Answer to Question #230594 in Mechanics | Relativity for Jakie Abedin

Question

Two vectors A→ and B→ have magnitude A = 2.90 and B = 3.02. Their vector product is A×B = -4.96k^ + 2.04 i^. What is the angle between A and B?

Answer 1

$\sin\alpha=\frac{|\vec A×\vec B|}{{\vec A}\cdot \vec B}=\frac{\sqrt{(-4.96)^2+(2.04)^2}}{2.90\cdot 3.02}=0.6123,$

$\alpha=37.8°.$