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Proof of a property of the cross product

  1. Dec 1, 2014 #1
    1. The problem statement, all variables and given/known data
    I could prove a, trying b now.

    2. Relevant equations
    The definition of the cross prod.?

    3. The attempt at a solution
    I did not manage to get a scalar times v and a scalar times w. (No need to point this out.)

    This is not an assignment, so you can give me the answer right away (LaTeX because I like it). But please, just point my mistake so that I can hopefully learn something.
  2. jcsd
  3. Dec 1, 2014 #2


    Staff: Mentor

    Your link requires us to signup for Dropbox so perhaps you could post it some other way.
  4. Dec 1, 2014 #3
    Really? I can access it without being logged in. I am pretty sure that is a shared link. Try again.
  5. Dec 1, 2014 #4
    Anyway, here it is.

    Attached Files:

  6. Dec 1, 2014 #5


    Staff: Mentor

    Ok got it, I was fooled by the Dropbox dialog that pooped up. At the bottom of the dialog was a no thanks link...
  7. Dec 1, 2014 #6
    Good to know.
  8. Dec 1, 2014 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    You were correct until the very last step. Back up a step to get the matrix:

    [itex]\left( \begin{array} & (u_2 w_2 + u_3 w_3) v_1 \\ (u_3 w_3 + u_1 w_1) v_2 \\ (u_1 w_1 + u_2 w_2) v_3 \end{array} \right) - \left( \begin{array} & (u_2 v_2 + u_3 v_3) w_1 \\ (u_3 v_3 + u_1 v_1) w_2 \\ (u_1 v_1 + u_2 v_2) w_3 \end{array} \right) [/itex]

    Look at just the top row:
    [itex](u_2 w_2 + u_3 w_3) v_1 - (u_2 v_2 + u_3 v_3) w_1[/itex]

    The expression [itex](u_2 w_2 + u_3 w_3)[/itex] is almost [itex]u \cdot w[/itex]. The expression [itex](u_2 v_2 + u_3 v_3)[/itex] is almost [itex]u \cdot v[/itex]. What's missing from the two expressions?
  9. Dec 1, 2014 #8


    Staff: Mentor

    Have you tried working it from the other side and then seeing where they meet.

    From looking at it but not actually doing the work it seems you might have to add and subtract some additional terms so that you get the u.w scalar instead of a row vector.

    In the first element, you need a u1w1 term and in the second you need a u2w2 term and in the third you need a u3w3 term right giving u.w scalar
  10. Dec 1, 2014 #9
    Starting at the end seemed easier. Thank you guys.

    Is this proof reasonable? Starting with the rhs is not a problem, right?

    Attached Files:

  11. Dec 1, 2014 #10


    Staff: Mentor

    It shouldn't matter you've proved that they are equal.

    I use the trick to get the ends to meet because sometimes you need a leap of creativity to see the next step or you come from the other side and it becomes obvious.
  12. Dec 1, 2014 #11
    I saw it.

    I am somewhat new to proofs and theorems and can't help but find this whole stuff useless as f***. I'd rather do math for the numbers, not symbols.
  13. Dec 1, 2014 #12


    Staff: Mentor

    You should reconsider your position. What you learn doing proofs can come in handy for a lot of things?

    From Math to Programming to Physics to Law, all of these use proof-like expositions to explain their conclusions.
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