Finding whether two lines intersect each other in 3dimensional space

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Homework Help Overview

The problem involves determining whether two lines in three-dimensional space intersect. The lines are defined parametrically with variables t and r, and the original poster seeks guidance on how to approach the problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to equate the components of the two lines to find potential intersection points. There is a question about the nature of the parameters t and r and how they relate to the lines.

Discussion Status

Some participants have suggested setting the equations equal to each other to explore intersections. There is a recognition that specific values for t and r may yield intersection points, and one participant has noted a potential intersection at (7, 5, 0) based on their calculations.

Contextual Notes

There is an acknowledgment that in three-dimensional space, most lines do not intersect, and further checks are needed to confirm any intersection found by solving the equations.

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Homework Statement


do the lines (x,y,z) = (5+2t, 3+2t,1-t) and (x,y,z) = (13-3r, 13-4r, 4-2r) intersect? If so, at what point? If not, how do we know?


Homework Equations





The Attempt at a Solution



I just do not know where to begin...
I mean what do you do with the variables t and r.
Do those values constanly change?
If you want to direct me to some useful information to help me understand this concept.. feel free to do so
 
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If they do intersect, then the x component of the first line must equal the x component of the second line, and similar for the y and z components.

So, perhaps you should set the equations equal to each other...
 
does this mean
5+2t=13-3r
3+2t=13-4r
1-t=4-2r ?
oh i guess when t=1 and r=2, those two lines intersect each other
 
Yes, it happens that when t= 1 and t= 2, all three equations are satisfied. And you can do more. By putting t= 1 into the equations for that line or by putting r= 2 into the equations for the second line, you get x= 5+2= 7, y= 3+ 2= 5, and z= 1-t= 0 or, equivalently, 13- 6= 7, y= 13- 8= 5, 4- 4= 0 so the two lines intersect at (7, 5, 0).

Of course, in three dimensions, "most" lines do NOT intersect. You could always solve two of the equations, say, the x and y equations, for s and t. But then you would have to check in the z equation to see if that also was satisfied.
 

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