July 14th, 2016, 12:10 AM  #1 
Senior Member Joined: Jul 2011 Posts: 395 Thanks: 15  vector proving
Let $\displaystyle \vec{a},\vec{b},\vec{c},\vec{d}$ are four distinct vectors satisfying $\displaystyle \vec{a}\times\vec{b}=\vec{c}\times\vec{d}$ and $\displaystyle \vec{a}\times\vec{c}=\vec{b}\times\vec{d}$. Then Show that $\displaystyle \vec{a}.\vec{b}+\vec{c}.\vec{d}\neq\vec{a}.\vec{c} +\vec{b}.\vec{d}$ Last edited by panky; July 14th, 2016 at 12:13 AM. 
September 8th, 2016, 07:58 PM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 
axb=cxd axc=bxd subtract (ad)x(bc)=0 assume a.b+c.d=a.c+b.d then (ad).(bc)=0 Contradiction. Two vectors can't be parallel and perpendicular simultaneously. 

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