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How to do cross product if I have got only two coordinate?

  1. Jul 25, 2013 #1
    a =(x,y), b =(h,k)
    a cross b =?

    I have idea what to type on google. Is that doing like matrices , a cross b = xk-hy?
    thanks.
     
  2. jcsd
  3. Jul 25, 2013 #2

    WannabeNewton

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    You can't. The cross product is only naturally defined for 3 dimensions and 7 dimensions. If you want to do the above cross product you would have to add a zero as a third coordinate so that those vectors are on the plane embedded in 3-space i.e. a = (x,y,0) and b = (h,k,0).
     
  4. Jul 25, 2013 #3

    symbolipoint

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    "...naturally defined for 3 ... and 7 dimension"


    Does that mean no cross product can occur in 4 or 5 dimension? I have been reviewing dot and cross product in the last recent few days, and although I have not yet tried to, one would imagine in four dimensions, one could have two vectors, and set up dot product relationships to find what new vector is orthogonal (dot products are zero) to both of these two vectors.
     
  5. Jul 25, 2013 #4
    Well...yes. There is a form of "cross product" in 4 or 5 dimensional Euclidean space. Kind of.

    Recall that the magnitude of the cross product ##\vec{u}\times\vec{v}## of two vectors ##\vec{u}## and ##\vec{v}## is equal to the area of the parallelogram outlined by the two vectors. That is also the magnitude of the corresponding bivector, which can be intuitively thought of as an oriented family of parallelograms. This bivector is the Hodge dual of the cross product, given by ##\vec{u}\wedge\vec{v}##. Thus, we can define the cross product by ##\vec{u}\times\vec{v}=\star\left(\vec{u}\wedge\vec{v}\right)##.

    However, if you think about it, what's the difficulty of this definition? For example, what would be the cross product in ##\mathbb{R}^4##? :wink:
     
    Last edited: Jul 25, 2013
  6. Jul 25, 2013 #5

    WannabeNewton

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    Indeed dimensions 3 and 7 are the only ones that admit non-degenerate cross products (dimension 1 admits a degenerate cross product). It's a result of Hurwitz's theorem, see here (section 5): http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/hurwitzlinear.pdf
     
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