Question

DUPLICATE WITH ATTACHMENT
Vector A⃗ has a magnitude of 3.00 and is directed parallel to the negative y-axis and vector B⃗ has a magnitude of 3.00 and is directed parallel to the positive y-axis. Determine the magnitude and direction angle (as measured counterclockwise from the positive x-axis) of vector C⃗ , if C⃗ =A⃗ −B⃗ .
C = 6.00; θ = 270˚
C = 3.00; θ = 270˚
C = 6.00; θ = 90˚
C = 3.00; θ = 90˚

Answer 1

C=A-B, according to the information, the vector A⃗ has a magnitude of 3.00 and is directed parallel to the negative y-axis: that means Vector OA (0, -3)

 and vector B⃗ has a magnitude of 3.00 and is directed parallel to the positive y-axis. 

that means Vector OB (0, 3), so C=A – B means  vectOC = vect OA  -  vect OB =(0,-3)-(0,3)=(0,-6), so the magnitude of OA is sqrt((-6)^2)=6, so C=6 

and according to the information (as measured counterclockwise from the positive x-axis), so the answer is C = 6.00; θ = 90˚