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Application of circularity of hybrid product

  1. Oct 16, 2016 #1
    1. The problem statement, all variables and given/known data
    http://photouploads.com/images/8ba21e.png
    8ba21e.png
    {moderator's not: Inserted image so it's visible without having to follow a link}

    2. Relevant equations

    a·(b x c) = b·(c x a)

    3. The attempt at a solution
    How did they get from the (b) to (c)? In particular, I am referring to the numerator and denominator of the fraction. I mean if you apply a·(b x c) = b·(c x a) then surely the terms inside the brackets (of both numerator and denominator) of part (c) should be the other way round?

    I mean for instance rCD·(k x rBA) = k·(rBA x rCD) and not rCD·(k x rBA) = k·(rCD x rBA) as they got?
     
    Last edited by a moderator: Oct 17, 2016
  2. jcsd
  3. Oct 17, 2016 #2

    gneill

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    Staff: Mentor

    Did you happen to notice that the prefixed "-" sign also disappeared from (b) to (c)?...
     
  4. Oct 20, 2016 #3
    Ah I never noticed that! So -rCD·(k x rBA) = -k·(rBA x rCD) but I'm still confused in terms of how they got from -k·(rBA x rCD) to k·(rCD x rBA)? I know that A x B = -B x A so -k·(rBA x rCD) can be written as -k·(-rCD x rBA) and I'm guessing the two negatives cancel somehow but don't know how..
     
  5. Oct 20, 2016 #4

    gneill

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    Staff: Mentor

    -rCD can be written as (-1)rCD, the scalar -1 being "pulled out" of the vector. That scalar in turn can be moved out of the parentheses and applied to the vector -k.
     
  6. Oct 20, 2016 #5
    So -k·((-1)rCD x rBA)=(-1)(-k)·(rCD x rBA) = k·(rCD x rBA)?

    I was thinking it would be something along these lines but I wasn't sure how to pull out the negative when dealing with vectors.
     
  7. Oct 20, 2016 #6

    gneill

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    Staff: Mentor

    Yes that's good. A "negative" is just a scalar constant, -1. So you can manipulate it as you would any scalar multiplyer of a vector.
     
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