5
$\begingroup$

I am told that, given two points $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$, the set of all points which are twice as far from $\mathbf{a}$ as they are from $\mathbf{b}$:

$$\{\mathbf{p} : |\mathbf{p} - \mathbf{a}| = 2(|\mathbf{p} - \mathbf{b}|)\}$$

is a sphere. I'm wondering why this is true intuitively. Given two points $\mathbf{a} = (a_x, a_y, a_z)$, $\mathbf{b} = (b_x, b_y, b_z)$, I can write down the equation:

$$\sqrt{(a_x - p_x)^2 + (a_y - p_y)^2 + (a_z - p_z)^2} = 2\sqrt{(b_x - p_x)^2 + (b_y - p_y)^2 + (b_z - p_z)^2}$$

go through the algebra, square both sides, collect like terms, and get the equation of a sphere. But I can't manage to pierce the fog of algebra with my intuition. What I'm trying to understand intuitively or geometrically is:

  • What determines the radius and center of the resulting sphere?
  • If I ask instead for the set of all points which are equidistant from $\mathbf{A}$ and $\mathbf{B}$, I get a plane. If I ask for the same in $\mathbb{R}^2$, I get a line. If I ask for all points equidistant from just one point in $\mathbb{R}^3$, though, I get a sphere. Is it true in general that the set of all points in $\mathbb{R}^n$ equidistant from $k$ selected points is a subspace of dimension $n - k$?
  • If I don't ask for equidistant points but instead ask for something like the above, i.e some distances are proportional to some other distances, what is the analogous result?
  • Is there a geometric way of understanding the above?

EDIT: It strikes me that a sphere isn't a subspace of $\mathbb{R}^3$ because it's not a vector space. Can I still talk about it having "dimension" 2?

$\endgroup$
  • 2
    $\begingroup$ Yes, $S^2$ (= the unit sphere in $Bbb R^3$) quite legitimately has dimension 2. The concept of dimension is much broader than vector spaces. Generally, if the natural structure of an object requires two distinct parameters to specify where something is, then the object is two-dimensional. $\endgroup$ – Paul Sinclair Sep 5 '15 at 20:50
  • 1
    $\begingroup$ Formally, it's a two-dimensional manifold. $\endgroup$ – Rahul Sep 5 '15 at 21:24
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Circles_of_Apollonius $\endgroup$ – Jack D'Aurizio Sep 5 '15 at 21:26
  • $\begingroup$ @Rahul It's also 2-dimensional under a number of other interpretations, like "Hausdorff dimension". $\endgroup$ – Mark S. Sep 5 '15 at 23:22
3
$\begingroup$

What determines the radius and center of the resulting sphere?

Because the condition is defined in terms of points $\mathbf a$ and $\mathbf b$ only, the points satisfying it have to be symmetrically distributed about the line through $\mathbf a$ and $\mathbf b$. Which means that the center of the sphere must lie on this line. There are two points on this line that satisfy the condition themselves: one is two thirds of the way from $\mathbf a$ to $\mathbf b$, and the other is on the opposite side of $\mathbf b$ from $\mathbf a$, at the same distance away. These two points have to be the opposite sides of a diameter, so the center is halfway between them, and the radius is half the distance between them, which is $2/3$ the distance from $\mathbf a$ to $\mathbf b$.

If I ask instead for the set of all points which are equidistant from A and B, I get a plane. If I ask for the same in $\mathbb R^2$, I get a line. If I ask for all points equidistant from just one point in $\mathbb R^3$, though, I get a sphere. Is it true in general that the set of all points in $\mathbb R^n$ equidistant from k selected points is a subspace of dimension $n−k$?

Look carefully again at your examples. Both examples in 3-space are sets of dimension 2, even though one is a distance from two points and the other from one point. And for the examples in 2-space, both are of dimension 1, even though one is a distance from two points and the other from one point.

There are some rules here, but not what you gave. Generally if you have a variable defined on a space and then ask for the set where that variable has a specific value, you can expect that set to be 1 dimension less than the space (we say "of codimension 1") if you have two independent variables and specify values for both of them at the same time, you get something of codimension 2, etc.

If I don't ask for equidistant points but instead ask for something like the above, i.e some distances are proportional to some other distances, what is the analogous result?

The ratio is a variable that you are proscribing a value for, so it still defines a set of codimension 1.

$\endgroup$
  • $\begingroup$ Thank you for this useful answer! I see that I was wrong -- all my examples are of dimension $n - 1$ for $\mathbb{R}^n$. But if I had talked about points equidistant from $3$ (non-colinear) points, isn't that a line? I don't see why the progression should be 1 point $\rightarrow 2$, 2 points $\rightarrow 2$, 3 points $\rightarrow 1$. $\endgroup$ – Eli Rose -- REINSTATE MONICA Sep 6 '15 at 15:23
  • $\begingroup$ No. the set of points equidistant from 3 fixed points is usually a single point. Take any pair of the fixed points, and the set equidistanct from them both is a line, so you have three such lines, and to be equidistant from all three, a point has to lie on all three of them. For three arbitrary lines, you can't expect them to have a common intersection at all, but in this case they all intersect at the centroid of the three points. $\endgroup$ – Paul Sinclair Sep 7 '15 at 4:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.