A) Vector A
The x-component of a vector can be found by using the formula
![v_x = v cos \theta](https://tex.z-dn.net/?f=v_x%20%3D%20v%20cos%20%5Ctheta)
where
v is the magnitude of the vector
is the angle between the x-axis and the direction of the vector
- Vector A has a magnitude of 50 units along the positive x-direction, so
. So its x-component is
![A_x = A cos \theta_A = (50) cos 0^{\circ}=50](https://tex.z-dn.net/?f=A_x%20%3D%20A%20cos%20%5Ctheta_A%20%3D%20%2850%29%20cos%200%5E%7B%5Ccirc%7D%3D50)
- Vector B has a magnitude of 120 units and the direction is
(negative since it is below the x-axis), so the x-component is
![B_x = B cos \theta_B = (120) cos (-70^{\circ})=41](https://tex.z-dn.net/?f=B_x%20%3D%20B%20cos%20%5Ctheta_B%20%3D%20%28120%29%20cos%20%28-70%5E%7B%5Ccirc%7D%29%3D41)
So, vector A has the greater x component.
B) Vector B
Instead, the y-component of a vector can be found by using the formula
![v_y = v sin \theta](https://tex.z-dn.net/?f=v_y%20%3D%20v%20sin%20%5Ctheta)
Here we have
- Vector B has a magnitude of 50 units along the positive x-direction, so
. So its y-component is
![A_y = A sin \theta_A = (50) sin 0^{\circ}=0](https://tex.z-dn.net/?f=A_y%20%3D%20A%20sin%20%5Ctheta_A%20%3D%20%2850%29%20sin%200%5E%7B%5Ccirc%7D%3D0)
- Vector B has a magnitude of 120 units and the direction is
, so the y-component is
![B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7](https://tex.z-dn.net/?f=B_y%20%3D%20B%20sin%20%5Ctheta_B%20%3D%20%28120%29%20sin%20%28-70%5E%7B%5Ccirc%7D%29%3D-112.7)
where the negative sign means the direction is along negative y:
So, vector B has the greater y component.