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.

- vishal_kothari

- schrodinger

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- anonymous

Is heuristic reasoning okay, or should I make a rigorous proof?

- vishal_kothari

easy proof...

- anonymous

Not for me. :P

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## More answers

- vishal_kothari

yes for u only..

- AravindG

this is easy

- AravindG

tell me which text u refer

- vishal_kothari

yeah...

- AravindG

do u have balagopal book?

- vishal_kothari

yeah..

- AravindG

this is there in it!!

- AravindG

refer page 116

- vishal_kothari

ya but the proof is very easy....

- anonymous

Proof by probabilistic assumption. :P There are no sets of regular non-planes that adhere to the rules you specify.

- vishal_kothari

kkk

- anonymous

Lol, is that really an answer? Laaame.

- vishal_kothari

no...it needs 3 d geometry....

- vishal_kothari

what are u typing?

- anonymous

A rigorous mathematical proof by extending vectors. XD

- anonymous

Not including non-Euclidian, of course, because I don't know that stuff.

- anonymous

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.

- AravindG

use balagopal book

- anonymous

\[AB = \vec{b} - \vec{a}\]\[BC= \vec{c} - \vec{b}\]
\[\angle ABC=90 \implies (\vec{b} - \vec{a})(\vec{c} - \vec{b})=0 \]
Something like this, and then applying the same procedure for every angle. In the end you might get something.

- anonymous

Gah, I wish I had a tablet.

- vishal_kothari

take it then...

- vishal_kothari

bye..

- anonymous

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.

- anonymous

We can probably determine the last part through annoying set elimination by contradiction or through matrix algebra.

- vishal_kothari

thanks....

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