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Dot and cross product vector

  1. Sep 21, 2003 #1

    Dx

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    [SOLVED] dot and cross product vector

    Hi!
    The quantity (absolute value vectorA x vectorB close absolute value)^2 + (absolute value vectorA * vectorB close absolute value)^2 is equivalent to:

    I am confused, a close example used numbers in place of the vectors so I gave it a try in a book I have. I know because its absolute that the numbers would be positive then all I would have to do is square those values and then solve. Nope! I must have done something wrong because it doesn't match any of the answers found. I thought the answer was vectorA^2 + 2AvectorA * vectorB but I really guessed so I am not to sure what to do for this problem. Anyone want to help me please?
    Dx :wink:
     
    Last edited by a moderator: Sep 21, 2003
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  3. Sep 21, 2003 #2

    HallsofIvy

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    I think the fact that you don't even tell us what "numbers" you are talking about indicates that you are not reading the problem very closely.

    Perhaps it would be simplest to use the fact that, for cross product, |u x v|= |u||v|sin(theta), where theta is the angle between u and v, and, for dot product, u*v= |u||v|cos(theta).

    |u x v|+ |u*v|= |u||v|(sin(theta)+ |cos(theta)|)

    You have to keep the absolute value on cosine since, if the angle between the vectors is greater than 90 degrees, it will be negative.

    I don't know why you are talking about squaring: the problem as you gave it said nothing about squaring.
     
  4. Sep 21, 2003 #3

    Dx

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    HOI! I re-edited my question, it does have ^2 in my problem now. The "NUMBERS" I refered to were generic, made-up, don't exist simply because the problem did not define the vectors.
    I always understood that if you (eg; |-1| = 1) had an absolute of anything it turns negatives to positive numbers. Can you show me what your talking about below with the |cos(theta) being greater than 90 degrees which would yeild a negative number[?]
    I appreciate your help.
    Dx :wink:
     
    Last edited by a moderator: Sep 21, 2003
  5. Sep 21, 2003 #4

    Dx

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    How about this?

    |u x v|^2+ |u*v|^2= |u||v|-(u*v)^2 + |u*v|^2
    Then cancel out the dot products leaving |u||v| as my answer.
    What do you think?
    Dx :wink:
     
  6. Sep 21, 2003 #5
    |A*B| = |A||B|cos θ
    |AxB| = |A||B|sin θ

    therefore

    |A*B|2 + |AxB|2 = |A|2|B|2(cos2 θ + sin2 θ) = |A|2|B|2
     
  7. Sep 21, 2003 #6

    HallsofIvy

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    What reason do you have for thinking that equation is true?
    Which part is |u x v|^2 and which part is |u*v|^2?

    You might do best by taking the formulas that both Lethe and I suggested and squaring them.
     
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