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.
Is heuristic reasoning okay, or should I make a rigorous proof?
easy proof...
Not for me. :P
yes for u only..
this is easy
tell me which text u refer
yeah...
do u have balagopal book?
yeah..
this is there in it!!
refer page 116
ya but the proof is very easy....
Proof by probabilistic assumption. :P There are no sets of regular non-planes that adhere to the rules you specify.
kkk
Lol, is that really an answer? Laaame.
no...it needs 3 d geometry....
what are u typing?
A rigorous mathematical proof by extending vectors. XD
Not including non-Euclidian, of course, because I don't know that stuff.
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.
use balagopal book
\[AB = \vec{b} - \vec{a}\]\[BC= \vec{c} - \vec{b}\] \[\angle ABC=90 \implies (\vec{b} - \vec{a})(\vec{c} - \vec{b})=0 \] Something like this, and then applying the same procedure for every angle. In the end you might get something.
Gah, I wish I had a tablet.
take it then...
bye..
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.
We can probably determine the last part through annoying set elimination by contradiction or through matrix algebra.
thanks....
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