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Param. Lines and Planes

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the point of intersection of the lines r(t)=< 2t+1, 3t+2, 4t+3> and
    x=s+2
    y=2s+4
    z=-4s-1
    Then, find the plane determined by these lines.

    2. Relevant equations
    Intersection is when points meet.
    So, just equating x,y, and z variables will yield the point of intersection.

    3. The attempt at a solution
    The point of intersection is P(1,2,3), but I'm clueless on how to find the plane with two parametric equations. It's very simple to find the plane, when they are not parametric.

    Could someone help?
     
  2. jcsd
  3. Feb 25, 2012 #2
    You need to find the normal vector n between the two vectorys r(t) and p(t)=<s+2, 2s+4,-4s-1> by taking their cross product. Then just plug that into the equation of a plane:

    [tex]n\mdot (r-p)[/tex] where r and p are the vectors you have...
     
  4. Feb 25, 2012 #3
    That's my problem. How do I find the cross product of two parametric equations?
     
  5. Feb 27, 2012 #4
    Could someone help, please?

    I've been working on it and looking for info on the net, but I have been unsuccessful.

    Thank you.
     
  6. Feb 27, 2012 #5
    You will have to change the format of the two equations. An easier way to look at it would be in vector form:

    L1(s)= (2,4,-1)+ s(1,2,-4)
    L2(t)= (1,2,3)+ t(2,3,4)

    then take the cross product of the vector portion of those two equations to find the normal line of the plane. Your cross product should look something like

    | i j k |
    | 1 2 -4 |= vector for line perpendicular to plane
    | 2 3 4 |

    Hope that helps a bit
     
  7. Feb 27, 2012 #6
    It does help. It's interesting, the vector directions are use as if they were in a plane to find the "normal" vector passing through them.
     
  8. Feb 27, 2012 #7
    That's right :)
     
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