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The product between quaternion

  1. Aug 4, 2006 #1
    For example:
    i^2=?
    j^2=?
    k^2=?
    ij=?
    jk=?
    ik=?
    ijk=?
    Is ij=ji?
    And how to prove them?

    And also,vector times vector, what is the product?
     
  2. jcsd
  3. Aug 4, 2006 #2
    This is from wikipedia http://en.wikipedia.org/wiki/Quaternions

    i2 = j2 = k2 = ijk = -1
    ij = k
    jk = i
    ki = j
    ji = -k
    kj = -i
    ik = -j

    The first one I think is the basic definition and the rest follow from that, some of the proofs are on the wikipedia page, and what do you mean by a vector times a vector? The above equalities allow a general definition of a product of quaternions.
     
  4. Aug 4, 2006 #3
    Just like:
    A=i+j+k
    B=2i+3j-4k
    They are both vector.
    I heard that a vector times a vector become a tensor(rank two).
    If ij=-k,ik=-j,jk=-i.
    The product is:
    (i+j+k)(2i+3j-4k)=-2-3k+4j-2k-3+4i-2j-3i+4=i+2j-5k-1
    I think it is still a vector, not a tensor(rank two).
     
  5. Aug 4, 2006 #4
    The product of two quaternions is still a quaternion (and quaternions are not vectors). You did the multiplication wrong up there, if A and B are as you had then AB is -7i + 6j + k - 1.

    You can define many products between vectors, for example in [itex]\mathbb{R}^3[/itex] you have the usual cross and dot products (and the dot product generalizes to other spaces of course). Those two products can be read off the result of quaternion multiplication:

    [tex]<1,1,1> \times <2,3,4> = <-7,6,1>[/tex]

    and

    [tex]<1,1,1> \cdot <2,3,4> = 1 = -(-1).[/tex]

    In general if the product of two quaternions [itex]A = a_1 i + a_2 j + a_3 k[/itex] and [itex]B = b_1 i + b_2 j + b_3 k[/itex] is [itex]AB = C = c_1 i + c_2 j + c_3 k - c_4[/itex], then [itex]<a_1, a_2, a_3> \times <b_1, b_2, b_3> = <c_1, c_2, c_3>[/itex] and [itex]<a_1, a_2, a_3> \cdot <b_1, b_2, b_3> = c_4.[/itex]
     
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