How Do You Calculate the Angle Between Two Vectors?

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To calculate the angle between vectors A and B, the formula used is θ = arccos[(A * B) / (magnitude of A * magnitude of B)]. The magnitudes of vectors A and B were correctly calculated as 4.58 and 3.16, respectively. The dot product of vectors A and B is -10, leading to the calculation θ = arccos(-10 / 14.47), which results in an angle of approximately 133.7 degrees. The user initially struggled with the arithmetic but ultimately found the correct solution after re-evaluating their calculations. This experience highlights the importance of careful computation in vector mathematics.
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Homework Statement



Let vectors \vec{A}=(2,1,-4) \vec{B}=(-3, 0, 1), and \vec{C}= (-1, -1, 2) .
Calculate the following:

What is the angle\theta_{AB} between \overline{A} and \overline{B} ?

Homework Equations



\vec{A} * \vec{B}= magnitude of A * Magnitude of B * Cos\theta

magnitide of X= \sqrt{a^{2}+b^{2}+C^{2}}

The Attempt at a Solution



magnitude of A= 4.58
Magnitude of B= 3.16

\theta=arccos[( A * B)/ magnitude of A * magnitude of B)

I tried that but the incorrect answer. Maybe i did the arithmetic wrong. Please Help..
 
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I can't tell you if you did the arithmetic wrong if you don't show your calculations...

Also, it seems you posted this problem in the forum 3 times. Please do not flood the forum with repeat posts.
 
sorry if i posted it three times. It is my first time using this program.

magnitude of a= \sqrt{2^{2}+ 1^{2}+ 4^{2}}= 4.58

Magnitude of b= \sqrt{-3^{2}+ 1^{2}}= 3.16

and the vector A * the Vector B gives me -10
 
Last edited:
OK, you calculations for the magnitudes are correct. Can you show me what you get for \theta and the full calculation involved?

I ask because you have the correct formula, so I think the error is probably in your calculation somewhere.
 
a ha! i just saw my mistake! But i will show you my calculations that i just re-did


\theta_{AB}= arccos[ -10/ (4.58* 3.16)]

= arccos(-10/ 14.47)
= 133.7

Thanks a lot!
 
Yahaira.Reyes said:
a ha! i just saw my mistake! But i will show you my calculations that i just re-did


\theta_{AB}= arccos[ -10/ (4.58* 3.16)]

= arccos(-10/ 14.47)
= 133.7

Thanks a lot!

;-)

I can't tell you the number of times I begin to spend the time typing up a problem here, and then all of a sudden I figure out the answer to my question.
 
Good Job!
 
i know i couldn't believe it! I was so frustrated trying the problem out over and over. I lauphed so hard when i figured it out lolz... I thought it was such a hard problem, turned out pretty easy
 
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