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Multiplying a Vector Product by Another Vector

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data

    The question asks to calculate (AxBC, where A's magnitude is 5.00, B's magnitude is 4.00, and they are both in the xy-plane. B is 37° counter clockwise from A. C has a magnitude of 6.00 and is in the +z-direction.

    2. Relevant equations

    (A×B) = ABsinθ = D; D·C = DCcosσ = R

    3. The attempt at a solution

    (A×B) = 5×4×sin37° = 12.363 = D. D·C = 12.363×6.00×sin90° = 72.2 = R which is the correct answer. But doesn't the dot in (A×BC imply a scalar product in which case the last part should be 12.363×6.00×cos90°?
     
  2. jcsd
  3. Jan 9, 2013 #2

    gneill

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    Does the cross product yield a vector or a scalar?
     
  4. Jan 9, 2013 #3
    You mean to say that [itex]\vec{A} \times \vec{B} = (5)(4)(sin37°) \widehat{n}[/itex], where n-hat is the unit vector orthogonal to both vectors.
     
  5. Jan 9, 2013 #4

    haruspex

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    ... and don't forget to use the right-hand rule to get the direction of the product in relation to the z axis sign.
     
  6. Jan 9, 2013 #5
    I don't know. It just asks to calculate (A×BC with the values that I gave above.
     
  7. Jan 9, 2013 #6
    We haven't seen that n yet, so I don't think we are suppose to use it in this question.
     
  8. Jan 9, 2013 #7

    Dick

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    Mandelbroth just means n to be the direction that AxB points. You have to know that to find the angle between AxB and C. How is the direction AxB points related to the directions A and B point?
     
  9. Jan 9, 2013 #8
    It would be in the positive direction, but isn't A×B perpindicular to C either way? And why does the answer seem to use sin90° instead of cos90°; doesn't the function (·) mean a scalar product?
     
  10. Jan 9, 2013 #9

    Dick

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    What do you mean by positive direction? Which positive? C points in the positive z direction. Which positive direction does AxB point? I don't think they are perpendicular. If the answer says sin(90), it really shouldn't. That's misleading. The answer should contain a cos. cos of what?
     
  11. Jan 9, 2013 #10

    haruspex

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    By definition, AxB is perpendicular to both A and B. What does that tell you about its direction in relation to C?
     
  12. Jan 9, 2013 #11
    Ah, I got it now. Oh ya, and the answer did not have sin90°, I just fudged that in there to try to make sense of it all, but I realize now that it is cos0° - thanks.
     
  13. Jan 9, 2013 #12
    Thanks, I go it now.
     
  14. Jan 9, 2013 #13

    haruspex

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    OK, but just to check... why cos(0) and not cos(180 degrees)?
     
  15. Jan 10, 2013 #14
    [Deleted]
     
    Last edited: Jan 10, 2013
  16. Jan 10, 2013 #15

    CAF123

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    @Puky The question haruspex asked was directed towards the OP to check his understanding.
     
  17. Jan 10, 2013 #16
    My mistake, I thought it was the OP who asked that question.
     
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