# Triple product vector problem

by chmate
Tags: product, triple, vector
 P: 37 Find $(\vec{a}\times \vec{b})\cdot \vec{c}$ if $\vec{a}=3\vec{m}+5\vec{n}$, $\vec{b}=\vec{m}-2\vec{n}$, $\vec{c}=2\vec{m}+7\vec{n}$, $|\vec{m}|=\frac{1}{2}$, $|\vec{n}|=3$, $\angle(\vec{m},\vec{n})=\frac{3\pi}{4}$ This is my approach: $(\vec{a}\times\vec{b})\cdot\vec{c}=[(3\vec{m}+5\vec{n})\times(\vec{m}-2\vec{n})]\cdot(2\vec{m}+7\vec{n})=[3\vec{m}\times\vec{m}-6\vec{m}\times\vec{n}+5\vec{n}\times\vec{m}-10\vec{n}\times\vec{n}]\cdot(2\vec{m}+7\vec{n})=\bf(-11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n})$ I stuck here. I don't know the coordinates of $\vec{m}$ and $\vec{n}$. Maybe the whole approach is wrong. I don't have any other idea on solving this problem so I need your help.
 Sci Advisor HW Helper Thanks P: 5,183 What are mxm and nxn? Go back and _look at the definition_ of a vector product. RGV
 P: 37 mxm and nxn by the definition gives us 0. That doesn't change anything.
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P: 5,183
Triple product vector problem

 Quote by chmate mxm and nxn by the definition gives us 0. That doesn't change anything.
OK, I see now that you got rid of those terms; the results were hidden behind a popup that appeared on my screen before (but not now). Anyway, you now want to evaluate 11 mxn . (2m + 7n). Now go back and look at the definition of mxn; in particular, pay attention to the directions in which various vectors are pointing.

RGV
 P: 37 Hello Ray, I did these operations: $(-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}$$=$ $=$$-22(\vec{m}\times\vec{n})\cdot\vec{m}-77(\vec{m}\times\vec{n})\cdot\vec{n}$ So, according to definition, the vector $\vec{m}\times\vec{n}$ is normal with the plane spanned by vectors $\vec{m}$ and $\vec{n}$, as a result I got 0 at the and. Is this correct? Are these operations I did allowed?
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P: 26,148
hello chmate!

erm

before embarking on long calculations, always check the obvious
hint: what is b + c ?
 Quote by chmate … I got 0 at the and. Is this correct?
see above!
 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
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P: 26,148
let's see …
 Quote by chmate $(-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}$$=$ $=$$-22(\vec{m}\times\vec{n})\cdot\vec{m}-77(\vec{m}\times\vec{n})\cdot\vec{n}$ So, according to definition, the vector $\vec{m}\times\vec{n}$ is normal with the plane spanned by vectors $\vec{m}$ and $\vec{n}$, as a result I got 0 at the and. Is this correct? Are these operations I did allowed?
yes, that's fine

(and (A x B).B is always zero)