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Calculating the phase of the sum of two vectors

  1. May 1, 2009 #1
    Hi,

    suppose I have two vectors, one in the i direction and the other in j direction, i is horizontal and j vertical.

    The first vector Vi = 5i and the second vector Vj = 2j.

    If we sum this vectos it results in one vector that have module and phase.

    I found the module by Vij = ( 5^2 + 2^2 ) ^ (1/2) = 5.3852.

    But what about the phase?

    I think it was atan(y/x), but in case y=0 or x=0, it let me in troubles.
     
  2. jcsd
  3. May 1, 2009 #2

    dx

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    Your vector is 5i + 2j, so x = 5 and y = 2.
     
  4. May 1, 2009 #3
    Hi dx,

    I want to know how to calculate the angle(phase) between 5i and 2j...
     
  5. May 1, 2009 #4

    HallsofIvy

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    The vector, with tail at (0,0) and head at (5, 2) forms a right triangle with "opposite side" 2 and "near side" 2 so [itex]tan(\theta)= 2/5[/itex], so [itex]\theta= arctan(2/5)[/itex] and lies between 0 and [itex]\pi/2[/itex] since the components are both positive.

    Yes, The vector xi+ yj has angle given by arctan(y/x) (with the quadrant determined by the sign of y and x). If x= 0, then obviously the vector lies along the y axis so [itex]\theta= \pi/2[/itex] or [itex]\theta= 3\pi/2[/itex] depending on whether y is positive or negative.
     
    Last edited: May 2, 2009
  6. May 1, 2009 #5
    When I have only 2j, what is the phase?

    -pi/2 or pi/2

    ?

    Why?
     
  7. May 2, 2009 #6

    HallsofIvy

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    [itex]\pi/2[/itex]. The "phase angle" is measured counter-clockwise from the positive [itex]\vec{i}[/itex] axis. [itex]\pi/2[/itex] points along the positive y-axis while [itex]-\pi/2[/itex] points along the negative y-axis.
     
  8. May 2, 2009 #7
    Thanks for your reply HallsofIvy,

    I have one more question,

    If I have [tex]-5i[/tex], so the phase vector is [tex]\pi[/tex], correct?

    And in case I have [tex]5i[/tex], so the phase vector is [tex]-\pi[/tex], or zero?
     
  9. May 2, 2009 #8

    HallsofIvy

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    NO! As I said before, the "phase" angle of 5i is [itex]\pi/2[/itex] and the "phase" angle of -5i is [itex]-\pi/2[/itex], not "[itex]\pi[/itex]" or "[itex]-\pi[/itex]".
     
  10. May 3, 2009 #9
    Hi HallsofIvy,

    can you provide me some referece, so I can study this by my self.

    I fill I need to study.

    If you can provide online reference it was better.

    Thanks
     
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