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Lines and Planes in Space

  1. Jun 3, 2013 #1
    find the general form of the equation of the plane with the given characteristics:

    passes through (0,2,4) and (-1,-2,0) and is perpendicular to yz-plane


    I know what the general form of an equation is, but I was wondering, do I set the direction vector to be <0,1,1> or <1,0,0>? How do I use this to determine the general form of the equation?

    I started by determining the vector given by the two points and then used this vector with the cross product of the two vectors mentioned above, but I am not sure what to do after that?...
     
  2. jcsd
  3. Jun 3, 2013 #2

    Mark44

    Staff: Mentor

    The displacement vector between your two given points can be crossed with <1, 0, 0> to produce a vector that is normal to your plane.
    When you post a problem, don't delete the template elements. They are there for a reason.
     
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