problem01_88

problem01_88 - The angle between B A and is 120 o since one...

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1.88: a) This is a statement of the law of cosines, and there are many ways to derive it. The most straightforward way, using vector algebra, is to assume the linearity of the dot product (a point used, but not explicitly mentioned in the text) to show that the square of the magnitude of the sum B A + is ( 29 ( 29 φ cos 2 2 2 2 2 2 2 AB B A B A + + = + + = + + = + + + = + + B A B A B B A A B B A B B A A A B A B A Using components, if the vectors make angles θ A and B with the x -axis, the components of the vector sum are A cos A + B cos B and A sin A + B sin B , and the square of the magnitude is ( 29 ( 29 2 2 sin sin cos cos B A B A θ B θ A θ B θ A + + + ( 29 ( 29 ( 29 ( 29 cos 2 cos 2 sin sin cos cos 2 sin cos sin cos 2 2 2 2 2 2 2 2 2 2 AB B A θ AB B A θ θ θ θ AB θ θ B θ θ A B A B A B A B B A A + + = - + + = + + + + + = where = A B is the angle between the vectors. b) A geometric consideration shows that the vectors B A , and the sum B A + must be the sides of an equilateral triangle.
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Unformatted text preview: The angle between B A and , is 120 o , since one vector must shift to add head-to-tail. Using the result of part (a), with , B A = the condition is that cos 2 2 2 2 2 A A A A + + = , which solves for 1 = 2 + 2 cos , cos = , 2 1-and = 120 o . c) Either method of derivation will have the angle replaced by 180 o – , so the cosine will change sign, and the result is . cos 2 2 2 AB B A-+ d) Similar to what is done in part (b), when the vector difference has the same magnitude, the angle between the vectors is 60 o . Algebraically, is obtained from 1 = 2 – 2 cos , so cos = 2 1 and = 60 o ....
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This homework help was uploaded on 02/14/2008 for the course PHYS 112 taught by Professor Wheeler during the Spring '08 term at Cornell.

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