# Answer to Question #4270 in Mechanics | Relativity for valeria

Question #4270

Two vectors A and B have magnitude A = 2.96 and B = 3.08. Their vector product is A×B = -4.90k + 2.00i . What is the angle between and A and B?

Expert's answer

Vector product of two vectors A and B is the vector S perpendicular to both A and B and which has magnitude that is equal to the area of

parallelogram build on A and B. (By definition)

Thus, the area of that parallelogram is |S|=|AxB|

On the other hand this area can be obtained as cos(a)|A||B|, where ‘a’ is the angle between A and B.

Then we can get cos(a):

Cos(a) = |AxB|/(|A||B|)

Cos(a) = (4.90^2+2.00^2)^0.5/(2.96*3.08)=0.58051591

A = acos(0.58051591) = 54.51316281 degrees

Thus, the approximate value is 54.5 degrees.

parallelogram build on A and B. (By definition)

Thus, the area of that parallelogram is |S|=|AxB|

On the other hand this area can be obtained as cos(a)|A||B|, where ‘a’ is the angle between A and B.

Then we can get cos(a):

Cos(a) = |AxB|/(|A||B|)

Cos(a) = (4.90^2+2.00^2)^0.5/(2.96*3.08)=0.58051591

A = acos(0.58051591) = 54.51316281 degrees

Thus, the approximate value is 54.5 degrees.

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