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## anonymous 5 years ago a x (b x a) . c ? how to proceed?

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1. anonymous

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

2. anonymous

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

3. anonymous

WTH????

4. anonymous

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

5. 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.

6. anonymous

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

7. 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?).

8. 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...

9. anonymous

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

10. 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...

11. anonymous

okay thanks. but can you please elaborate on tensor?

12. 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$?

13. anonymous

no.

14. anonymous

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

15. anonymous

Is this for school or university?

16. anonymous

university. freshman.

17. anonymous

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

18. 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

19. anonymous

Is there a full question, with words?

20. anonymous

i tried using the assumed vectors but it gets messy

21. 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)

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

yea i am doin it that way..

27. anonymous

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

28. anonymous

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

29. anonymous

Yes

30. anonymous

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

31. anonymous

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

32. anonymous

wat to do abt a(a.b) ?

33. 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

34. anonymous

I'm glad *you* typed it out

35. anonymous

hm.. ok thanks a lot !

36. anonymous

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

37. 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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