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 3 years ago
A,B,C,D are four distinct points in three space. Suppose each
of the angles ABC, BCD, CDA, and DAB are right angles.
Show that all four points lie in the same plane.
 3 years ago
A,B,C,D are four distinct points in three space. Suppose each of the angles ABC, BCD, CDA, and DAB are right angles. Show that all four points lie in the same plane.

This Question is Closed

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.4Is heuristic reasoning okay, or should I make a rigorous proof?

vishal_kothari
 3 years ago
Best ResponseYou've already chosen the best response.1yes for u only..

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0tell me which text u refer

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0do u have balagopal book?

vishal_kothari
 3 years ago
Best ResponseYou've already chosen the best response.1ya but the proof is very easy....

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.4Proof by probabilistic assumption. :P There are no sets of regular nonplanes that adhere to the rules you specify.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.4Lol, is that really an answer? Laaame.

vishal_kothari
 3 years ago
Best ResponseYou've already chosen the best response.1no...it needs 3 d geometry....

vishal_kothari
 3 years ago
Best ResponseYou've already chosen the best response.1what are u typing?

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.4A rigorous mathematical proof by extending vectors. XD

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.4Not including nonEuclidian, of course, because I don't know that stuff.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0I am not sure about this, It's been time since I have done 3d. If you can take A,B,C,D to be following vectors \(\vec{a},\vec{b},\vec{c} \) and \(\vec{d}\), and then apply the dot product rule.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0\[AB = \vec{b}  \vec{a}\]\[BC= \vec{c}  \vec{b}\] \[\angle ABC=90 \implies (\vec{b}  \vec{a})(\vec{c}  \vec{b})=0 \] Something like this, and then applying the same procedure for every angle. In the end you might get something.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.4Gah, I wish I had a tablet.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.4lol Ishaan has the right idea. We take what we know... the vector sum being zero, and each individual vector extending in another component from the previous, alternating and opposite. We can determine then, through mathematics that I tried HTML typing, that the only possible unit vector components are ones that cancel out.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.4We can probably determine the last part through annoying set elimination by contradiction or through matrix algebra.
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