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Easy problem that I am stuck on...
Two vectors A and B have magnitude of A = 3.02 B=3.00. Their vector product is AXB = -4.90k + 2.01i. What is the angle between vector A and vector B?
since I know the vector product of A and B, I was thinking that Cx = 2.01 and Cz = -4.90. I would then add them together to get -2.89. Then could i use the equation C = ABsin(theta). That would be like
-2.89 = (3.02)(3.00)sin(theta)
-.319=sin(theta)
-18.6 = theta
I don't know if this is the right path to follow but this is what I could come up with.
THANKS for any help.
Homework Statement
Two vectors A and B have magnitude of A = 3.02 B=3.00. Their vector product is AXB = -4.90k + 2.01i. What is the angle between vector A and vector B?
Homework Equations
since I know the vector product of A and B, I was thinking that Cx = 2.01 and Cz = -4.90. I would then add them together to get -2.89. Then could i use the equation C = ABsin(theta). That would be like
-2.89 = (3.02)(3.00)sin(theta)
-.319=sin(theta)
-18.6 = theta
I don't know if this is the right path to follow but this is what I could come up with.
The Attempt at a Solution
THANKS for any help.