a x (b x a) . c ?
how to proceed?

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- anonymous

a x (b x a) . c ?
how to proceed?

- katieb

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- anonymous

is that how your problem starts? and is a*(b*a)*c the expression?

- anonymous

No. the expression is a x (b x a) . c

- anonymous

WTH????

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## More answers

- anonymous

and yes it starts like that. i am not sure what to do first: the dot product or the cross product.

- anonymous

In algebra, the dot and the cross express multiplication. In don't get it. What I can see there is a simple product. So it will result in (a*a)bc.

- anonymous

i am talking about dot and cross product
x = cross product
. = dot product

- anonymous

You have to take the cross product first. You don't know what (b x a) is yet, so you can't take the dot product (unless you know about tensors...do you?).

- anonymous

Once you have that result, you have to take the cross product with a (but keep the order of the vectors). You can't dot product until the end because the dot product punches out a scalar...

- anonymous

So (b x a) first. Say it equals Z. Then a x Z = W, say. Then W . c

- anonymous

see here a,b,c are indeed vectors otherwise there is no meaning of . and X. here obviously u have to do X product first. if you dont do that then dot product will give you a scaler and you can't make X product of a vector with scaler...

- anonymous

okay thanks. but can you please elaborate on tensor?

- anonymous

Have you ever seen the dot and cross product defined in this form:\[a.b \iff (a.b)_i=a_ib_i\]\[a \times b \iff (a \times b)_i =\epsilon_{ijk}a_jb_k\]?

- anonymous

no.

- anonymous

They're the tensor definitions for dot and cross product. If you haven't seen them, I wouldn't worry about it.

- anonymous

Is this for school or university?

- anonymous

university. freshman.

- anonymous

Okay. Just punch the vectors out the way you're used to for now.

- anonymous

but the thing is we are just given the vectors as a and b
we have to prove or disprove a result
unless i know how to go about it i would not be able to

- anonymous

Is there a full question, with words?

- anonymous

i tried using the assumed vectors but it gets messy

- anonymous

a x (b x a) = (a.a)b - (a.b)a
so
[a x (b x a)].c = [(a.a)b - (a.b)a].c
= (a.a)(b.c) - (a.b)(a.c)

- anonymous

okay. the question is prove or disprove the following:
a x (a x(a x b)).c = -|a|^2 a.b x c

- anonymous

Ah, okay, that's a full question...

- anonymous

yea. just give me a clue as to how to solve it.

- anonymous

In general we have
a x (b x c) = (a.c)b - (a.b)c
so try it slowly

- anonymous

yea i am doin it that way..

- anonymous

a x (b(a.a)-a(a.b)) . c what now?

- anonymous

can a.a be written as |a|^2 ?

- anonymous

Yes

- anonymous

the last part, i assume it is a.(bxc)

- anonymous

a x (b|a|^2 - a(a.b)) . c

- anonymous

wat to do abt a(a.b) ?

- anonymous

To solve this, use a x (b x c) = (a.c)b - (a.b)c and the scalar triple product a.(bxc) = (axb).c
[a x (a x (a x b))].c
= [(a.(axb)) a - (a.a)(axb)].c
=[a.(axb)](a.c) - (a.a)(axb).c
=[(axa).b](a.c) - |a|^2[a.(bxc)]
= -|a|^2[a.(bxc)] because axa =0

- anonymous

I'm glad *you* typed it out

- anonymous

hm.. ok thanks a lot !

- anonymous

just one more question. i want to practice these questions. can you suggest some website with answers?

- anonymous

Sorry, don't know of any. I'm sure you can find something, though. Schaum's Outlines are usually pretty good for practicing skills.

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