/5

Answer: L1 and L2 are skew lines
hope it helps :)

Two lines can either be:
A) Parallel
B) Intersecting
C) Skew

I'll prove that they're neither A nor B, and therefore they must be C.

Part A: The lines are not parallel to each other
L1 is parallel to the vector (1, 3, -1), by reading off the "slopes" of the parameterization. L2 is parallel to the vector (2, 1, 4). Since (1,3,-1) is not equal to k * (2,1,4), they can't be parallel.

Part B: The lines have no intersection
If we set the x coordinates equal, we get 1 + t = 2s.
If we set the y coordinates equal, we get -2 + 3t = 3 + s
Solve that system yourself, finding t and s.
This tells you that there's some point where both the x coordinate AND the y coordinate of both lines are equal.
Check the z coordinate of each line. They should be different.

(Think of it this way...you're solving for x and y, which tells you that you're at the same point on a flat map. But when you plug in for z, you find that one line might be above ground, while the other point may be at a different altitude, or even below ground.)

I used vertical bars to show a vector. They do not mean absolute value.

For example"
|a|
|b|
|c|
is the vector (a,b,c).

Using vector notation:


L1

x | 1 | | 1|
y = | - 2 | + t * | 3|
z | 4 | |-1|


and similarly for L2.

Notice that there is no s that multiplied by

2
1
4

can generate

1
3
-1