How to calculate the end point of a vector?

[Morgoth]

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]

 

[micromass]

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

$$Y_2=Y_1+\tan(\theta)(X_2-X_1)$$

Now plug this value for $Y_2$ in the equation

$$(X_2-X_1)^2 + (Y_2-Y_1)^2=R^2$$

and solve for $X_2$ (you will get two values, why?, which value you want depends on $\theta$).

 

[Fredrik]

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.

 

[micromass]

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

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.

 

[Morgoth]

thanks everyone. I think i thought of a faster way, like using:
(x2-x1)= r cosθ
(y2-y1)= r sinθ

 

[Fredrik]

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.

 

[Morgoth]

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.

 

[Studiot]

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.