Vector problem

**patzer** · August 24th 2011, 05:48 AM

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

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

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

Thank you

**alexmahone** · August 24th 2011, 06:00 AM

Originally Posted by patzer

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

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

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

Thank you

$\vec{a}-\vec{b}=\vec{c}$

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

**patzer** · August 24th 2011, 06:10 AM

Originally Posted by alexmahone

$\vec{a}-\vec{b}=\vec{c}$

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

That's ok, but is there any other way to solve it. I mean, imagine if $\vec{c}$ is another vector p.s $\vec{c}=2\vec{m}+6\vec{n}$ .

**alexmahone** · August 24th 2011, 06:15 AM

Originally Posted by patzer

That's ok, but is there any other way to solve it. I mean, imagine if $\vec{c}$ is another vector p.s $\vec{c}=2\vec{m}+6\vec{n}$ .

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.

**patzer** · August 24th 2011, 06:20 AM

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.

**alexmahone** · August 24th 2011, 06:23 AM

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.

**patzer** · August 24th 2011, 06:51 AM

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.

**patzer** · August 24th 2011, 09:08 AM

Again, this is my approach to this problem:

$(\vec{a}\times\vec{b})\cdot\vec{c}=[(3\vec{m}+5\vec{n})\times(\vec{m}-2\vec{n})]\cdot(2\vec{m}+7\vec{n})$ $=$
$=[3\vec{m}\times\vec{m}-6\vec{m}\times\vec{n}+5\vec{n}\times\vec{m}-10\vec{n}\times\vec{n}]\cdot(2\vec{m}+7\vec{n})$

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

so... $(-11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n})$ 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.

**Ackbeet** · August 24th 2011, 05:24 PM

Originally Posted by patzer

Again, this is my approach to this problem:

$(\vec{a}\times\vec{b})\cdot\vec{c}=[(3\vec{m}+5\vec{n})\times(\vec{m}-2\vec{n})]\cdot(2\vec{m}+7\vec{n})$ $=$
$=[3\vec{m}\times\vec{m}-6\vec{m}\times\vec{n}+5\vec{n}\times\vec{m}-10\vec{n}\times\vec{n}]\cdot(2\vec{m}+7\vec{n})$

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

so... $(-11\vec{m}\times\vec{n})\cdot(2\vec{m}+7\vec{n})$ 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.

**patzer** · August 25th 2011, 01:14 AM

Hello Ackbeet,

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 we get 0 at the and. Is this correct?

**Ackbeet** · August 25th 2011, 01:17 AM

Originally Posted by patzer

Hello Ackbeet,

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 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.

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