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Finding the equation of a plane

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Find an equation of the plane that passes through the point (6,0,-2) and contains the line x=4-2t, y=3+5t, z=7+4t.


    2. Relevant equations



    3. The attempt at a solution
    Here's what I did:
    I substituted t=2 to get:
    x=0, y=17, z= 15
    So in total, we have three points that the plane contains: P1(6,0,-2), P2(4,3,7), P3(0,17,15)
    Vector P1P2=<-2,3,9>
    Vector P1P3=<-6,17,17>
    I took a cross product of these vectors to get <-102,-20,-16>
    I used the point P1(6,0,-2) to obtain:

    -102(x-6)-20(y-0)-16(z+2)=0

    But the solution is 33(x-6)-10(y-0)-4(z+2)=0
    I'm unable to understand what I did wrong. I'd really appreciate any help!
     
  2. jcsd
  3. Sep 19, 2011 #2

    Dick

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    Homework Helper

    Substituting t=2 gives me x=0, y=13, z= 15. Check the y value again.
     
  4. Sep 20, 2011 #3

    dynamicsolo

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    In fact, why bother finding another point on the line for which you have the parametric equation? You already have its direction vector and you know a point on that line by using t = 0 .

    Find the vector from that t = 0 point to (6, 0, -2). You now have two vectors in the plane that you can find the cross product for. That would be less work than dealing with a third point.
     
  5. Sep 20, 2011 #4
    Thank you! I see what I did wrong in substitution now!
     
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