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Cross product / inner product

  1. Nov 16, 2011 #1
    I know the cross product and dot product of euclidean space R^3.

    But I wanted to know if there is a way of thinking the cross product "in terms of" the dot product.
    That is because the dot product can be generalized to an inner product, and from R^3 to an arbitrary inner vector space (and Hilbert space). In that process of generalization, where does the cross product fit?

    (1) I mean, is there a way (at least in R^3) of calculating a cross product using only the dot product?

    (2) Moreover, I want to know, the definition of euclidean space R^n, is the vector space along with the usual inner product (dot product), or with any other inner product also is considered an eucliden space?

  2. jcsd
  3. Nov 16, 2011 #2


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    in Rn, one can generalize the cross product to be the unique vector z in Rn such that for all w:

    [tex]\langle w,z\rangle = \det\begin{pmatrix}v_1\\ \vdots\\v_{n-1}\\w\end{pmatrix}[/tex]

    note that this means we need n-1 "multiplicands" to define a cross product.

    normally, we think of a product as a binary operation, so n-1 = 2 means that a "two-term" cross product can only be defined in R3.

    in 3 dimensions the above formula becomes:

    [tex]\bf{a}\cdot(\bf{b}\times\bf{c}) = \det\begin{pmatrix}\bf{a}\\ \bf{b}\\ \bf{c}\end{pmatrix}[/tex]
  4. Nov 16, 2011 #3
    Hi Deveno,
    Thanks, but that doesn't answer any of my questions.
    Can (even in R^3) the cross product be defined using the inner product only?
    What makes me unconfortable is that the dot product can be generalized to inner product spaces and hilbert spaces, while the cross product doesn't generalize to any kind of space?
  5. Nov 19, 2011 #4
    How can you relate the cross product and the dot product? One gives a vector and the other gives a scalar...
  6. Nov 19, 2011 #5
    You can't generalize the cross product. Ask yourself this simple question: which dimension is perpendicular to all x, y and z planes?

    If you say which dimension is perpendicular to x and y planes, the answer is the z plane but there is no such analog for higher dimensions.
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