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Easy problem that I am stuck on

  1. Aug 26, 2007 #1
    Easy problem that I am stuck on....

    1. The problem statement, all variables and given/known data
    Two vectors A and B have magnitude of A = 3.02 B=3.00. Their vector product is AXB = -4.90k + 2.01i. What is the angle between vector A and vector B?


    2. Relevant equations
    since I know the vector product of A and B, I was thinking that Cx = 2.01 and Cz = -4.90. I would then add them together to get -2.89. Then could i use the equation C = ABsin(theta). That would be like

    -2.89 = (3.02)(3.00)sin(theta)
    -.319=sin(theta)
    -18.6 = theta
    I don't know if this is the right path to follow but this is what I could come up with.


    3. The attempt at a solution

    THANKS for any help.
     
  2. jcsd
  3. Aug 26, 2007 #2
    it's
    [tex]\left| c\right| =\left| a\right| \,\left| b\right| \,sin\left( p\right) [/tex]
    where c = a x b
     
  4. Aug 26, 2007 #3
    so am I solving it incorrectly because i am using the same equation, its just that I don't know if the value of c that I have is correct. Also the answer I get is not right. So, what should I change? thanks for the help though.
     
  5. Aug 26, 2007 #4
    get the absolute value of c.
    c^2=x^2+y^2+z^2
     
  6. Aug 26, 2007 #5
    Thank You So Much! I Realized What I Was Doing Wrong Was That I Was Taking The Sin When I Should Have Been Taking The Arcsin. Thanks Again!!
     
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