Vector calculus need help please

  • Thread starter lijet13
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  • #1
Ok here is the problem:

Given two lines in space, either they are parallel, or they intersect or they are skew. Determine whether the lines taken two at a time, are parallel, intersect or are skew. If they intersect find the point of intersection.

line 1: x=1+2t, y=-1-t, z=3t; -infiniti<t<infiniti
line 2: x=2-s, y=3s, z=1+2; -infiniti<s<infiniti
line 3: x=5+2r, y=1-r, z=8+3r; -infiniti<r<infiniti

I'm not really sure where to go with this. I found the normal vector forms of the equations and I don't think any of them are parallel using the cross product=0 when vectors are parallel but i have no idea how to find itnersection or skew when given the parametric equations. Did I do the parallel part right? and where would I begin for the other parts. Do you use each part (x,y,z) from the parametric as three points to find the equation of the plane formed?

Thanks so much for any help

Answers and Replies

  • #2
Homework Helper
Pick some axis, say z, (make sure none of the lines are perpendicular to this axis first; I haven't checked) and solve for x and y in terms of z for each line. Then solve for the z where two lines have the same x, say. There are a few possibilities:

1. This equation is inconsistent (reduces to somethine like 1=2).
2. The equation holds for all z (reduces to something like 1=1).
3. The y values are also the same at this z.
4. The y values are different at this z.

I'll let you figure out which each case means, but as a clue, I'll tell you that 2 of the above cases tell you they don't intersect, one tells you they do and one requires a little more work to get an answer.
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