Not Determinated
Show that the equidistant set of two points in R^3 is a plane.?

Show that the equidistant set of two points in R^3 is a plane. a) Show that the plane passes through O if the two points are both at distance 1 from O.
b) Deduce from (a) that the equidistant set of two points on S^2 is a "line" (great circle) on S^2. Next, we establish that there is a unique point on S^2 at given distances from three points not in a "line".
c) suppose that two points P,Q belong to S^2 have the same distances from three points A,B,C belong to S^2 not in a "line". Deduce from (b) that P=Q.
d) deduce from (c) that an isometry of S^2 is determinated by the images of three points A,B,C not in a "line".
Thus, it remains to show that the following. Any three points A,B,C beong to S^2 not in a "line" can be mapped to any other three points A',B',C' belong to S^2, which are separated by the same respective distances, by one, two ore three reflections.
Complete this proof o the three reflections theorem

That's a long homework problem. I gather you're already stuck on the part that you put into the title of the question?

The proof of that is pretty simple. Suppose a point (x, y, z) has distance d from each of (a,b,c) and (e,f,g). The (x-a)^2 + (y-b)^2 + (z-c)^2 = d^2 = (x - e)^2 + (y-f)^2 + (z-g)^2.

But the quadratic terms cancel out, so you have the equation of a plane.

Technically, that just shows the solution set lies IN a plane, but it isn't hard to reverse the process and show that any point on the plane is also equidistant from the two points given -- i.e., that the solution set is the WHOLE plane.

As for why it passes through the origin in part a -- that's VERY simple. The origin has distance 1 from both the points, and hence is equidistant from them!



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