Find the angle 𝜃 (in degrees) between the vectors. (Round your answer to two decimal places.)
u = 6i − 4j
v = −4i − 4j
To find the angle 𝜃 between the vectors u and v, we can use the dot product formula:
u • v = ||u|| ||v|| cos(𝜃)
First, let's find the magnitudes of the vectors:
||u|| = √(6^2 + (-4)^2) = √(36 + 16) = √52 = 2√13
||v|| = √((-4)^2 + (-4)^2) = √(16 + 16) = √32 = 4√2
Next, let's find the dot product of u and v:
u • v = 6(-4) + (-4)(-4) = -24 + 16 = -8
Now plug these values back into the dot product formula:
-8 = 2√13 * 4√2 * cos(𝜃)
-8 = 8√26 * cos(𝜃)
cos(𝜃) = -8 / 8√26 = -1 / √26 = -√26 / 26
Now, use the arccos function to find 𝜃:
𝜃 = arccos(-√26 / 26) ≈ 131.81 degrees
Therefore, the angle 𝜃 between the vectors u and v is approximately 131.81 degrees.
You answer is correct, except for the last part. The arccos of -√26 / 26 is approximately 101.31, not 131.81.
Apologies for the error in my response. Thank you for pointing it out.
Therefore, the correct angle 𝜃 between the vectors u and v is approximately 101.31 degrees.