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Physics: Multiplying Unit vectors

  1. Oct 18, 2015 #1
    [Moderator note: Post moved from New Member Introductions forum, so no template]

    I am having trouble understanding how to multiply unit vectors. I know that: (please excuse the notation)
    i^×j^ = k^
    j^×k^ = i^
    k^×i^ = j^

    The question I am stuck on is: What is (i^×j^)×k^?
    So far I have (i^×j^) = k^, which would mean the above equation turns into k^×k^. I am supposed to represent the answer in unit vector form, but the only answers I can come up with are k^ squared or -1... What am I missing? All help is appreciated!
     
    Last edited by a moderator: Oct 18, 2015
  2. jcsd
  3. Oct 18, 2015 #2

    SteamKing

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    Don't ask technical questions in your Introduction post.

    You should learn some basic facts about dot products and cross products of unit vectors.

    https://en.wikipedia.org/wiki/Cross_product
     
  4. Oct 18, 2015 #3
    What can I ask in my introduction post?
     
  5. Oct 18, 2015 #4

    Mark44

    Staff: Mentor

    You shouldn't ask anything. The purpose of an introduction post is to introduce yourself.
     
  6. Oct 18, 2015 #5

    HallsofIvy

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    If that is all you know then you don't know enough to calculate anything else! You would also need to know that the "cross product" of two vectors is "bilinear" meaning (au^+ bv^) x w^= a(u^xw^)+ b(v^xw^) and that the cross product is "anti-commutative" meaning that u^xv^= -v^xu^

    There are three different kinds of "multiplication" defined for vectors, the "scalar product", of a scalar with a vector, returning a vector, the "dot product" (also called "inner product"), of two vectors, returning a scalar, and the "cross product", that you are talking about here, of two vectors, returning a vector. The terminology "k^ squared" is ambiguous because that would be a product of two vectors which could be either the dot product or the cross product. To find the cross product of k with itself, use the fact that "u^xv^= -v^xu^" with both u^ and v^ equal to k^ that gives k^xk^= -k^xk^. What does that tell you?
     
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