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Equation of plane parallel vectors

  1. Jun 4, 2013 #1
    1. The problem statement, all variables and given/known data
    The vectors a= <-4,3,3> and b = <2,-6,-5> are parallel to a plane PI and R is a point on
    with position vector <104,8,-6> . Find the Cartesian equation of the plane. What is the
    distance of the plane from the origin?


    2. Relevant equations



    3. The attempt at a solution

    This is my thinking.

    Since we are told the vectors a and b are parallel to the plane if we cross them we can get a vector perpendicular to the plane. We can turn that into a unit vector and dot it with the point r.

    So,

    a cross b = <3,-14,18> : call this vector v
    v = √(3^2) + (-14^2)+(18)^2 = 23
    So v(hat) = 1/23<3,-14,18>
    Then to get the equation of the plane we can do

    x dot v(hat) = r dot v(hat)

    <x,y,z> dot 1/23<3,-14,18> = <104,8,-6> dot <1/23<3,-14,18>

    So I got
    3/23x - 14/23y + 18/23z = 92/23
    Then just multiply through to clear out the fraction

    3x - 14y +18z = 92

    and the distance from the plane to the origin is 4 units because r dot v hat is 4
    Did I do this correct?
     
  2. jcsd
  3. Jun 4, 2013 #2
    Looks good to me
     
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