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Discrete Mathematics 12

  1. Jan 2, 2005 #1
    Find the scalar eq'n of a plane that is perpendicular to the plane with normal vector [3,1,2] and passes through points A(2,-6,-1) and B(1,2,-4).

    I think that the normal vector can be the direction vector of this new plane. But then, in order to find the scalar eq'n I need a normal vector of this new plane. How do I find this?
  2. jcsd
  3. Jan 2, 2005 #2


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    Strictly speaking a plane doesn't have a "direction vector". What is true is that the vector [3,1,2] is in the plane you want. You also know that the vector from A(2,6,-1) to B(1,2,-4) (which is, of course, [1-2,2-6,-4-(-1)]= [-1, -4, -3] is in the plane. Do you know how to find a vector that is perpendicular to both [3,1,2] and [1,4,-3]?
  4. Jan 2, 2005 #3
    I understand what you just found. But no, I do not know how to find a vector's that perpendicular to both those vectors.
  5. Jan 3, 2005 #4
    do u know the direction of a vector that is a cross product of two vectors?

    -- AI
  6. Jan 3, 2005 #5


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    TenaliRaman's point: the cross product of two vectors is always perpendicular to both.

    The cross product of [a1,a2,a3] and [b1,b2,b3] is the vector [a2b3-a3b2,a3b1-a1b3,a1b2-a2b1].
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