Given two non-zero vectors A and B, how do you find the angle between them? Take, for example, A = (2, -1, 2), B = (1, -1, 0) and find the angle between them.
To find the angle between two vectors A and B, you can use the dot product formula:
A · B = |A| |B| cosθ
where A · B is the dot product of A and B, |A| and |B| are the magnitudes of A and B respectively, and θ is the angle between them.
Let's calculate it using the given vectors A = (2, -1, 2) and B = (1, -1, 0):
1. Find the magnitudes of A and B:
|A| = sqrt((2)^2 + (-1)^2 + (2)^2) = sqrt(9) = 3
|B| = sqrt((1)^2 + (-1)^2 + (0)^2) = sqrt(2)
2. Calculate the dot product of A and B:
A · B = (2)(1) + (-1)(-1) + (2)(0) = 2 + 1 + 0 = 3
3. Substitute the values into the dot product formula to find the angle:
3 = (3)(sqrt(2))cosθ
4. Solve for θ:
cosθ = 3 / (3√2) = 1 / √2 = √2 / 2
Now, to find θ, we take the inverse cosine (arc cosine) of this value:
θ = cos^(-1)(√2 / 2) ≈ 45°
Therefore, the angle between vectors A = (2, -1, 2) and B = (1, -1, 0) is approximately 45°.
To find the angle between two vectors A and B, you can use the dot product formula and trigonometry.
1. Calculate the dot product of vectors A and B using the formula: A · B = |A| |B| cos(theta), where |A| and |B| are the magnitudes of vectors A and B, respectively, and theta is the angle between them.
Let's calculate the dot product of vectors A and B:
A · B = (2 * 1) + (-1 * -1) + (2 * 0) = 2 + 1 + 0 = 3.
2. Calculate the magnitudes of vectors A and B using the formula: |A| = sqrt(Ax^2 + Ay^2 + Az^2) and |B| = sqrt(Bx^2 + By^2 + Bz^2), where Ax, Ay, Az are the components of vector A and Bx, By, Bz are the components of vector B.
Let's calculate the magnitudes of vectors A and B:
|A| = sqrt(2^2 + (-1)^2 + 2^2) = sqrt(4 + 1 + 4) = sqrt(9) = 3.
|B| = sqrt(1^2 + (-1)^2 + 0^2) = sqrt(1 + 1 + 0) = sqrt(2).
3. Substitute the dot product and magnitudes into the dot product formula to solve for cos(theta):
3 = (3)(sqrt(2)) cos(theta).
Simplify: cos(theta) = 3 / (3sqrt(2)) = 1 / sqrt(2) = sqrt(2) / 2.
4. Use the inverse cosine (arccos) function to find theta:
theta = arccos(sqrt(2) / 2) = pi / 4 radians.
Therefore, the angle between vectors A = (2, -1, 2) and B = (1, -1, 0) is pi / 4 radians or 45 degrees.