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Simplification of Cross Product Expression

  1. Nov 19, 2005 #1
    Hey,

    I have the following question,

    Simplify
    (au + bv) x (cu + dv) where a,b,c,d are scalars and u,v are vectors.

    I know that we can take ab ab and cd outside to make the expression
    ab(u +v) x cd(u + v) but I am unsure on where to go from here.

    Thanks in advance
     
  2. jcsd
  3. Nov 19, 2005 #2
    I don't really know how you should go about simplifying this expression yet, but I do believe your first step is wrong.

    See, the way you factored it, when you expand it again the expression would be : (abu + abv) x (cdu + cdv)
     
  4. Nov 19, 2005 #3

    robphy

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    You can't take ab outside (or cd outside), as you have done.
    Instead, use the "FOIL" method, as you would for expanding out the ordinary product of two binomials (a+b)*(c+d). When doing this with the cross product, you have to keep order of the factors.
     
  5. Nov 19, 2005 #4

    Hurkyl

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    What would you do if u and v were just numbers, and the cross product was just ordinary multiplication?

    Can't you do most of the same thing in exactly the same way with cross products? (yes -- but you will have to pay attention to what won't work)
     
  6. Nov 22, 2005 #5
    Hmm. I thought of something else. Is this right?

    (au+bv) X (cu+dv)
    = (au X cu) + (au X dv) + (bv X cu) +(bv X dv)
    = 0 + ad(u X v) + bc(u X v) + 0
    = ad(u X v) + bc(u X v)
     
  7. Nov 22, 2005 #6

    robphy

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    almost.... can you justify the second equal sign?
     
  8. Nov 22, 2005 #7
    It would just be any vector crossed by itself gives 0. u X u = 0 and v X v = 0.

    For the last line, can it be further simplified to

    abcd(u X v)?

    or should it remain ad(u X v) + bc(u X v)?

    Thanks again
     
  9. Nov 22, 2005 #8

    robphy

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    Is the second term bc(u X v)?
     
  10. Nov 22, 2005 #9
    ohh bc(v X u)

    So,

    ad(u X v) + bc(v X u)

    or

    ad(u X v) -bc(u Xv) (because u X v = -(v X u) )

    Hopefully I am now finally right :)
     
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