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Scalar triple product, volume, and ordering

  1. Nov 10, 2008 #1
    Greetings all,

    I'm reading about a way to solve for the volume of a "parallelepiped" in 3 space, which is determined by vectors u, v and w. The volume is apparently the absolute value of the determinant given by the matrix

    u1 u2 u3
    v1 v2 v3
    w1 w2 w3

    which is the same as the scalar triple product given by: u . (v x w)

    My question is, the cross product of two vectors is not communitive. v x w
    does not equal w x v.
    Yet, the vectors for a parallelepiped could be given in any order. How do I know how to set up the matrix for an object like this, as it will affect the volume?

    Thanks!
     
    Last edited: Nov 10, 2008
  2. jcsd
  3. Nov 10, 2008 #2

    Office_Shredder

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    [u,v,w] = [v,w,u] = [w,u,v] so you can 'shift' them so to speak, but once you flip the order you get the negative

    [u,v,w] = -[v,u,w]

    What do you know about determinants? You should be able to consider changing the order in terms of row operations (particularly swapping rows). So different orders will only give you possibly the negative of the answer you want.

    If you're calculating volume, obviously if you get a negative answer you can just make it positive by taking absolute value (this would correspond to the case where uxv, instead of pointing into the parallelepiped, points out from them)
     
  4. Nov 10, 2008 #3

    HallsofIvy

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    Notice the crucial phrase in your first post: "The volume is apparently the absolute value of the determinant given". (Emphasis added)

    Changing the order of the vectors changes the sign but since you are taking the absolute value, that does not matter.
     
  5. Nov 10, 2008 #4
    Hmm... ok I understand what your saying with regards to the signs when the equation is already set. But, are you saying if I'm given three random vectors, I can insert them in any order into u . (v x w) and every answer will be the same?
     
  6. Nov 10, 2008 #5

    Office_Shredder

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    Except for a sign change, yes. You posted the determinant method of calculating the triple product, do you know how the determinant changes when you swap rows?
     
  7. Nov 10, 2008 #6
    Luckily, it is a special case of not being commutative. It is anticommutative: a single transposition of the list of arguments gives you the negation of the untransposed result. Thus, you don't have to worry about the magnitude changing.
     
  8. Nov 10, 2008 #7
    Thanks everyone
     
  9. Nov 11, 2008 #8

    HallsofIvy

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    No, for two reasons. First your reference to three random vectors is ambiguous: your question could be interpreted as asking "If I pick three vectors x, y, and z and calculate [itex]x\cdot (y\times z)[/itex] will that be the same as if I pick three vectors a, b, and c and calculate [itex]a\cdot (b\times c)[/itex]?" and the answer to that is clearly "no". Second, the word "answer" is ambiguous. It would be reasonable to interpret "answer" as referring to [itex]u\cdot (v\times w)[/itex] and that may differ in sign if you change the order of u, v, and w. It is the absolute value that does not change.

    [itex]u\times v[/itex] is not commutative but it is anti-commutative. [itex]u\times v= -(v\times u)[/itex].
     
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