wo vectors A⃗ and B⃗ have magnitude A = 3.02 and B = 3.03. Their vector product is A⃗ ×B⃗ = -5.05k^ + 2.06 i^. What is the angle between A⃗ and B⃗ ?
To find the angle between vectors A⃗ and B⃗, we can use the formula:
A⃗ · B⃗ = |A⃗| |B⃗| cos(θ)
Where A⃗ · B⃗ is the dot product of A⃗ and B⃗, |A⃗| and |B⃗| are the magnitudes of A⃗ and B⃗ respectively, and θ is the angle between the two vectors.
Since we are given the magnitudes of A⃗ and B⃗, we can substitute the values into the formula:
-5.05 = 3.02 * 3.03 * cos(θ)
Now we can solve for the angle θ:
cos(θ) = -5.05 / (3.02 * 3.03)
Using a calculator, we can evaluate the cosine:
cos(θ) = -0.557
To find the angle θ, we can take the inverse cosine (or arccos) of -0.557:
θ = arccos(-0.557)
Using a calculator, we find:
θ ≈ 128.79 degrees
Therefore, the angle between vectors A⃗ and B⃗ is approximately 128.79 degrees.
To find the angle between two vectors, A⃗ and B⃗, you can use the dot product formula:
A⃗ · B⃗ = |A⃗ | * |B⃗ | * cos(θ)
where A⃗ · B⃗ is the dot product of A⃗ and B⃗, |A⃗ | and |B⃗ | are the magnitudes of A⃗ and B⃗, respectively, and θ is the angle between A⃗ and B⃗.
In this case, the dot product is not given, but the vector product A⃗ × B⃗ is provided as -5.05k^ + 2.06i^. However, we can still find the angle using the fact that the vector product is related to the magnitudes and the sine of the angle between the two vectors:
|A⃗ × B⃗ | = |A⃗ | * |B⃗ | * sin(θ)
Given that |A⃗ × B⃗ | is -5.05, |A⃗ | is 3.02, and |B⃗ | is 3.03, we can rearrange the equation to solve for sin(θ):
sin(θ) = |A⃗ × B⃗ | / (|A⃗ | * |B⃗ |)
sin(θ) = -5.05 / (3.02 * 3.03)
Now, we can use the inverse sine function (arcsine) to find the angle θ:
θ = arcsin(-5.05 / (3.02 * 3.03))
Evaluating this expression, we find that the angle θ is approximately -0.401 radians or -22.98 degrees.
Therefore, the angle between A⃗ and B⃗ is approximately 22.98 degrees.