thekidwhogames
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The vectors a,b,c are such that b x a = (2 1 0) and a x c = (0 3 -2)

Find (-2b-c) x (-3a - 4b + 2c)

What's the fastest solution?
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DFranklin
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Distributive law: (p + q) x (r + s) = p x r + p x s + q x r + q x s.
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thekidwhogames
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So it comes down to:

(-2b x -3a) + (-2b x - 4b) + (-2b x 2c) + (-c x -3a) + (-c x -4b) + (-c x 2c)?
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DFranklin
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(Original post by thekidwhogames)
So it comes down to:

(-2b x -3a) + (-2b x - 4b) + (-2b x 2c) + (-c x -3a) + (-c x -4b) + (-c x 2c)?
Use p x p = 0 and p x q = - q x p to simplify, but yes.
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thekidwhogames
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(Original post by DFranklin)
Use p x p = 0 and p x q = - q x p to simplify, but yes.
Is -a x -b = b x a?
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DFranklin
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(Original post by thekidwhogames)
Is -a x -b = b x a?
No. It's a x b. \lambda {\bf a} \times \mu {\bf b} = \lambda \mu ({\bf a} \times {\bf b}) for any \lambda, \mu \in \mathbb{R}. Here you have \lambda = \mu = -1.
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thekidwhogames
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(Original post by DFranklin)
No. It's a x b. \lambda {\bf a} \times \mu {\bf b} = \lambda \mu ({\bf a} \times {\bf b}) for any \lambda, \mu \in \mathbb{R}. Here you have \lambda = \mu = -1.
Is this right?

This makes it (-2b x -3a) + (-2b x 2c) + (-c x -3a) + (-c x -4b)

-2b x -3a = 6(bxa) = (12 6 0)
-2bx2c = -4(bxc)
-c x -3a = 3(cxa)
-c x -4b = 4(cxb)

So it becomes (12 6 0) - 4(bxc) + 4(cxb) + 3(cxa)

If b x c = k then c x b = -k which means that -4(bxc) = -4k and 4(cxb) = -4k

So -4(bxc) + 4(cxb) = -8(bxc)

So now its (12 6 0) - 8(bxc) + 3(cxa)

We know a x c = b then c x a = -b which means that c x a = (0 -3 2) so 3(cxa) = (0 -9 6)

So it becomes (12 6 0) - 8(cxb) + (0 -9 6)

How would I find c x b? I know b x a + a x c so a ( b x c) I can get by adding the 2 given
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DFranklin
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Hmm... I confess I made a sign error and thought the bxc and cxb terms would cancel. I'm a bit flummoxed.
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thekidwhogames
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(Original post by DFranklin)
Hmm... I confess I made a sign error and thought the bxc and cxb terms would cancel. I'm a bit flummoxed.
Yeah I attempted it earlier and thought I forgot how the cross product worked but I got the same issue with the end.

Can you try it please?
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DFranklin
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(Original post by thekidwhogames)
Yeah I attempted it earlier and thought I forgot how the cross product worked but I got the same issue with the end.

Can you try it please?
I'm not seeing anything quick.

What am I missing Zacken...?
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thekidwhogames
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(Original post by DFranklin)
I'm not seeing anything quick.

What am I missing Zacken...?
Perhaps an issue in the question? The resource had issues before. Thanks for the help by the way!
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ghostwalker
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(Original post by DFranklin)
I'm not seeing anything quick.

What am I missing Zacken...?
I suspect you meant to tag Zacken, rather than Zackenn

(Original post by thekidwhogames)
Perhaps an issue in the question? The resource had issues before. Thanks for the help by the way!
Can't see anything wrong with the working, and am inclined to agree - issue with the question.

At a guess, meant to be:

(-2b-c) x (-3a - 4b - 2c)

or perhaps

(-2b+c) x (-3a - 4b + 2c)

If the answer they're looking for is available, it may be possible to back track to one of these.
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thekidwhogames
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(Original post by ghostwalker)
I suspect you meant to tag Zacken, rather than Zackenn



Can't see anything wrong with the working, and am inclined to agree - issue with the question.

At a guess, meant to be:

(-2b-c) x (-3a - 4b - 2c)

or perhaps

(-2b+c) x (-3a - 4b + 2c)

If the answer they're looking for is available, it may be possible to back track to one of these.
Yeah, suspected so - thanks!
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