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Is it a scalar product? I'm kind of lost

  1. Nov 28, 2013 #1

    The Vector A points 17° counterclockwise from the positive x axis. Vector B lues in the first cuadrant of the xy plane. The magnitudes of the cross product and the dot product are the same:
    i.e, |AXB|= |A(times)B|
    What Angle does B make with the positive x axis?



    2. Is ti a scalar product? I'm kind of lost



    3. I was thinking of using cos(σ)= (AtimesB)/ (|A||B|)
     
  2. jcsd
  3. Nov 28, 2013 #2

    tiny-tim

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    Hi Dan350! :smile:

    The dot product and the scalar product (of two vectors) are the same thing. :wink:

    (and are both written A.B not AtimesB)
     
  4. Nov 28, 2013 #3

    CompuChip

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    The magnitude of the cross product is |AxB| = |A| |B| sin(theta)
    The magnitude of the dot product is |A.B| = |A| |B| cos (theta).

    What does this tell you when they are equal?
     
  5. Nov 29, 2013 #4
    Is A condition given in the problem,
    How do I reach this problem?
     
  6. Nov 29, 2013 #5
    Look at what CompuChip is saying.

    The magnitude of the cross product is |A||B|sin(theta). The scalar yielded by the dot product is |A||B|cos(theta).

    You know they are equal. So what does that say about theta?
     
  7. Nov 29, 2013 #6
    I'm not quite sure, does that mean that they are 90° apart?

    or that I can substitute vector "A" bye sin(17)??
     
  8. Nov 29, 2013 #7

    tiny-tim

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    solve the equations!!! :rolleyes:

    show us how you do it :smile:
     
  9. Nov 29, 2013 #8
    So I have

    |AxB|sinθ= |A*B|cosθ

    Well since they are giving that Vector A is 17 counter clockwise form the x axis that meas is Acos(17) and vector B is , well I'm stuck there,, wouldnt that be vector Bsin(73)?
     
  10. Nov 29, 2013 #9

    tiny-tim

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    no, it's …
    ok, you have two equations, and you want to solve for one unknown (the angle, θ) …

    show us how you do it​
     
  11. Nov 29, 2013 #10
    |A| |B| sin(theta)= |A| |B| cos(theta)

    sinθ=cosθ
    sin^-1(cosθ)=θ

    We know cosθ=cos(17)
    so θ= 73
    Am I right?
     
  12. Nov 29, 2013 #11

    tiny-tim

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    where does that come from?? :confused:
    how does that help?

    sinθ=cosθ … if you can't think of a way of solving it, draw a diagram or a graph
     
  13. Nov 29, 2013 #12
    I thought that since is counterclockwise form the x axis,, is going to be cos(17)
    and I solved for theta in "sin^-1(cosθ)=θ"

    I drew it, i only have my Vector A 17 raising counterclockwise form the x axis,, the vector B is inthe same cuadrant

    Any clue?

    Thanks
     
  14. Nov 29, 2013 #13

    tiny-tim

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    no, just draw sinθ = cosθ, and then solve it
     
  15. Nov 29, 2013 #14
    At what angle does sinθ = cosθ, you know that at cos(0) = 1 and sin(0) = 0, and cos(90) = 0, sin(90) = 1, so since they are continuous functions, you can estimate at what θ they will be =.
     
  16. Nov 29, 2013 #15

    haruspex

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    You're confusing two angles.
    The question asks you to find the angle B makes with the positive x axis. The angle θ in |A.B| = |A| |B| cos(θ) and |AxB| = |A| |B| sin(θ) is the angle between A and B. So, first find θ (which will not involve the 17 degrees), then use θ and the 17 degrees to find the angle B makes with the positive x axis.
     
  17. Nov 29, 2013 #16

    So for sinθ=cosθ the angle will be 45°
    Now how do I find angle B using the 17°??
     
  18. Nov 29, 2013 #17
    that's equal to 45°
     
  19. Nov 29, 2013 #18

    tiny-tim

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    correct :smile:

    (btw, the quickest way of solving that would be sinθ = cosθ, so tanθ = 1, so θ = 45°)

    ok, so the angle θ is 45° …

    between what and what is θ the angle?​
     
  20. Nov 29, 2013 #19
    How did I miss that! haha

    and for the other part, isnt between 45° and 17° ?? we are know looking for VEctor B
     
  21. Nov 29, 2013 #20

    tiny-tim

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    i think i know what you mean, but that doesn't actually make sense, does it? :redface:

    an angle is between two lines

    what are the lines? :smile:
     
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