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| 2024-10-09 21:29:27 +0200 | CiaoSen | (~Jura@2a05:5800:2e5:2400:ca4b:d6ff:fec1:99da) (Ping timeout: 246 seconds) |
| 2024-10-09 21:22:03 +0200 | tromp | (~textual@92-110-219-57.cable.dynamic.v4.ziggo.nl) |
| 2024-10-09 21:21:54 +0200 | AlexZenon | (~alzenon@178.34.163.165) |
| 2024-10-09 21:19:13 +0200 | <tomsmeding> | I guess the scaling is superfluous, but it helps intuition |
| 2024-10-09 21:18:49 +0200 | <tomsmeding> | "so that B-A is (1,0)" -> "so that B-A is (1,0,0)" |
| 2024-10-09 21:17:24 +0200 | tromp | (~textual@92-110-219-57.cable.dynamic.v4.ziggo.nl) (Quit: My iMac has gone to sleep. ZZZzzz…) |
| 2024-10-09 21:17:02 +0200 | <tomsmeding> | convoluted, though |
| 2024-10-09 21:16:36 +0200 | <tomsmeding> | that feels like an argument |
| 2024-10-09 21:16:23 +0200 | <tomsmeding> | and angles in the projection are preserved by rotations around the z-axis and scalings |
| 2024-10-09 21:16:06 +0200 | <tomsmeding> | it's not parallel to the z-axis because that would mean that OAB is parallel to the x-y plane, which it isn't, so it has a nonzero y component |
| 2024-10-09 21:15:34 +0200 | <tomsmeding> | ah: B-A lies in the x-y plane, so rotate and scale the whole system so that B-A is (1,0); then the (rotated) cross product is still orthogonal, so it's in the y-z plane |
| 2024-10-09 21:15:22 +0200 | briandaed | (~root@185.234.210.211.r.toneticgroup.pl) (Remote host closed the connection) |
| 2024-10-09 21:14:28 +0200 | <tomsmeding> | it is relevant that B-A lies in the x-y plane, because without that restriction, it's easy to think of two orthogonal vectors in R^3 that are not orthogonal after projection to the x-y plane |
| 2024-10-09 21:14:24 +0200 | AlexNoo_ | (~AlexNoo@178.34.163.165) (Ping timeout: 246 seconds) |
| 2024-10-09 21:13:30 +0200 | <dolio> | Yeah, I'm not sure how to see geometrically that the projection is still orthogonal. |
| 2024-10-09 21:13:14 +0200 | weary-traveler | (~user@user/user363627) user363627 |
| 2024-10-09 21:13:00 +0200 | <tomsmeding> | but that's cheating again |
| 2024-10-09 21:12:53 +0200 | <tomsmeding> | you can do that algebraically (the cross product is orthogonal to B-A, but the z-component of B-A is zero, hence it's orthogonal regardless of the z component of the cross product, hence also at zero, hence also when projected to the x-y plane) |
| 2024-10-09 21:12:16 +0200 | AlexNoo | (~AlexNoo@178.34.151.120) (Ping timeout: 252 seconds) |
| 2024-10-09 21:11:35 +0200 | <tomsmeding> | right |
| 2024-10-09 21:11:32 +0200 | AlexNoo__ | (~AlexNoo@178.34.163.165) |
| 2024-10-09 21:11:14 +0200 | AlexZenon | (~alzenon@178.34.151.120) (Ping timeout: 260 seconds) |
| 2024-10-09 21:11:11 +0200 | <tomsmeding> | it was a \qed with a ? inside |
| 2024-10-09 21:11:06 +0200 | <tomsmeding> | when I did a bachelor in maths I had some fun with a fellow student and wrote a macro which, translated to English, would I guess be \maybeBoxy |
| 2024-10-09 21:10:48 +0200 | wootehfoot | (~wootehfoo@user/wootehfoot) (Read error: Connection reset by peer) |
| 2024-10-09 21:10:44 +0200 | <dolio> | Well, you need some fact about projections back to the x-y plane. |
| 2024-10-09 21:10:12 +0200 | <tomsmeding> | \qed? |
| 2024-10-09 21:10:09 +0200 | <tomsmeding> | the cross product is normal to the plane spun up by the triangle OAB, hence it is normal to all vectors in that plane, in particular B-A |
| 2024-10-09 21:09:40 +0200 | <tomsmeding> | okay fair |
| 2024-10-09 21:09:30 +0200 | <tomsmeding> | well it always was |
| 2024-10-09 21:09:25 +0200 | <tomsmeding> | yes ... oh |
| 2024-10-09 21:09:08 +0200 | <dolio> | It's normal to a plane containing the line between the a and b points. |
| 2024-10-09 21:08:39 +0200 | AlexNoo_ | (~AlexNoo@178.34.163.165) |
| 2024-10-09 21:08:38 +0200 | <tomsmeding> | so I'm not sure about your "contains the line you care about" |
| 2024-10-09 21:07:55 +0200 | <tomsmeding> | the cross product is going to be _normal_ to that plane |
| 2024-10-09 21:07:45 +0200 | <tomsmeding> | that's just the same plane as we've been looking at before ("the triangle OAB"), only shifted down by z :p |
| 2024-10-09 21:07:24 +0200 | <dolio> | Yes. |
| 2024-10-09 21:07:17 +0200 | <tomsmeding> | and "the plane" is the one generated by {(a1,a2,0), (b1,b2,0), (0,0,-z)}? |
| 2024-10-09 21:06:48 +0200 | <dolio> | Yeah. |
| 2024-10-09 21:06:37 +0200 | <tomsmeding> | dolio: which "those vectors"? (a1,a2,z) and (b1,b2,z)? |
| 2024-10-09 21:05:19 +0200 | <tomsmeding> | Lears: well, it's orthogonal to b-a. :P |
| 2024-10-09 21:01:46 +0200 | <dolio> | tomsmeding: Maybe this is a better thing to consider. Those vectors are what you get by going between your actual points and the point (0,0,-z). And that plane actually contains the line you care about. |
| 2024-10-09 21:00:42 +0200 | caconym | (~caconym@user/caconym) caconym |
| 2024-10-09 21:00:41 +0200 | <Lears> | The projection should be orthogonal to a+b, not b-a? |
| 2024-10-09 21:00:25 +0200 | <Lears> | tomsmeding, dolio: Distributivity. (a1, a2, z) x (b1, b2, z) = ((a1, a2, 0) + (0, 0, z)) x ((b1, b2, 0) + (0, 0, z)) = ((a1, a2, 0) + (b1, b2, 0)) x (0, 0, z) + (0, 0, <junk>) |
| 2024-10-09 21:00:05 +0200 | caconym | (~caconym@user/caconym) (Quit: bye) |
| 2024-10-09 20:58:21 +0200 | drdo8 | drdo |
| 2024-10-09 20:58:21 +0200 | drdo | (~drdo@bl9-110-63.dsl.telepac.pt) (Ping timeout: 276 seconds) |
| 2024-10-09 20:56:47 +0200 | drdo8 | (~drdo@bl9-110-63.dsl.telepac.pt) drdo |
| 2024-10-09 20:54:00 +0200 | jinsun | (~jinsun@user/jinsun) (Read error: Connection reset by peer) |