Question: Find the angle between the vectors A = (3, -2, 1) and B = (1, 1, 1) if A · B = |A||B|cos(θ).
Options:
Correct Answer: 60°
Solution:
A · B = 3*1 + (-2)*1 + 1*1 = 3 - 2 + 1 = 2. |A| = √(3^2 + (-2)^2 + 1^2) = √14, |B| = √3. cos(θ) = 2/(√14 * √3). θ = 60°.