How Do You Calculate the Angle Between Two Vectors?

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To calculate the angle between two vectors A and B with magnitudes of 3.02 and 3.00, and a vector product of AXB = -4.90k + 2.01i, the correct approach involves using the equation |C| = |A| |B| sin(theta). The user initially miscalculated by directly adding the components of the vector product instead of finding its magnitude. The correct magnitude of C should be calculated as c^2 = x^2 + y^2 + z^2, leading to the correct value for |C|. The final realization was that the arcsin function should be used instead of sin to find the angle theta.
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Easy problem that I am stuck on...

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.
 
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it's
\left| c\right| =\left| a\right| \,\left| b\right| \,sin\left( p\right)
where c = a x b
 
so am I solving it incorrectly because i am using the same equation, its just that I don't know if the value of c that I have is correct. Also the answer I get is not right. So, what should I change? thanks for the help though.
 
get the absolute value of c.
c^2=x^2+y^2+z^2
 
Thank You So Much! I Realized What I Was Doing Wrong Was That I Was Taking The Sin When I Should Have Been Taking The Arcsin. Thanks Again!
 
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