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How to calculate the end point of a vector?

  1. Aug 11, 2012 #1
    Knowing the initial point of a vector (X1,Y1) and its magnitude and angle (R,θ)
    HOW CAN I CALCULATE ITS FINAL POINT (X2,Y2)?

    like I know
    (X2-X1)^2 = R^2 - (Y2-Y1)^2
    tanθ= [Y2-Y1]/[X2-X1]
     
  2. jcsd
  3. Aug 11, 2012 #2

    micromass

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    Re: Vector

    The last equation gives us

    [tex]Y_2=Y_1+\tan(\theta)(X_2-X_1)[/tex]

    Now plug this value for [itex]Y_2[/itex] in the equation

    [tex](X_2-X_1)^2 + (Y_2-Y_1)^2=R^2[/tex]

    and solve for [itex]X_2[/itex] (you will get two values, why?, which value you want depends on [itex]\theta[/itex]).
     
  4. Aug 11, 2012 #3

    Fredrik

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    Re: Vector

    Hint: Do you also know what (Y2-Y1)/R and (X2-X1)/R are?

    Edit: I wonder how many times I've made a post only to find that micromass replied a minute earlier. :smile:
     
  5. Aug 11, 2012 #4

    micromass

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    Re: Vector

    You snooze, you lose :tongue2:

    But your to this question answer is way better than mine!! I didn't even think of something like that.
     
  6. Aug 11, 2012 #5
    thanks everyone. I think i thought of a faster way, like using:
    (x2-x1)= r cosθ
    (y2-y1)= r sinθ
     
  7. Aug 11, 2012 #6

    Fredrik

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    That's exactly what I suggested. :smile:
     
  8. Aug 11, 2012 #7
    lol fair enough :p the x/r and y/r thing didn't work in my mind that fast...

    Now about the question to "why"
    prolly because a vector showing to the -x or +x makes no difference for measuring its magnitude (r=+/- sqrt[Δx^2 + Δy^2, however - has no meaning, for you define r>=0). it's a problem of θ to solve.
     
  9. Aug 12, 2012 #8
    An interesting variation of this problem is if θ is the bearing from North as used by surveyors and navigators.
     
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