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What is the correct method of finding the direction of a Vector

  1. Sep 15, 2006 #1
    How the direction of a vector is found ?
    In figure "a" we want to find the direction of the vector AB.In order to find the direction of vector AB,why we considered the angle between AB and y-axis.Why we did not considered this angle between AB and x-axis.
    On the other hand the figure ''b".
    Also in second example we want to find the direction of the resultant vector of A and B.While A=3cm and B=4cm.
    Its resultant will be A+B=5cm.
    Now we find the direction of A+B.
    I see in my book that the angle is found between resultant vector (A+B) and A i.e (x-axis) the answer is 53 degree.that is correct.
    but answer of the first example is also correct.
    1) Please tell me what is the correct mathod of finding the direction of a vector.

    Attached Files:

  2. jcsd
  3. Sep 15, 2006 #2

    Andrew Mason

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    The direction of a vector is relative to the origin and the reference frame. The choice of the origin and reference frame is arbitrary. So long as you specify which axis it is relative to, there is no problem.

  4. Sep 15, 2006 #3

    Doc Al

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    Staff: Mentor

    Just to add to Andrew's explanation: It doesn't matter which axis you use to specify the direction of a vector—either one will do. And if you know the angle made with the y-axis, you should be able to figure out the angle made with the x-axis.
  5. Sep 16, 2006 #4
    Please give me another example to explain the vector direction method.
  6. Sep 16, 2006 #5


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    It is irrelevant which from which axis you chose to take the angle, so long as you state this in your answer. For example say you have a 2D vector [itex]l = 3i + 4j[/itex] where i and j are unit vectors in the positive x and y direction respectively. Now, the magnitude of the vector would of course be;

    [tex]\left| l \right| = \sqrt{3^2 + 4^2} = 5[/tex]

    Now, it is up to use from which axis (x or y) we wish to take the angle. You may at this point wish to draw yourself a diagram. First lets take the angle anti-clockwise from the x axis;

    [tex]\theta = \tan^{-1} \left( \frac{4}{3} \right) \approx 53.1^{o}[/tex]

    Therefore, the vector would have a magnitude of 5 acting 53.1o anti-clockwise from the x axis.

    Alternatively, we could take the angle clockwise from the y axis, in which case the angle would be;

    [tex]\theta = 90 - \tan^{-1} \left( \frac{4}{3} \right) \approx 36.9^{o}[/tex]

    And therefore, our vector would still have a magnitude of 5 and is directed 36.9o clockwise from the y axis.
  7. Sep 16, 2006 #6
    Thank you Hootenanny .This example is proved helpful for me.
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