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Vector Equation of a Plane

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the vector equation of a plane that contains the following line L1: (x,y,z) = (1,2,3) + t(2,4,3) and does not have a z-intercept.

    3. The attempt at a solution
    This question was on the test and I think I might have gotten it wrong.
    The correct answer is: (x,y,z) = (1,2,3) + t(2,4,3) + s(0,0,1)

    It makes sense to me now because (0,0,1) is parallel to the z-axis. Hence, it does not intercept the z-axis.

    What I did was... think about the question in Cartesian Equation form.
    Ax + By + Cz + D = 0 (where A,B,C are normal to the plane)

    I realized that the co-efficient should be 0 since there is no z-intercept. So Ax + By + D = 0.
    So the normal would be (A, B, 0).

    Therefore,
    (2,4,3) CROSS (x, y, z) = (A, B, 0) [where (x,y,z) is the second direction vector]

    I did the cross product formula and ended up with:
    (4z-3y, 3x-2z, 2y-4x) which should equal to (A, B, 0)
    2y-4x = 0
    2y=4x --> y = 2x and value of z doesn't matter

    Therefore, (2,4,3) CROSS (x, 2x, 0) = (A, B, 0)
    Let x = 1
    So: (2,4,3) and (1,2,0) are the direction vectors.

    So the equation of the plane is: (x,y,z) = (1,2,3) + t(2,4,3) + s(1,2,0)

    Could this be an acceptable answer?
    Thanks
     
    Last edited: Oct 28, 2011
  2. jcsd
  3. Oct 28, 2011 #2

    CompuChip

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    Unfortunately, if you set t = -1 and s = 1 (or t = 1, s = -3, and probably many other combinations) you will find a point on the z-axis, so this plane does intercept it.
     
  4. Oct 28, 2011 #3

    HallsofIvy

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    At this point, a perfectly valid way to solve the probleme would be to note that when t=0, (x,y,z)= (1,2,3) and when t= 1, (x, y, z)= (1, 2, 3)+ (2, 4, 3)= (3, 6, 6) are on the given line and so on the plane. If the equation of the plane is Ax+ By+ D= 0, then A+ 2B+ D= 0 and 2A+ 4B+ D= 0 give two equations to solve (of course, any multiple of (A,B,D) is also a solution). Subtracting the two equations eliminates D and gives A+ 3B= 0 so that A= -3B. Taking B= 1 gives A= -3 and then -3+ 2+ D= 0 gives D= 1: -3x+ y= 1.

     
  5. Oct 28, 2011 #4
    I need a vector equation.

    So is my answer correct?
    (x,y,z) = (1,2,3) + t(2,4,3) + s(1,2,0)
     
  6. Oct 29, 2011 #5

    HallsofIvy

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    Same answer as you were given before: if you take t= -1/2, s= 0, you get (x,y,z)= (1- 1,2- 2, 3- 3/2)= (0, 0, 3/2) which is a "z-intercept". Your plane was to have NO z-intercept which means the z-axis must be parallel to it.
     
  7. Oct 29, 2011 #6
    Oh, so this would be the correct answer, right?

    (x,y,z) = (1,2,3) + t(2,4,3) + s(0,0,1)


    Thanks for your help everyone,
     
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