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Equation of plane from 2 lines

  1. Feb 1, 2010 #1
    OK here is the question.

    Find an equation of the plane that contains the lines given by

    (x-1)/-2 = y = z+1


    (x+1)/-2 = y-1 = z-2

    This is what I did.

    I found 2 points for the first line and 2 points for 2nd line

    A= (0 , 0.5, -0.5)
    B=(1 , 0 , 1)

    C= (0, 0.5, 1.5)
    D= (1, 0, 1)

    then i got their vector equations
    AB = <1, -0.5, 1.5>
    CD = <1, -0.5, -0.5>

    What now? Do i get their cross product? I did cross product and ended up with x+2y = 1
  2. jcsd
  3. Feb 1, 2010 #2
    The cross product is the normal vector of the plane.
  4. Feb 2, 2010 #3
    ok, then how do i obtain the vector of the plane?
  5. Feb 2, 2010 #4
    OK so you have the normal vector by cross product of the vectors. Now you just need a point on that vector to define the plane, subsitute the intersection point of the lines.

    http://en.wikipedia.org/wiki/Plane_(geometry [Broken])
    Last edited by a moderator: May 4, 2017
  6. Feb 2, 2010 #5


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    What do you mean by "the vector of the plane"? Did you mean "equation"? If <A, B, C> is a normal vector to the plane and [itex](x_0, y_0, z_0)[/itex] a point in the plane, then the equation of the plane is [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex].

    However, be careful! The cross product of two vectors is normal to the plane containing the two vectors but that is NOT necessarily a plane containing the two lines. First you have to be certain that the two lines lie in one plane! That is, that they either intersect or are parallel, that they are not skew lines. (Of course, you probably checked that without mentioning it because it is so easy.)
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