Vector Subtraction: A^{→} - B^{→}

[defiledxhalo]

Homework Statement

For the vectors $A^{→}$ and $B^{→}$, calculate the vector difference $A^{→}$ - $B^{→}$. Magnitude of vector $A^{→}$ is 12 meters, with an angle of 180°. Magnitude of vector $B^{→}$ is 18 meters, with an angle of 37°.

Homework Equations

$A{y}$ = Asinθ; $B{y}$ = Bsinθ
$A{x}$ = Acosθ; $B{x}$ =Bcosθ
Resultant vector = $\sqrt{Rx^{2} + Ry^{2}}$

The Attempt at a Solution

I know not providing a graph might make this problem a bit more difficult. I just really desperately need help on how to calculate vector subtraction because I'm not sure if I'm doing it right.

I found that the x-component of vector $A^{→}$ is -12 meters and the y-component is 0 meter. The x-component of vector $B^{→}$ is 14.4 meters and the y-component is 10.8 meters. From that, the $R{x}$ would be 2.4 meters and $R{y}$ would be 10.8 meters.

If I were to just do vector $A^{→}$ + $B^{→}$, I know how to calculate that. I would use the Resultant vector formula $\sqrt{Rx^{2} + Ry^{2}}$, which would give me of R = 11.06. But if I'm doing what this problem is doing, I don't know if that's the right formula to use.

I understand that $A^{→}$ - $B^{→}$ is the same thing as -$B^{→}$ + $A^{→}$. I was wondering how the negative part translated into the calculation. For example, do I make vector $B^{→}$'s x-component negative (to -14.4 meters), and have the $R{x}$ = -26.4 meters and the $R{y}$ = -10.8 meters (because vector $B^{→}$'s y-component would then be -10.8 mters)? And if all that is correct, would I carry on with the same resultant vector formula $\sqrt{Rx^{2} + Ry^{2}}$, plugging in the numbers to get R = 28.5 meters?

Thank you so much for helping me!

 

[ehild]

The problem asks the vector difference, not only the magnitude of the difference vector. The components of the resultant difference vector are Ax-Bx and Ay-By.

ehild