problem01_82

# problem01_82 - z-direction From Eq(1-30 with the...

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1.82: a) The angle between the vectors is , 140 70 210 = - and so Eq. (1.18) gives ( 29 ( 29 2 m 62 . 6 140 cos m 40 . 2 m 60 . 3 - = = B A Or, Eq. (1.21) gives ( 29 ( 29 ( 29 ( 29 2 m 62 . 6 210 sin m 4 . 2 70 sin m 6 . 3 210 cos m 4 . 2 70 cos m 60 . 3 - = + = + = y y x x B A B A B A b) From Eq. (1.22), the magnitude of the cross product is ( 29 ( 29 2 m 55 . 5 140 sin m 40 . 2 m 60 . 3 = , and the direction, from the right-hand rule, is out of the page (the
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Unformatted text preview: + z-direction). From Eq. (1-30), with the z-components of B A and vanishing, the z-component of the cross product is ( 29 ( 29 ( 29 ( 29 2 m 55 . 5 210 cos m 40 . 2 70 sin m 60 . 3 210 sin m 40 . 2 70 cos m 60 . 3 =-=- x y y x B A B A...
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