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Direction of Vectors

  1. Aug 29, 2004 #1
    Vector A has components Ax=1.50cm, Ay =2.05cm
    Vector B has components Bx=4.40cm, By=-3.65cm

    Find the Direction of Vector A + B. (Let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise from the axis)

    Hi, can someone help me with this problem. i know that it's easy, but i dont remeber how to do it. thanks
  2. jcsd
  3. Aug 29, 2004 #2
    Vectors add just like integers. Add the x components and the y components. Draw yourself a picture for it to make more sense.

    Once you have your Cx and Cy, you can use your choise of arctan, arcsin with corresponding sides to find the angle.
    Last edited: Aug 29, 2004
  4. Aug 29, 2004 #3
    yea i know how to add vectors, but i dont know what the question is asking me to do. "Find the Direction of Vector A + B. (Let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise from the axis)" i'm very confused, what is the question asking me to do extactly? sorry, i really suck when it comes to physics.
  5. Aug 29, 2004 #4


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    You need to find the angle the vector C=A+B encloses with the x axis. Remember that
    [tex] tan(\gamma ) = c_y/c_x[/tex]
    From this, you get gamma back, but decide if the angle is in the first, second, third or fourth quadrant.
    If both components are positive
    [tex] 0 < \gamma < \pi/2 [/tex]
    if the x component is negative and the y component is positive
    [tex] \pi/2 < \gamma < \pi[/tex]
    If both components are negative
    [tex]\pi <\gamma <3 \pi /2[/tex]
    if the x component is positive and the y component is negative
    [tex] 3\pi /2 < \gamma < 2 \pi [/tex]

    So calculate:
    [tex] c_x = a_x + b_x [/tex] and [tex] c_y = a_y + b_y [/tex]
    [tex]tan(\gamma ) = c_y/c_x [/tex]
    find gamma, you get an angle between 0 and [tex] \pi [/tex] (or between 0 and [tex]180^o [/tex]). Check if [tex] c_y [/tex] is positive or negative. If it is negative add [tex]\pi (180^o ) [/tex] to the value of the angle you got.

    Last edited: Jun 29, 2010
  6. Aug 29, 2004 #5
    Not sure what class this is for but I remember for my first physics class we would write a vector as C [[itex]\theta[/itex] deg. above positive x-axis]
  7. Aug 29, 2004 #6


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    This is actually a problem in mathematics. Don't blame Physics! :mad: :tongue2:

    After doing the addition of vectors, as described in the other posts here, you are essentially converting from rectangular coordinates to polar coordinates.
    This is probably in your trig or pre-calc text.

    In physics, one merely uses that mathematical idea.
  8. Aug 29, 2004 #7

    ok here's what i got:
    [tex] c_x = 5.9cm [/tex]
    [tex] c_y = -1.6cm [/tex]

    [tex](\gamma ) = -15.17291 [/tex]

    the angle is negative, so i added 180 to it, which equals .... 164.827 <-- is that the answer?
  9. Aug 29, 2004 #8
    Add 360, not 180. Remember, the A+B vector is in the 4th quad so to get the CCW angle from the x-axis you have go through the first three quads which is 270 degrees. And then you need to go an additional 75'ish degrees to find the angle from the x-axis to the vector. Or the easier route of 360+[itex]\gamma[/itex]
  10. Aug 29, 2004 #9


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    Ooops...Sorry, I was wrong, faust9 is right.
    The calculator returns an angle between [tex]-90^o [/tex] and [tex]+90^o[/tex]. Add 180 degrees if [tex]c_x <0[/tex] but 360 degrees if [tex] c_x>0 \mbox{ and } c_y<0[/tex].
    So your angle is [tex]344.8^o[/tex] Remember, you have to check the sign of both components, it is not enough to check if the angle is negative or positive.

    Try the vectors {1,1); (-1,1); (-1,-1); (1,-1). They make the angles with the positive x axis counter-clockvise [tex] 45^o, 135^o, 225^o, 315^0 [/tex]. The tangents are: 1, -1, 1, -1. The function arctan returns 45, -45, 45 -45. You have to add 0, 180, 180, 360, respectively, to get back the original angles.

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