# Homework Help: Determine if two lines are parallel, intersecting, or skewed.

1. Jan 30, 2012

### getty102

1. The problem statement, all variables and given/known data
Determine if the lines r1=<1+2s,3+2s,2-2s> and r2=<2+t,6-t,-2+t> are parallel, intersecting, or skew. If they intersect, find the point of intersection.

2. Relevant equations
x: 1+2s=2+t
y: 3+2s=6-t
z: 2-2s=-2+t

3. The attempt at a solution
I set the components of r1=r2 then solved for s and t using substitution.
s=1
t=2
I then used 1+2(1)=2+(2) to get 3≠4
therefore it is skewed? or intersecting?

I'm not sure how to find the point of intersection if it does intersect.

2. Jan 30, 2012

### Joffan

The initial process for finding intersection is good (although you didn't tell us which equations you initially used to get candidates for s & t, so checking was impossible). But in essence we need the same two values to satisfy all three equations.

If there is no solution as here, you then need to get direction vectors for the two lines, which consists of the multipliers of the variable parameter here. Then if those vectors are a simple multiple of one another, the lines are parallel, otherwise skewed.

There's also the possibility that the lines are coincident, which will come out of your check for intersection as a relationship between the two parameters (choose any s and you can find a suitable t to satisfy the equations).

3. Jan 31, 2012

### getty102

I determined that the two planes stated above are not parallel because the planes are not multiples of each other, therefore they are skewed and intersecting. I'm not what the next step is to find the intersecting line.

4. Jan 31, 2012

### Joffan

The lines you specify are not intersecting, because you couldn't find values for t and s that made all three equations true. (Or, more accurately, you showed that no such s & t values exist). Finding a unique s & t would have also given you the intersection point.

I don't know about any planes.