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Equation of a plane?

  1. Aug 26, 2009 #1
    equation of a plane??

    1. The problem statement, all variables and given/known data

    Find an equation of the plane through the point and perpendicular to the given line.
    (-2, 8, 10)
    x = 4 + t, y = 3t, z = 4 - 4t


    3. The attempt at a solution

    i know the point (-2,8,10) is the starting positon (call it R0)
    i know that N (normal) is perpendicular to R - R0 where R is ending position
    R = (x,y,z) so i said
    R= (4+t,3t,4-4t)
    the N (normal) = (a,b,c) and i dont know how to get these numbers. if i can find out these i can plug it into the scalor equation of a plane formula and be set.

    any help would be great!!
    thanks
     
  2. jcsd
  3. Aug 26, 2009 #2

    rock.freak667

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    Homework Helper

    Re: equation of a plane??

    You need to find the direction of x = 4 + t, y = 3t, z = 4 - 4t, this direction is parallel to the normal of your plane

    eg. x=2t,y=3t,z=t

    [x] [0+2t] [0] [2t] [0] [2]
    [y] = [0+3t] = [0] + [3t] = [0] =t[3]
    [z] [0+t] [0] [t] [0] [1]

    so <2,3,1> is the direction of that line

    Do the same for your line.
     
  4. Aug 26, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: equation of a plane??

    In general, if vector <A, B, C> is perpendicular the plane and (p,q,r) is a point in the plane, then, for a general point (x,y,z) in the plane, <x- p, y- q, z- r> is a vector in the plane and so <A, B, C>.<x- p, y- q, z- r>= A(x- p)+ B(y- q)+ C(z- r)= 0.

    In your case, a vector in the direction of the line x = 4 + t, y = 3t, z = 4 - 4t is <1, 3, -4>, the coefficients of t in each component.
     
  5. Aug 26, 2009 #4
    Re: equation of a plane??

    thanks alot everyone. i got in a little study group with a few of my friends and we strugled for a while but finally found it out.
    (5 hours for 30 problems... not too bad) haha
     
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