1. ## Vector problem

Find $\displaystyle (\vec{a}\times \vec{b})\cdot \vec{c}$ if $\displaystyle \vec{a}=3\vec{m}+5\vec{n}$, $\displaystyle \vec{b}=\vec{m}-2\vec{n}$, $\displaystyle \vec{c}=2\vec{m}+7\vec{n}$, $\displaystyle |\vec{m}|=\frac{1}{2}$, $\displaystyle |\vec{n}|=3$ and angle between $\displaystyle \vec{m}$ and $\displaystyle \vec{n}$ is $\displaystyle \frac{3\pi}{4}$.

I've tried to substitute $\displaystyle [(3\vec{m}+5\vec{n})\times({\vec{m}-2\vec{n})]\cdot(2\vec{m}+7\vec{n})$ and by doing a cross product leading me to the expression $\displaystyle (11\vec{n}\times\vec{m})\cdot(2\vec{m}+7\vec{n})$.

My book gives this solution: $\displaystyle (\vec{a}\times \vec{b})\cdot \vec{c}=0$.

Thank you

2. ## Re: Vector problem

Originally Posted by patzer
Find $\displaystyle (\vec{a}\times \vec{b})\cdot \vec{c}$ if $\displaystyle \vec{a}=3\vec{m}+5\vec{n}$, $\displaystyle \vec{b}=\vec{m}-2\vec{n}$, $\displaystyle \vec{c}=2\vec{m}+7\vec{n}$, $\displaystyle |\vec{m}|=\frac{1}{2}$, $\displaystyle |\vec{n}|=3$ and angle between $\displaystyle \vec{m}$ and $\displaystyle \vec{n}$ is $\displaystyle \frac{3\pi}{4}$.

I've tried to substitute $\displaystyle [(3\vec{m}+5\vec{n})\times({\vec{m}-2\vec{n})]\cdot(2\vec{m}+7\vec{n})$ and by doing a cross product leading me to the expression $\displaystyle (11\vec{n}\times\vec{m})\cdot(2\vec{m}+7\vec{n})$.

My book gives this solution: $\displaystyle (\vec{a}\times \vec{b})\cdot \vec{c}=0$.

Thank you
$\displaystyle \vec{a}-\vec{b}=\vec{c}$

So, $\displaystyle \vec{a}, \vec{b}, \vec{c}$ are linearly dependent and their triple product $\displaystyle (\vec{a}\times \vec{b})\cdot \vec{c}=0$.

3. ## Re: Vector problem

Originally Posted by alexmahone
$\displaystyle \vec{a}-\vec{b}=\vec{c}$

So, $\displaystyle \vec{a}, \vec{b}, \vec{c}$ are linearly dependent and their triple product $\displaystyle (\vec{a}\times \vec{b})\cdot \vec{c}=0$.
That's ok, but is there any other way to solve it. I mean, imagine if $\displaystyle \vec{c}$ is another vector p.s $\displaystyle \vec{c}=2\vec{m}+6\vec{n}$.

4. ## Re: Vector problem

Originally Posted by patzer
That's ok, but is there any other way to solve it. I mean, imagine if $\displaystyle \vec{c}$ is another vector p.s $\displaystyle \vec{c}=2\vec{m}+6\vec{n}$.
In that case I would do it the way you tried in post #1.

PS: On second thoughts, the vectors would still be linearly dependent and so their triple product would be zero.

5. ## Re: Vector problem

Sir, it was an attempt and didn't know what to do there. Can you write the solution for me so I can see where I did it wrong.

6. ## Re: Vector problem

Originally Posted by patzer
Sir, it was an attempt and didn't know what to do there. Can you write the solution for me so I can see where I did it wrong.
I don't know why you're trying to complicate matters when I've given you a simple solution in post #2.

7. ## Re: Vector problem

I don't know why you're trying to complicate matters even more, while I've asked a simple question in the first post. I said, given these facts, help me to solve algebraicaly this problem.

8. ## Re: Vector problem

Again, this is my approach to this problem:

$\displaystyle (\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})$$\displaystyle = \displaystyle =[3\vec{m}\times\vec{m}-6\vec{m}\times\vec{n}+5\vec{n}\times\vec{m}-10\vec{n}\times\vec{n}]\cdot(2\vec{m}+7\vec{n}) \displaystyle [(\vec{m}\times\vec{m}=0, \vec{n}\times\vec{n}=0, (\vec{m}\times\vec{n})=-(\vec{n}\times\vec{m})] so... \displaystyle (-11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n}) I stopped here because I didn't know what to do. So, here I need your help on this. I appreciate the alexmahone's help, but I'm looking to solve the problem using the facts that were mentioned in problem. 9. ## Re: Vector problem Originally Posted by patzer Again, this is my approach to this problem: \displaystyle (\vec{a}\times\vec{b})\cdot\vec{c}=[(3\vec{m}+5\vec{n})\times(\vec{m}-2\vec{n})]\cdot(2\vec{m}+7\vec{n})$$\displaystyle =$
$\displaystyle =[3\vec{m}\times\vec{m}-6\vec{m}\times\vec{n}+5\vec{n}\times\vec{m}-10\vec{n}\times\vec{n}]\cdot(2\vec{m}+7\vec{n})$

$\displaystyle [(\vec{m}\times\vec{m}=0, \vec{n}\times\vec{n}=0, (\vec{m}\times\vec{n})=-(\vec{n}\times\vec{m})]$

so... $\displaystyle (-11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n})$ I stopped here because I didn't know what to do.
So, here I need your help on this. I appreciate the alexmahone's help, but I'm looking to solve the problem using the facts that were mentioned in problem.
Try distributing the cross product over the sum.

10. ## Re: Vector problem

Hello Ackbeet,

I did these operations: $\displaystyle (-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}$$\displaystyle = \displaystyle =$$\displaystyle -22(\vec{m}\times\vec{n})\cdot\vec{m}-77(\vec{m}\times\vec{n})\cdot\vec{n}$

So, according to definition, the vector $\displaystyle \vec{m}\times\vec{n}$ is normal with the plane spanned by vectors $\displaystyle \vec{m}$ and $\displaystyle \vec{n}$, as a result we get 0 at the and. Is this correct?

11. ## Re: Vector problem

Originally Posted by patzer
Hello Ackbeet,

I did these operations: $\displaystyle (-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}$$\displaystyle = \displaystyle =$$\displaystyle -22(\vec{m}\times\vec{n})\cdot\vec{m}-77(\vec{m}\times\vec{n})\cdot\vec{n}$

So, according to definition, the vector $\displaystyle \vec{m}\times\vec{n}$ is normal with the plane spanned by vectors $\displaystyle \vec{m}$ and $\displaystyle \vec{n}$, as a result we get 0 at the and. Is this correct?
Sounds good to me, assuming all your work up until that point is correct, which I haven't checked.