Question

Two vectors A⃗ and B⃗ have magnitude A = 2.93 and B = 3.00. Their vector product is A⃗ ×B⃗ = -4.95k^ + 1.94 i^. What is the angle between A⃗ and B⃗ ?

Answer 1

Answer:

Angle is 37.21 degree

Step-by-step explanation:

Given Data

|A|=2.93

|B|=3.00

A×B=-4.95k+1.94i

Angle=?

Solution

|A×B|=|A|×|B|Sinα

First for |A×B|

|A×B|=$\sqrt{x^{2}+y^{2} }$

|A×B|=$\sqrt{(-4.95)^{2}+(1.94)^{2} }$

|A×B|=5.316

|A×B|=|A|×|B|Sinα

Sinα=$\frac{|A*B|}{|A||B|}$

Sinα=$\frac{5.316}{8.79}$

α=37.21 degree