# Physics I: Vector Subtraction

1. May 12, 2013

### x31fighter

1. The problem statement, all variables and given/known data
C = B - A
A = 16 Units
A-theta = 42 degrees from the y-axis
Ai is pointing in second quadrant

B = 7 units
B-theta = 31 degrees from the x-axis
B is pointing in the third quadrant
2. Relevant equations

3. The attempt at a solution
I am suppose to find the magnitude of C from the components given. When I attempted it, I did it using trigonometric. That got marked wrong. Just a random question, am I able to do it using cross-product or something like that?

Va-x = -16*sin(42)
va-y = 16*cos(42)

Vb-x = - 7.0*sin(31)
Vb-y = - 7.0cos(31)

Did Pythagorean theorem and combined the x and y and got the following result:
C =15.45

2. May 12, 2013

### rude man

C is also a vector, so you need to come up with an angle as well as the correct magnitude.

Va_x and Va_y are correct.
Vb_x and Vb_y are incorrect. 31 deg. is from the x, not the y, axis ...

3. May 13, 2013

### x31fighter

Oops...

I fixed that. I am getting the correct answer now.

4. May 13, 2013

### SteamKing

Staff Emeritus
Vector subtraction has nothing to do with the cross product.