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Vector product and vector product angles

  1. Sep 8, 2014 #1
    Hello! I have a problem in my calculus based physics class regarding vectors. The problem says:

    Vectors A and B have a scalar product -6.00 and their vector product has magnitude 9.00 what is the angle between these two vectors?

    Here is how I approached it:

    -6=|A||B|cos (theta)
    9=|A||B|sin (theta)
    tan (theta)= sin (theta)/cos (theta)
    tan (theta)=9/-6=-56.31 degrees
    since the sine is positive and cosine is negative the angle lies in the second quadrant.
    180 degrees -56.31 degrees= 123.69 degrees which is approximately 124 degrees.

    Now, why does the angle between the scalar product and the vector product of A and B give us the angle between A and B?

    Thanks!
     
  2. jcsd
  3. Sep 8, 2014 #2

    BvU

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    Scalar product is a number. There is no angle between a number and a vector...
     
  4. Sep 8, 2014 #3
    Then how would I explain how I got the right answer?
     
  5. Sep 9, 2014 #4

    BvU

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    |vector product| / |scalar product| gives you sin/cos, so the tangent. That is enough to extract the angle in ##[0, \pi]##
     
  6. Sep 9, 2014 #5
    Thanks! That clarifies a lot of things. ^-^
     
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