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Homework Help: Find all points with same distance to (A) and (B)

  1. Dec 12, 2009 #1
    I have two given points.

    A=(-1,2,3) and B=(3,4,-3).

    I want to find all points (x1,x2,x3) that have the same distance to A and B.

    Attempt:

    I think finding the line they both lie on is a good first step. To do this, I've drawn a vector between A and B, and found the line's equation. L=(3,4,-3)+t(2,1,-3). I've come to the conclusion that I have to find a perpendicular bisector of some sort. And I guess cross product wont work, cause I only have one line(?). What is the next step?
     
  2. jcsd
  3. Dec 12, 2009 #2
    A perpendicular bisector would be appropriate in two dimensions, but in three dimensions we are looking for a plane with normal L.

    We can find it by bruteforce:
    [itex](x,y,z)[/itex] is
    [tex]\sqrt{(x+1)^2+(y-2)^2+(z-3)^2}[/tex]
    from A and
    [tex]\sqrt{(x-3)^2+(y-4)^2+(z+3)^2}[/tex]
    from B so you're looking for all triples (x,y,z) such that:
    [tex]\sqrt{(x+1)^2+(y-2)^2+(z-3)^2} = \sqrt{(x-3)^2+(y-4)^2+(z+3)^2}[/tex]
    Square, expand and cancel a bit and you should get the equation of a plane that describes all points equally far from A and B.

    You could also have noted that [itex]C=(A+B)/2[/itex] is equally far from A and B, so you want to find the plane with point C and normal CA. From analytic geometry you should be able to find this plane (the same as before), but in my opinion this is more complicated than bruteforce.
     
  4. Dec 13, 2009 #3
    Of course! What was I thinking...

    But if that is true, which it is, wouldn't it be easier to just use the fact that
    and pull out the normal from L's equation, (2,1,-3), and establish that the plane's equation is 2x1+x2-3x3?

    If so, I would need a right hand side as well. And the only point I know for sure in the plane is C.

    You say that C=(A+B)/2. Why is that? Is it plain ol' division on each coordinate? (-1,2,3)+(3,4,-3)=(2,6,0) // division // (1,3,0)

    Gives me: 2x1+x2-3x3=5 which is a correct answer according to the textbook. But the last step confused me.
     
    Last edited by a moderator: May 4, 2017
  5. Dec 13, 2009 #4
    Yes it would be possible, and what you find easiest is up to you. Neither is particularly complicated.

    Yes. We use (A+B)/2 because that is the mid-point of the line segment connecting A and B. For one way of seeing this consider the vector AB. Adding half of this to A we get half-way to B from A so:
    C = A + 1/2 * AB = A + 1/2(B-A) = 1/2(A+B)
     
  6. Dec 13, 2009 #5
    Ahh, perfect. Now I see it - thanks a bunch!
     
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