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3d graphing: equidistant

  1. Jun 5, 2008 #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
    dot product
    cross product

    3. The attempt at a solution
    the book found the line: 14x - 6y - 10z = 9 which is perpendicular and i guess cuts in half way between the two dots.
    i know that u take 2 lines and multiply them using dot product and it needs to equal 0 for them to be perpendicular. but i only know how to do that with 2 lines with endpoints... idk how to do this since i have 2 end points and i need to figure out an equation.

    if any1 can explain how the book got that equation or even explain how to get any correct answer, that would definitely help. thanks
  2. jcsd
  3. Jun 5, 2008 #2


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    Homework Helper

    That looks more like a plane in 3D space rather than a line. It's easy to see why it's a plane and not a line, considering two points in space and a line joining them. A plane situated equidistant from the two points and normal to the connecting line would represent the locus of points which are equidistant from the 2 points. You can't represent a line in 3D space with a single equation unless it's a vector equation.

    A plane is associated with a normal vector to it and an arbitrary point on the plane. Can you see how to get both?
  4. Jun 5, 2008 #3
    ok im not sure what calculations im supposed to make to get this answer. im a little clueless right now

    does it have anything to do with vector project or scalar project? idk what formulas im supposed to be using on this 1
  5. Jun 5, 2008 #4


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    Homework Helper

    If you want to take the direct route just write an equation that says that the distance squared from a point (x,y,z) to (-1,5,3) equals the distance to (6,2,-2) and simplify it.
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