anonymous
  • anonymous
a x (b x a) . c ? how to proceed?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
is that how your problem starts? and is a*(b*a)*c the expression?
anonymous
  • anonymous
No. the expression is a x (b x a) . c
anonymous
  • anonymous
WTH????

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anonymous
  • anonymous
and yes it starts like that. i am not sure what to do first: the dot product or the cross product.
anonymous
  • 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
  • anonymous
i am talking about dot and cross product x = cross product . = dot product
anonymous
  • 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
  • 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
  • anonymous
So (b x a) first. Say it equals Z. Then a x Z = W, say. Then W . c
anonymous
  • 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
  • anonymous
okay thanks. but can you please elaborate on tensor?
anonymous
  • 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
  • anonymous
no.
anonymous
  • anonymous
They're the tensor definitions for dot and cross product. If you haven't seen them, I wouldn't worry about it.
anonymous
  • anonymous
Is this for school or university?
anonymous
  • anonymous
university. freshman.
anonymous
  • anonymous
Okay. Just punch the vectors out the way you're used to for now.
anonymous
  • 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
  • anonymous
Is there a full question, with words?
anonymous
  • anonymous
i tried using the assumed vectors but it gets messy
anonymous
  • 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
  • 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
  • anonymous
Ah, okay, that's a full question...
anonymous
  • anonymous
yea. just give me a clue as to how to solve it.
anonymous
  • anonymous
In general we have a x (b x c) = (a.c)b - (a.b)c so try it slowly
anonymous
  • anonymous
yea i am doin it that way..
anonymous
  • anonymous
a x (b(a.a)-a(a.b)) . c what now?
anonymous
  • anonymous
can a.a be written as |a|^2 ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
the last part, i assume it is a.(bxc)
anonymous
  • anonymous
a x (b|a|^2 - a(a.b)) . c
anonymous
  • anonymous
wat to do abt a(a.b) ?
anonymous
  • 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
  • anonymous
I'm glad *you* typed it out
anonymous
  • anonymous
hm.. ok thanks a lot !
anonymous
  • anonymous
just one more question. i want to practice these questions. can you suggest some website with answers?
anonymous
  • 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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