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Triple product vector problem 
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#1
Aug2411, 05:07 PM

P: 37

Find [itex](\vec{a}\times \vec{b})\cdot \vec{c}[/itex] if [itex]\vec{a}=3\vec{m}+5\vec{n}[/itex], [itex]\vec{b}=\vec{m}2\vec{n}[/itex], [itex]\vec{c}=2\vec{m}+7\vec{n}[/itex], [itex]\vec{m}=\frac{1}{2}[/itex], [itex]\vec{n}=3[/itex], [itex]\angle(\vec{m},\vec{n})=\frac{3\pi}{4}[/itex]
This is my approach: [itex](\vec{a}\times\vec{b})\cdot\vec{c}=[(3\vec{m}+5\vec{n})\times(\vec{m}2\vec{n})]\cdot(2\vec{m}+7\vec{n})=[3\vec{m}\times\vec{m}6\vec{m}\times\vec{n}+5\vec{n}\times\vec{m}10\vec{n}\times\vec{n}]\cdot(2\vec{m}+7\vec{n})=\bf(11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n})[/itex] I stuck here. I don't know the coordinates of [itex]\vec{m}[/itex] and [itex]\vec{n}[/itex]. Maybe the whole approach is wrong. I don't have any other idea on solving this problem so I need your help. 


#2
Aug2411, 05:58 PM

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What are mxm and nxn? Go back and _look at the definition_ of a vector product.
RGV 


#3
Aug2411, 06:12 PM

P: 37

mxm and nxn by the definition gives us 0. That doesn't change anything.



#4
Aug2411, 07:33 PM

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P: 5,183

Triple product vector problem
RGV 


#5
Aug2511, 09:19 AM

P: 37

Hello Ray,
I did these operations: [itex](11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n})=(11\vec{m}\times\vec{n})\cdot2\vec{m}+(11\vec{m}\times\vec{n})\cdot7\vec{n}[/itex][itex]=[/itex] [itex]=[/itex][itex]22(\vec{m}\times\vec{n})\cdot\vec{m}77(\vec{m}\times\vec{n})\cdot\vec{n}[/itex] So, according to definition, the vector [itex]\vec{m}\times\vec{n}[/itex] is normal with the plane spanned by vectors [itex]\vec{m}[/itex] and [itex]\vec{n}[/itex], as a result I got 0 at the and. Is this correct? Are these operations I did allowed? 


#6
Aug2511, 10:16 AM

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hello chmate!
erm … before embarking on long calculations, always check the obvious … hint: what is b + c ? 


#7
Aug2511, 10:25 AM

P: 37

Hi tinytim,
I see now that the vectors are lineary dependent so 0 is the right answer. I just want to have this question answered, are these operations I did legal? Thank you 


#8
Aug2511, 10:33 AM

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let's see …
(and (A x B).B is always zero) 


#9
Aug2511, 11:32 AM

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P: 5,183

RGV 


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