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