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July 14th, 2016, 12:10 AM   #1
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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.
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September 8th, 2016, 07:58 PM   #2
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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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