How to calculate the end point of a vector?

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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]
 

Answers and Replies

  • #2
22,129
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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]

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]).
 
  • #3
Fredrik
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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:
 
  • #4
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3,297


Edit: I wonder how many times I've made a post only to find that micromass replied a minute earlier. :smile:

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.
 
  • #5
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thanks everyone. I think i thought of a faster way, like using:
(x2-x1)= r cosθ
(y2-y1)= r sinθ
 
  • #6
Fredrik
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thanks everyone. I think i thought of a faster way, like using:
(x2-x1)= r cosθ
(y2-y1)= r sinθ
That's exactly what I suggested. :smile:
 
  • #7
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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.
 
  • #8
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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)?

An interesting variation of this problem is if θ is the bearing from North as used by surveyors and navigators.
 

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