Parallel, intersecting, or skew

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Homework Help Overview

The discussion revolves around determining the relationship between two lines, L1 and L2, given their parametric equations. Participants are exploring whether these lines are the same, parallel, intersecting, or skew based on their parameterizations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are setting the components of the two lines equal to each other to analyze their relationships. There is a focus on the implications of matching z components while questioning the conditions under which x and y components may or may not match.

Discussion Status

Some participants have provided guidance on the necessity of checking all components for intersection and have highlighted the importance of parameterization. There is an ongoing exploration of whether the lines can intersect based on the conditions derived from the equations.

Contextual Notes

Participants are considering the implications of the lines being skew and questioning if there exists a value of s that could make the y components equal. There is a reference to a test scenario where the original poster's interpretation was deemed incorrect, prompting further inquiry into the reasoning.

dswatson
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Let L1 be the line parametrized by r1(t) = <t+1,2t-1,-t+2> and L2 be the line parametrized by r2(t)=<2t+6,-t-1,-2t-3>. Determine if L1 and L2 are the same, parallel, intersecting, or skew. I set x1=x2,y1=y2, and z1=z2 and the t does not equal ct so they are not parallel or the same line.

I then changed the parameters of L2 from t to s and set the components equal.

2s+6=t+1
t=2s+5
then plugged back into the equations.
I got 2s+5,4s+9,-2s-3=2s+6,-s-1,-2s-3

the z terms are the same so does that mean they intersect at some point?

I put this on a test and it was counted as incorrect and I was wondering how to correct it. Thank you in advance.
 
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dswatson said:
Let L1 be the line parametrized by r1(t) = <t+1,2t-1,-t+2> and L2 be the line parametrized by r2(t)=<2t+6,-t-1,-2t-3>. Determine if L1 and L2 are the same, parallel, intersecting, or skew. I set x1=x2,y1=y2, and z1=z2 and the t does not equal ct so they are not parallel or the same line.

I then changed the parameters of L2 from t to s and set the components equal.
(This is the correct way to solve this problem.) There is no guarantee the the "t" in one parametrization is the same as the "t" in the other.

2s+6=t+1
t=2s+5
then plugged back into the equations.
I got 2s+5,4s+9,-2s-3=2s+6,-s-1,-2s-3

the z terms are the same so does that mean they intersect at some point?
When the z terms are the same, are the x & y terms also the same? If not, then that shows that the lines do NOT intersect

I put this on a test and it was counted as incorrect and I was wondering how to correct it. Thank you in advance.
Hello dswatson.

Some comments are in red above.

When you found t=2s+5 from the x (or x1 ) components, the the x should also match for the two lines.

If t=2s+5, then r1(t) = <2s+5+1,2(2s+5)-1,-(2s+5)+2> = <2s+6, 4s+9, -2s-5>.

Compare with: r2(s)=<2s+6,-s-1,-2s-3>

Both the x & the z match, but NOT the y. What do you suppose that means?
 
I would assume that the lines match in the x and z if looking at a graph but the lines would be at different points on the y-axis? is this why they are skew?
 
Is there any value of s that makes y match for the two lines?
 

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