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Cross product vectors problem

  1. Aug 6, 2013 #1
    a and b are two vectors and x is the angle between them.

    ||axb|| = ||a||||b||sinx ------(1)
    ||axb|| = ||a||||b||||sinx|| ------(2)

    which one is correct? why?
     
  2. jcsd
  3. Aug 6, 2013 #2

    SteamKing

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    1. That is one definition of cross product.
     
  4. Aug 6, 2013 #3

    CAF123

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    x is the smallest angle between vectors a and b. You should convince yourself that this implies both forms are equivalent.
     
  5. Aug 6, 2013 #4
    if x is the smallest angle between a and b, when n is the unit vector perpendicular to both a and b,

    axb = ||axb||n

    am i correct?
     
  6. Aug 6, 2013 #5

    Zondrina

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    Your first one is correct :

    ##||axb|| = ||a|| ||b|| sin(θ)##
     
  7. Aug 6, 2013 #6

    CAF123

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    Yes

    I think the OP also wanted to know why. I would say the two forms are equivalent. What is your argument for why the other form is incorrect?
     
  8. Aug 6, 2013 #7
    if two forms are equivalent then ||sinx||= sinx (where x is the smallest angle between a and b)
    how can you explain this?
     
  9. Aug 6, 2013 #8

    CAF123

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    Given that x is the smallest angle between a and b what is the possible values of x?
     
  10. Aug 6, 2013 #9

    lurflurf

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    This has to do with what values of x we allow. We need the absolute value if we are to allow x such that sin x<0
     
  11. Aug 6, 2013 #10
    assume x= 11∏/6 , then smallest angle between a and b ∏/6, but when applying it to x it should be -∏/6. shouldn't it? (if it shouldn't then why?)

    are ||sin(-∏/6)|| and sin(-∏/6) equivalent?
     
  12. Aug 6, 2013 #11

    CAF123

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    No, x is restricted to lie within the interval ##[0, \pi]##. Check your book. This condition will come with the definition of cross product.
     
  13. Aug 6, 2013 #12

    lurflurf

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    Who cares if x is in [0,pi) or not. We have |sin x| if sin x is already positive sin x will do.
     
  14. Aug 6, 2013 #13

    CAF123

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    Yes, so if ##x \in [0,\pi]## then ##|a \times b| = |a| |b| \sin x## since sin x is positive. If x is not in this interval, then ##|a \times b| = |a| |b| |\sin x|## (the magnitude of ##a\times b## is not negative so we take the modulus of sin x)

    In any book I have read, they put the condition that x is in [0,π], so you see it written ##|a \times b| = |a| |b| \sin x## more commonly.
     
  15. Aug 6, 2013 #14
    According to the way of my thinking ||axb|| = ||a||||b||sinx is enough to represent magnitude of axb for all x angle.

    let n be an unit vector perpendicular to both a and b vectors. therefore we get,

    ||axb|| = || ||a||||b||sinx n ||

    when x > 0, ||axb|| = ||a||||b||sinx ||n||
    since ||n|| = 1 , ||axb|| = ||a||||b||sinx

    when x<0, ||axb|| = || ||a||||b||sin(-x) n || = ||a||||b||sinx ||-n|| (with right hand rule)
    since ||-n|| = 1 , ||axb|| = ||a||||b||sinx

    although the other form (2) gives the same magnitude, it is theoretically incorrect.
    and only this imaginary of mine is not confusing me. I wanna know does this really happen there?
     
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