Find the angle θ between the vectors v = - ((3)^1/2)/2 I +(-((3)^1/2) j and W = 4(3)^1/2 i+ 4 j

Answer 1

vector v is [ √3/2 , -√3] parallel to [ 1 , -2]

vector w = [4√3 , 4] parallel to [ √3 , 1 ]

using dot product definition

√3 - 2 = (√5)(√4)cosØ
cosØ = (√3-2)/(2√5)
Ø = 93.4°

or doing it as is:
cosØ = ( (√3/2)(4√3 - √3(4) )/( (√15/2)(8))
= (6 - 4√3)/(4√15
Ø = 93.4
same as above

Answer 2

well, since v•w = |v| |w| cosθ,
cosθ =

12-4√3 = √(15/4)(8)cosθ
cosθ = 2(3-√3)/√15