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### Re^2: The intersection of M hyperplanes (Ndim)

by bliako (Abbot)
 on Jul 18, 2024 at 15:59 UTC Need Help??

So, I understand that if a1=0 and a2=0 then the vectors are "parallel on that axis", and so I proceed to check the other coefficients in order to determine if the planes are parallel. If one of a1 OR a2 is zero then I conclude the vectors (and the planes) are not parallel and the test can stop right there.

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Re^3: The intersection of M hyperplanes (Ndim)
by LanX (Saint) on Jul 19, 2024 at 09:42 UTC
Look, similar to the rule you already stated:

taking two vectors, for them to be collinear there must exist one k != 0 such that for all coordinate d: d1 = k * d2

Obviously d1 == 0 iff d2 == 0 because k != 0 !!!

But if one coordinate is 0 for both it doesn't matter which k is chosen for the other ones.

Isn't it obvious now?

You check all coordinates in a loop and break if

• either only one is 0 (next if both are)
• or the ratio k is not like before (unless undef)

##### Edit
FWIW: There is one edge case left, the zero vector. But a normal vector with all coordinates being 0 looks like a bug to me and should raise an exception.

Cheers Rolf
(addicted to the Perl Programming Language :)
see Wikisyntax for the Monastery

got it, thanks LanX

fyi, I googled it a lot before asking this particular question. "when are 2 vectors / 2 planes are parallel". What I got was lots and lots of the ratio test but not mentioning at all any of the edge cases. I finally asked here for a clarification : https://math.stackexchange.com/questions/3925150/find-if-2-planes-are-parallel-or-not, the answeree has a huge XP but fails to complete their answer with common sense question.

FYI, like many others, I did it in "high school" for 3D but most never apply it again. (When studying Math at uni this was covered in just a week at best)

So while most think they still master it they are quite rusted. Or reluctant like I am.°

I'm quite confident about 3D because it's very visual. The abstraction to higher dimensions is a door I never really actively passed.

(It's "too trivial" for mathematicians ;)

I hope my explanation was of help, do you need the algorithm as explicit code?

Cheers Rolf
(addicted to the Perl Programming Language :)
see Wikisyntax for the Monastery

°) instead of remembering the rules like engineers do, we tend to deduce it from basic axioms

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