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October 4th, 2018, 10:51 PM  #1 
Member Joined: Jun 2009 Posts: 83 Thanks: 1  Generalized Pythagorean theorem
Hi, let S be bounded piece of a plane in the space E3 and let's note Si an orthogonal projection of S into xy, xz and yz planes respectively. Then it can be proved that (1) $\displaystyle area(S)^2=area(S1)^2+area(S2)^2+area(S3)^2$. But there is also a general theorem, that in a vector space with dot product where u,v,w are orthogonal vectors the identity (2) $\displaystyle u+v+w^2=u^2+v^2+w^2$ is true. There is great similarity between (1) and (2) here so my question is  can (1) be proved with help of (2), ie can S,Si be somehow interpreted as some vectors of some vector space (such that Si are orthogonal and S=S1+S2+S3)? Thank you for any suggestions. Last edited by skipjack; October 5th, 2018 at 12:13 AM. 
October 5th, 2018, 09:30 AM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
Interesting question. Any three vectors can be interpreted as orthogonal components of an area vector A=u+v+w. But that doesn't prove the area projection formula. You have to prove that the projection of an area vector A onto a plane whose normal is n is A.n. Not that easy as I recall, a little subtle. 
October 5th, 2018, 02:26 PM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
It’s interesting to note that you can project an area vector onto 3 nonorthogonal planes in which case A=u+v+w is still true but A^2=(u+v+w)^2 is u.u+u.v+u.w....... 

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dot product, generalized, pythagorean, pythagorean theorem, theorem, vector space 
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