How Do You Calculate the Angle Between Two Vectors?

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In summary, the conversation discusses finding the angle between two vectors A and B using their vector product, AXB, which has been given as -4.90k + 2.01i. The individual components of AXB, 2.01 and -4.90, were used to calculate the magnitude of C, which is equal to the magnitude of A multiplied by the magnitude of B and multiplied by the sine of the angle between them. However, it was later realized that the inverse sine (arcsin) should have been used instead of the sine in the equation.
  • #1
azila
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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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  • #2
it's
[tex]\left| c\right| =\left| a\right| \,\left| b\right| \,sin\left( p\right) [/tex]
where c = a x b
 
  • #3
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.
 
  • #4
get the absolute value of c.
c^2=x^2+y^2+z^2
 
  • #5
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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