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

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

well, since v•w = |v| |w| cosθ,

cosθ =

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

To find the angle θ between two vectors, you can use the dot product formula. The dot product of two vectors v and w is given by the equation:

v ⋅ w = |v| |w| cos(θ)

Where |v| and |w| are the magnitudes of the vectors v and w, respectively, and θ is the angle between them.

To get started, we need to find the magnitudes |v| and |w| of the given vectors v and w.

Magnitude of vector v:
|v| = √((- ((3)^1/2)/2)^2 + (-((3)^1/2))^2)
= √((3/4) + (3/4))
= √(3/2)
= √3/√2
= √3/2

Magnitude of vector w:
|w| = √((4(3)^1/2)^2 + 4^2)
= √(48 + 16)
= √64
= 8

Now that we have the magnitudes, we can calculate the dot product v ⋅ w:

v ⋅ w = (- ((3)^1/2)/2)(4(3)^1/2) + (-((3)^1/2))(4)
= -2 + (-12)
= -14

Using the dot product formula, we can solve for the angle θ:

-14 = (√3/2)(8) cos(θ)
cos(θ) = -14 / (√3 * 8)
cos(θ) = -7 / (4√3)

To find the angle θ, we take the inverse cosine (arccos) of the above value:

θ = arccos( -7 / (4√3) )

Using a calculator, we approximate the value of θ to be approximately 120.96 degrees.