Timeline for Is there a ruler and compass construction of the common perpendicular of two geodesics in H^3?
Current License: CC BY-SA 4.0
12 events
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| Jul 16, 2022 at 16:49 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://math.meta.stackexchange.com/a/34713/228959
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| Apr 25, 2013 at 20:45 | history | edited | ε-δ |
edited tags
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| Jan 1, 2011 at 20:21 | answer | added | Will Jagy | timeline score: 4 | |
| Dec 18, 2010 at 13:57 | comment | added | Sam Nead | @Alex - But the question was asked in $H^3$, not in $R^3$... | |
| Dec 4, 2010 at 5:53 | comment | added | Alex B. | @Will Jagy Can you post this as an answer, so Henry can accept it and the question doesn't get bumped at irregular intervals? | |
| Nov 20, 2010 at 4:58 | answer | added | Robert Haraway | timeline score: 6 | |
| Oct 30, 2010 at 13:05 | comment | added | Henry Segerman | Ah yes, that works for $R^3$. | |
| Oct 30, 2010 at 4:37 | comment | added | Will Jagy | Now that I see what you are prepared to use, it is easy in $R^3.$ Take the two direction vectors for your skew lines and take their cross product, call that $N.$ Construct the plane normal to $N$ that contains one of the lines. Take two distinct points on the other line and drop their perpendiculars to the plane, thses being parallel to $N.$ Draw the line between the two image points, mark where it connects with the first line. Draw another line parallel to $N$ through that point. | |
| Oct 29, 2010 at 23:35 | comment | added | Henry Segerman | A solution to the $R^3$ version is not immediately apparent to me (although I think it should be possible), I'll have a think. For the Poincaré ball model embedded in $R^3$ there is more to work with: you have the endpoints of the $H^3$ geodesics to start from. | |
| Oct 29, 2010 at 23:30 | comment | added | Henry Segerman | @Will: Yes, this is a practical problem with drawing a diagram in a 3d computer model. I don't know if an exhaustive list of operations would be useful, but I can do things like draw lines between two points, circles or planes through three non-colinear points, spheres through 4 points, find intersections and I can do Euclidean isometries to objects (by distances and angles that I already know). | |
| Oct 29, 2010 at 22:22 | comment | added | Will Jagy | I can't tell what your axioms might be. If you gave me two skew lines in $R^3$ and a compass and straightedge to wave about in midair, but no plane on which to draw, I'm not sure what I could accomplish. So, as I think you are probably envisioning a computer model of some kind, perhaps you could solve the $R^3$ problem and describe what was involved. Given enough detail on that I can probably solve your problem in $H^3$ or show it cannot always be done. | |
| Oct 29, 2010 at 8:23 | history | asked | Henry Segerman | CC BY-SA 2.5 |