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

[reedy]

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 won't work, cause I only have one line(?). What is the next step?

 

[rasmhop]

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:
(x,y,z) is
\sqrt{(x+1)^2+(y-2)^2+(z-3)^2}
from A and
\sqrt{(x-3)^2+(y-4)^2+(z+3)^2}
from B so you're looking for all triples (x,y,z) such that:
\sqrt{(x+1)^2+(y-2)^2+(z-3)^2} = \sqrt{(x-3)^2+(y-4)^2+(z+3)^2}
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 C=(A+B)/2 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.

 

[reedy]

A perpendicular bisector would be appropriate in two dimensions, but in three dimensions we are looking for a plane with normal L.

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

http://en.wikipedia.org/wiki/Normal_vector]For[/PLAIN] a plane given by the equation ax + by + cz = d, the vector (a,b,c) is a normal.

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.

 

[rasmhop]

and pull out the normal from L's equation, (2,1,-3), and establish that the plane's equation is 2x₁+x₂-3x₃?

Yes it would be possible, and what you find easiest is up to you. Neither is particularly complicated.

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)

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)

 

[reedy]

C = A + 1/2 * AB = A + 1/2(B-A) = 1/2(A+B)

Ahh, perfect. Now I see it - thanks a bunch!