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Cross product vs dot product headache

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Show (A x B) dot (C x D) in terms of dot products only.


    2. Relevant equations
    A x B = ABsin(theta)
    A dot B = AB cos (theta)


    3. The attempt at a solution
    Subbing both those formulas into the top I got
    [ABsin(AtoB) times CDsin(CtoD) ] cos(between the two resultant vectors)

    Yet then when I try to prove it, plugging in random numbers for the four beginning vectors, it doesn't come out right... any help?
     
  2. jcsd
  3. Oct 6, 2008 #2

    gabbagabbahey

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    I don't think that is what the question is wanting you to do. I'm pretty sure they want you to prove Lagrange's Identity;

    [tex](\vec{A} \times \vec{B}) \cdot (\vec{C} \times \vec{D})= (\vec{A} \cdot \vec{C})(\vec{B} \cdot \vec{D})-(\vec{B} \cdot \vec{C})(\vec{A} \cdot \vec{D})[/tex]

    by decomposing each vector into its components (eg. [itex]\vec{A}=(A_x,A_y,A_z))[/itex] and carrying out the cross products and dot product and then rearranging terms to show that it equals the above expression.
     
  4. Jun 15, 2009 #3
    Question from a multivariable calc newbie.
    I don't see how (A subx, Asuby, Asubz) is a vector. That is a point 3D space.
    This is a vector,
    (A1subx, A1suby, A1subz, A2subx, A2suby, A2subz) plus all the infinite vectors that are parallel with it. Yet I always see vectors with three variables not six. Is that because you can assume it is coming from the origin?
    So in that case, you would never need more than one point to define a vector in 3D space?
     
  5. Jun 15, 2009 #4

    tiny-tim

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    Hi Billybob! :smile:
    You left out the triple product identity … (A x B). E = (E x A).B :wink:
    Hi rockyshephear! :smile:

    Yes!!

    Technically, parallel vectors are the same

    a vector only has direction and magnitude.

    (I know we say force is a vector, and obviously force also has a line of application, which does matter when we calculate torques, though not when we calculate linear components … but calling force a vector is technically a bit of an understatement :wink:)
     
  6. Jun 15, 2009 #5
    Thanks Tiny-Tim. You get an extra drumstick for seeing that I'm really getting it...a bit anyway. :)
    Question 1
    What I'm working at, my long term goal, is a deeper appreciation of Maxwell's Equations. They're good to the last drop. ha!
    So an electric field is a bunch of vectors that nobody really knows where they're at in 3D space?
    Question 2
    Is it difficult to do cross product and dot product calculations if you are NOT assuming that the origin is the start of your vectors and you actually have two defined points for each vector?

    R
     
  7. Jun 15, 2009 #6
    Oh and Question 3
    Since you only need one point and the origin to define a vector, are there a million ways to write a vector? I've seen a few
    (x1, x2, x3)
    (3i hat, 4j hat 1k hat)
    (x, y, z)

    (3i hat, 4j hat 1k hat) I like this one best because it give exact position of the point. The others do not.
    I would think that there would always be a ref to each coord plane for a vector to be defined. What does (x1, x2, x3) mean???

    Thx
     
  8. Jun 15, 2009 #7

    tiny-tim

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    Hi Rocky! :smile:
    No an electric field is a vector field … a different ordinary vector (ie just a direction and a magnitude) at each point. :wink:
    Cross product and dot product only work for "ordinary" vectors :smile:
    (x1, x2, x3) and (x, y, z) both presume that a basis (of three unit vectors) has already been specified (and then the x1 or y etc is just any number).

    (3i hat, 4j hat 1k hat) is wrong, I'm afraid …

    it makes no sense … you have to write 3i hat + 4j hat + k hat.
     
  9. Jun 15, 2009 #8
    One last thing then I'll leave you alone for a while.
    You said "ordinary vectors". What is the difference between an ordinary one (ordinate? or trite?) and a non-ordinary one?
     
  10. Jun 15, 2009 #9

    tiny-tim

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    ah, i was calling the official vector, with just magnitude and direction, an "ordinary" vector …

    a vector with a line of application added is often also called a vector, but it should really have a longer name … I think of it as a vector-with-bells-on :biggrin:
     
  11. Jun 15, 2009 #10
    Well, I lied. lol
    So a vector field IS, as I have said, a field of vectors, each of which is a vector, as a single vector is a vector. No distinction except there are many of them and all together they give a pictogram of the motion of particles, etc.
    Thanks for all your time answer my questions.
    R
     
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