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(edit:solved) Vector Triple Product, Components Parallel and Perpendicular

  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data
    By considering A x (B x A) resolve vector B into a component parallel to a given vector A and a component perpendicular to a given vector A.

    2. Relevant equations
    a x (b x c) = b (ac) - c (ab)

    3. The attempt at a solution
    I've applied the triple product expansion and reached

    A x (B x A) = B (AA) - A (AB) = |A|2 B - |A||B|cos(θ) A

    and hit a brick wall. I'm not entirely sure what the question is asking me to do, and I feel like I'm missing crucial information. Should I be splitting vectors A and B into their cartesian components?


    EDIT:

    I got it. Taking A x (B x A) = |A|2 B - A (AB) , I can rearrange to find

    |A|2 B = A x (B x A) + A (AB)

    Dividing through by |A|2 results in

    B = A x (B x A) |A|-2 + A ((AB)/|A|2)

    So B is given as a component perpendicular (first term) and parallel (second term) to A.

    I'm fairly sure this is right anyway. Stupid I suddenly work this out after posting here after looking at it for 30 mins before!
     
    Last edited: Apr 4, 2015
  2. jcsd
  3. Apr 5, 2015 #2

    FactChecker

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    Gold Member

    FYI, The second term is called the "projection" of B on A and the first term is called the "rejection" of B on A.
     
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