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Equation of a place from a point and parametrics

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane that passes through the point (1,−2,−1) and contains the line x(t)=−1−3ty(t)=−3−2tz(t)=4+4t.


    2. Relevant equations

    none

    3. The attempt at a solution

    I used vector <-3,-2,4> from the parametrics and crossed it with <2,1,-5> which i got by doing given point (1-2,-1)-(-1,-3,4). I got the 2nd point by puting t=0 into the parametric. When I crossed those i got <6, -7, -7> which is the normal I think? Then I just did 6(x-1) - 7(y+2) - 7 (z+1) =0

    so that seems ok tome... but i checked and thats not correct!! where am i going wrong?
     
  2. jcsd
  3. Sep 23, 2009 #2

    Office_Shredder

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    Gold Member

    I think you did the cross product wrong.

    (-3,-2,4) dotted with (6,-7,-7) is:

    -3*6 + (-2)(-7) + 4(-7) = -18 + 14 - 28

    which is not 0. So the vector you got is not normal to your first vector
     
  4. Sep 23, 2009 #3
    ok thanks! i got it :)
     
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