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Plane in 3-space not parallel to xy- zx- or zy- plane

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Find an equation of the set of all points equidistant from the points A(-1,5,3) and B(6,2,-2). Describe the set.

    2. Relevant equations
    ##|A_2 B_1|/2## = ## \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}/2 ##
    This would be the distance to the midpoint along the line connecting A and B
    ##(\frac{x_ 1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})##
    Coordinate of midpoint between A and B
    3. The attempt at a solution

    For ##|A_2 B_1|/2## I got ##\frac{\sqrt{83}}{2}##
    and the coordinates of the mid point are
    ##(\frac{5}{2},\frac{7}{2},\frac{1}{2})##
    I can visualize that this set of points would be a plane bisecting a line connecting A and B at its mid point.

    The books answer is ##14x-6y-10z=9##
    Not sure how they got this. Any help appreciated thanks.
     
  2. jcsd
  3. Oct 5, 2014 #2

    Mark44

    Staff: Mentor

    Do you know about the dot product for vectors? If so, the vector that joins A and B is perpendicular to every vector that lies in the plane. That means that the dot product of these vectors will be zero.
     
  4. Oct 5, 2014 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Or another way, if you let ##P=(x,y)## you could try setting ##d^2(A,P) = d^2(B,P)##. Using the squares eliminates the square roots in the distance formulas.
     
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