How to Calculate the Scalar Triple Product in Vector Calculus?

[chmate]

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.

 

[Ray Vickson]

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

RGV

 

[chmate]

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

 

[Ray Vickson]

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

 

[chmate]

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?

 

[tiny-tim]

hello chmate!

erm …

before embarking on long calculations, always check the obvious …

hint: what is b + c ? ​

… I got 0 at the and. Is this correct?

see above!

 

[chmate]

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

 

[tiny-tim]

let's see …

$(-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)

 

[Ray Vickson]

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

Yes, everything is legal. You just used things like A*(B+C) = A*B + A*C and A*(rB) = r(A*B) for scalar r; these are true if * is either the dot product or the cross-product. You also used AxB = -BxA which is specific to the cross product. Of course, you could have written the answer right away, without any calculations, because you had (UxV).W, where U, V and W are all linear combinations of m and n, so all lie in the same plane containing m and n---hence UxV is perpendicular to W.

RGV