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Vector image on another vector

  1. Mar 1, 2013 #1
    Hello
    I want to demonstrate a equation of Vector image on another vector (A on B)
    gif.latex?A_{B}=\frac{(\bar{A}.\bar{B}).\bar{B}}{|B|^{2}}.gif
    So i go this steps
    gif.gif
    and as we know
    gif.gif
    so change the equation
    a}_{B}|Cos\Theta%20_{AB}=\bar{A}.\hat{a}_{B}=|\bar{A}|.\hat{a}_{B}.\hat{a}_{B}=\frac{(A.\bar{B}).gif
    but in my equation A is (vector scale) not a (Vector)!!!!
    what is my problem?
     
  2. jcsd
  3. Mar 1, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have [itex]\overline{A}[/itex], [itex]A[/itex], and [itex]\left|\overline{A}\right|[/itex]. I understand that [itex]\overline{A}[/itex] is a vector and that [itex]\left|\overline{A}\right|[/itex] is its length but what is [itex]A[/itex]?
     
  4. Mar 1, 2013 #3
    Thanks for your replay
    A and |A| both of them are length of A vector (in electromagnetic books we assume that A=|A| for easy to write)
     
  5. Mar 1, 2013 #4

    pwsnafu

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    Science Advisor

    Then pick one and stick with it. Similarly, don't swap between upper and lower case.
     
  6. Mar 1, 2013 #5
    I rewrite the Steps.
    main formula:
    gif.latex?A_{B}=\frac{%28\bar{A}.\bar{B}%29.\bar{B}}{|B|^{2}}.gif

    steps:
    gif.gif
    as we know
    gif.gif
    and
    gif.latex?\bar{A}.gif
    so
    heta%20_{AB}=\bar{A}.\hat{a}_{B}=|\bar{A}|.\hat{a}_{B}.\hat{a}_{B}=\frac{%28|\bar{A}|.\bar{B}%29.gif

    but in my equation |A| is (vector length) not a (Vector)!!!!
    what is my problem?
     
  7. Mar 2, 2013 #6
    Oh Baby,

    What are you doing? You are taking the dot product of vector A and vector B which results in a scalar. Then you are taking the dot product of that scalar with vector B. That makes no sense.

    I think you want the dot product of vector A and vector B/|B|. B/|B| is the unit vector in the B direction.

    Ratch
     
  8. Mar 3, 2013 #7
    Thanks friends
    i found my probelm
    Image of Vector of A on B is a Vector and we should define the direct of it when write the image equation
    gif.latex?A_{B}=ACos\Theta%20_{AB}.gif
     
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