Finding whether two lines intersect each other in 3dimensional space

  • Thread starter Thread starter hangainlover
  • Start date Start date
  • Tags Tags
    Lines Space
hangainlover
Messages
77
Reaction score
0

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
 
Physics news on Phys.org
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top