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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.
 2 years ago
 2 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.
 2 years ago
 2 years ago

This Question is Closed

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

vishal_kothariBest ResponseYou've already chosen the best response.1
yes for u only..
 2 years ago

AravindGBest ResponseYou've already chosen the best response.0
tell me which text u refer
 2 years ago

AravindGBest ResponseYou've already chosen the best response.0
do u have balagopal book?
 2 years ago

vishal_kothariBest ResponseYou've already chosen the best response.1
ya but the proof is very easy....
 2 years ago

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

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

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

vishal_kothariBest ResponseYou've already chosen the best response.1
what are u typing?
 2 years ago

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

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

Ishaan94Best ResponseYou've already chosen the best response.0
I 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.
 2 years ago

Ishaan94Best 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.
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.4
Gah, I wish I had a tablet.
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.4
lol 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.
 2 years ago

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