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 July 14th, 2016, 01:10 AM #1 Senior Member   Joined: Jul 2011 Posts: 399 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 01:13 AM.
 September 8th, 2016, 08:58 PM #2 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,230 Thanks: 93 axb=cxd axc=bxd subtract (a-d)x(b-c)=0 assume a.b+c.d=a.c+b.d then (a-d).(b-c)=0 Contradiction. Two vectors can't be parallel and perpendicular simultaneously.

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