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?
1
Expert's answer
2011-10-04T08:44:47-0400
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.
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Comments
Leave a comment