0

One hypersphere at $x^2 + y^2 + z^2 + w^2 = 1$ intersects another hypersphere at $(x - 1)^2 + y^2 + z^2 + w^2 = 1$.

(EDIT to address comments)

The intersection results in a (3-d) sphere with radius $ \frac{\sqrt 3}{2}$ contained within the hyperplane $x = \frac{1}{2}$

Why?

I can see the derivation of $x = 1$ through simultaneously solving both equations. I assume the radius comes from using the Pythagorean theorem in multiple dimensions, but I just can't imagine it in order to derive it.

Any explanation would be appreciated.

RTF
  • 424
  • 3
  • 9
  • 4
    I think you mean the plan $x = \frac12$? – Kroki May 05 '23 at 02:55
  • Not according to the text. Two normal spheres would intersect in a plane (or circle to be precise), so two hyperspheres would intersect in a hyperplane, no? – RTF May 05 '23 at 03:10
  • From your comment: "(or circle to be precise)". Yes! Be precise! The intersection of two spheres is a circle. The circle is contained within a plane, but the intersection is just the circle. – nickgard May 05 '23 at 07:29
  • If the text says the intersection is a sphere within the hyperplane $x = 1,$ the text is wrong. The earlier comment said $x = 1/2$ because that's the correct hyperplane. The $x$ coordinate is $x=1$ at the center of the second hypersphere. Did the answer that someone posted below resolve this for you? If so, consider accepting it in a day or two, otherwise you can ask for clarification. – David K May 06 '23 at 00:40
  • @Youem et al Apolgies. I thought the original comment was regarding the plane vs. hyperplane distinction. Yes, oversight on my part--the text says $x = \frac{1}{2}$. – RTF May 06 '23 at 02:36

1 Answers1

4

I can see the derivation of $x=1$ through simultaneously solving both equations.

How do you get $x = 1$ from that? If you take the differences between the two equations: $$\begin{array}{rrr}&x^2 + y^2 + z^2 + w^2 = 1\\ -[&(x-1)^2 + y^2 + z^2 + w^2 = 1&]\\\hline &x^2-(x-1)^2 = 0\\&2x-1 = 0\\&x=\frac 12\end{array}$$

And even without solving the equation, just geometrically, it is obvious by symmetry that the intersection of two hyperspheres of the same radius must lie in the hyperplane halfway between their centers, not a hyperplane passing through one of the centers.

As for the rest, once you realize the sphere has $x = \frac 12$, just substitute that value into one of the equations (both give the same result): $$\left(\frac 12\right)^2 + y^2 + z^2 + w^2 = 1\\ y^2 + z^2 + w^2 = \frac 34 = \left(\frac{\sqrt 3}2\right)^2$$

which is the equation of a sphere of radius $\frac{\sqrt 3}2$.

Paul Sinclair
  • 45,932