Hi Roy,

I'm going to use the distance formula in 3-space. If (x₁,y₁,z₁) and (x₂,y₂,z₂) are points in 3-space then the distance between them is

Sqrt[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

If P is a point on the y-axis then P has coordinates (0,y,0) for some number y. If P is equidistant from (2,-1,1) and (0,1,3) then

Sqrt[(2 - 0 )² + (-1 - y)² + (1 - 0 )²] = Sqrt[(0 - 0 )² + (1 - y )² + (3 - 0 )²]

Squaring both sides I get

5 + (y + 1)² = 9 + (y-1)²

Expand, simplify and solve for y.

The starting point for the second problem, is the image (or to imagine) the points that are equal distance from the two given points? They form a locus, from which you will do the rest of the reasoning.

In the plane, the points equal distance from two points A, B are a straight line (the right bisector of the segment (A,B)). It is clear that the midpoint of the segment C is such a point. If you take another point P on the right bisector, and form triangles with PAC, and PBC, you have two congruent triangles, and |PA|=|PB|.

You can also see this line, by taking a paper with the points A, B marked on it, and folding A onto B. The crease is the set of points equal distance from A and B! You have overlapped the plane in two layers, so that the crease points must be the same distance along the two layers from the point on the top layer (A) and the point on the bottom layer (B).

All of that is warm up you brain and give you a context. We can't quite fold 3-space (unless we step mentally into 4-space). However some of the thinking extends. One point equal distance from A, B, is the mid point of the segment joining them. If we take a plane containing the two points (and this segment) then in that plane, you can apply the previous reasoning, to find a right bisector within the plane (or the fold line within the plane - same set of points). So this reminds us that the midpoint of the segment is ONE of the possible answers - there are a lot. (That is one of the points in the answers from the back of the book below.)

Since every plane through the two points contains one of these lines, we can image the set of all possible points as a plane. This is the plane through the midpoint, at right angles to the segment (the spatial right bisector of the segment!)

If you really prefer to use algebra, you can now do that. Write down an unknown point (x,y,z). Write its distance from the two given points, and set them equal. After you square both sides (simplify) and shuffle things around, you will see that the x², y², z² terms cancel, and you do have the equation of a plane. There is an image that goes with this as well. You have two spheres (equal radius) which are intersecting in a plane circle. The center of this circle is the midpoint of the segment joining the points A and B. (That is the way two spheres always intersect). As the radius changes, this circle changes - but always in the same plane with the same center.

While this is a bit hard to imagine unless you have played, you can make the plane version is Geometer's SketchPad. The two circles with centers A and B will meet in two points. As the radius changes, these two points sweep out a line as their trace or locus: another image for the right bisector.

So the final step (a point on the y-axis) is now the intersection of this plane and the y-axis: where a line meets a plane (one of the standard type of questions in the vectors part of the course you are taking).

It is good to play with these ideas so you have a 'picture' and web of connections to organize which algebra, or which number calculations to do, and to check whether your 'answers' make sense in terms of the image. Line meeting plane: expect one point (though you can, on special occasions have none - if parallel, or multiple, if the line is in the plane.) Points equal distance - a whole plane in 3-space. Some points may be easier to find (the midpoint) but they are not unique.

Walter Whiteley
York University