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Product of Vectors (cross)

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    ^ ^ ^
    i = i; j = j; k = k;

    i = x j = y k = z
    So this is the result after cross multiplication.


    -2ij - 8ji = 6k

    -ik + 4ki = 5j

    -4jk +4kj = -8i

    Unfortunately I am a little confused with the 5j
    It seems to me that it would be

    -ik + 4ki = -j -4j

    because I comes first and so it would be -5j

    I know this is not the case but it ruins the logic of the whole ordeal for me ; /. How are you supposed to know which sign you need to flip? I thought it was if it was ij you would leave it alone but if it was ji it would be -ji because order changed and i is multiplied first.
     
  2. jcsd
  3. Sep 13, 2009 #2

    rl.bhat

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    Post your problem.
    Cross multiplication of which vectors?
     
  4. Sep 13, 2009 #3
    It is a . (b x c)
    http://img7.imageshack.us/img7/7537/problemd.th.png [Broken]

    Basically if

    a x b = (aybz - byaz)i + (azbx - bzax)j + (axby - bxay)k

    I assumed that the second part was negative because a x b = -b x a
    and so (aybz - byaz)i + (axby - bxay)k seemed alright but then the (azbx - bzax)j messed everything up for me because I thought it would be (axbz - bzax)j.
    Then after that idea failed miserably I had no clue. I have memorized the end result at this point but I havent a clue how to get there ; (
     
    Last edited by a moderator: May 4, 2017
  5. Sep 13, 2009 #4

    rl.bhat

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    Your a.(bxc) is correct.
    -ik + 4ki = -j -4j
    ixk = -j and kxi = j
    so -(-j) + 4J = 5J
     
  6. Sep 13, 2009 #5
    But what makes the ixk change signs as oppose to the kxi ?
    I am missing how the kxi takes precedence over the ixk. Wouldn't it be the other way around?
    You multiple the ixk first and i comes before k in equation :(.

    Is there a rule to determine which flips?
     
  7. Sep 13, 2009 #6

    rl.bhat

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    Yes.
    AxB = - BxA
     
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