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˚

 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˚

well my understanding of this question is they want you to use the y=14.1x -2.3 and put the coordinates they gave you onto the line segment of   y=14.1x -2.3

step-by-step explanation:

c) yes, because of asa or aas