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Finding the z component of vectors that form a triangle?

  1. Mar 27, 2013 #1
    For the vectors in the figure, with a = 1.1 and b = 2.6, what are (a) the z component of a x b, (b) the z component of a x c, and (c) the z component of b x c?


    Everything I've tried to look up involves vectors that are in unit notation, etc. I just don't understand how to do it when all you have for the vector is one number.
     

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  2. jcsd
  3. Mar 27, 2013 #2
    With the information given, you can work out the angle between a and c.
    Then use the formula for the cross product which uses the magnitudes and angle.
     
  4. Mar 28, 2013 #3
    So using a x b = |a| |b| sinθ told me that the z component of a x b = (1.1)(2.6)sin90 = 2.9.

    And for b x c = (2.6)(2.823)sin(90-67.07)=2.9.

    These were both correct, but then when I tried to do a x c = (1.1)(2.823)sin67.06 = 2.9 this was the wrong answer.

    Am I missing something completely obvious?

    Thank you for your help!
     
  5. Mar 28, 2013 #4

    haruspex

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    That should be |a x b| = |a| |b| sinθ. If you want a x b, not just its magnitude, you need to worry about direction. Yes, it's in the z direction, but is it positive or negative? You need to apply the convention for a x b (as distinct from b x a) to determine that.
    From the diagram, you have a + b + c = 0. So 0 = a x (a + b + c) = a x a + a x b + a x c = a x b + a x c. It follows that a x b and a x c must have opposite signs.
     
  6. Mar 28, 2013 #5
    Perfect THANK you that was certainly what I was missing.
     
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