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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    is that how your problem starts? and is a*(b*a)*c the expression?

  2. anonymous
    • 5 years ago
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    No. the expression is a x (b x a) . c

  3. anonymous
    • 5 years ago
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    WTH????

  4. anonymous
    • 5 years ago
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    and yes it starts like that. i am not sure what to do first: the dot product or the cross product.

  5. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    i am talking about dot and cross product x = cross product . = dot product

  7. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    So (b x a) first. Say it equals Z. Then a x Z = W, say. Then W . c

  10. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    okay thanks. but can you please elaborate on tensor?

  12. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    no.

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

  15. anonymous
    • 5 years ago
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    Is this for school or university?

  16. anonymous
    • 5 years ago
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    university. freshman.

  17. anonymous
    • 5 years ago
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    Okay. Just punch the vectors out the way you're used to for now.

  18. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Is there a full question, with words?

  20. anonymous
    • 5 years ago
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    i tried using the assumed vectors but it gets messy

  21. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Ah, okay, that's a full question...

  24. anonymous
    • 5 years ago
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    yea. just give me a clue as to how to solve it.

  25. anonymous
    • 5 years ago
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    In general we have a x (b x c) = (a.c)b - (a.b)c so try it slowly

  26. anonymous
    • 5 years ago
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    yea i am doin it that way..

  27. anonymous
    • 5 years ago
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    a x (b(a.a)-a(a.b)) . c what now?

  28. anonymous
    • 5 years ago
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    can a.a be written as |a|^2 ?

  29. anonymous
    • 5 years ago
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    Yes

  30. anonymous
    • 5 years ago
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    the last part, i assume it is a.(bxc)

  31. anonymous
    • 5 years ago
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    a x (b|a|^2 - a(a.b)) . c

  32. anonymous
    • 5 years ago
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    wat to do abt a(a.b) ?

  33. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    I'm glad *you* typed it out

  35. anonymous
    • 5 years ago
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    hm.. ok thanks a lot !

  36. anonymous
    • 5 years ago
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    just one more question. i want to practice these questions. can you suggest some website with answers?

  37. anonymous
    • 5 years ago
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    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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