Please help me. I can't seem to get the answer of 165 degrees.
Find the angle ϴ between the vectors.
u=3i-3j, v=-2i+2√3j
The magnitude of u is sqrt 18.
The magnitude of v is 4
The dot product of u and v is
u*v = -6 -6sqrt3
= -6(1+sqrt3)
The cosine of the angle between the vectors is
(u*v)/|u||v|
To find the angle between two vectors u and v, you can use the dot product formula. The dot product of two vectors u and v is defined as the magnitude of u multiplied by the magnitude of v, multiplied by the cosine of the angle between them.
The dot product formula is given as: u · v = |u| |v| cos(ϴ)
In this case, to find the angle ϴ between vectors u=(3i-3j) and v=(-2i+2√3j), we need to calculate their dot product.
First, let's find the magnitudes of the vectors:
|u| = √(3^2 + (-3)^2) = √(18) = 3√2
|v| = √((-2)^2 + (2√3)^2) = √(4 + 12) = √(16) = 4
Now, we can calculate the dot product of u and v:
u · v = (3i-3j) · (-2i+2√3j)
= 3(-2) + (-3)(2√3) [since i · i = j · j = -1]
= -6 - 6√3
= -6(1 + √3)
Next, plug in the values into the dot product formula to find the cosine of the angle:
-6(1 + √3) = 3√2 * 4 * cos(ϴ)
-6(1 + √3) = 12√2 * cos(ϴ)
Now, solve for cos(ϴ):
cos(ϴ) = (-6(1 + √3)) / (12√2)
cos(ϴ) = (-3(1 + √3)) / (6√2)
cos(ϴ) = -(1 + √3) / (2√2)
cos(ϴ) = -1/2 - √3/2
To find the actual angle ϴ, we need to take the arccosine of cos(ϴ):
ϴ = arccos(-1/2 - √3/2)
Using a calculator, we can find:
ϴ ≈ 165.19 degrees
So, the angle ϴ between the vectors u and v is approximately 165.19 degrees.