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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˚

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b