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PHYS218_CHP01_51-60

# PHYS218_CHP01_51-60 - 1.51 a From Eq(1.21 A B b A B AB cos...

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1.51: a) From Eq.(1.21), ( 29 ( 29 ( 29 ( 29 . 00 . 14 00 . 2 00 . 3 00 . 5 00 . 4 = - + = B A b) ( 29 ( 29 [ ] ( 29 . 7 . 58 .5195 arccos 39 . 5 00 . 5 00 . 14 arccos so , cos AB = = × = = θ θ B A 1.52: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle φ as . arccos arccos + = = AB B A B A AB y y x x B A φ In the intermediate calculations given here, the significant figures in the dot products and in the magnitudes of the vectors are suppressed. a) , 13 , 40 , 22 = = - = B A B A and so 165 13 40 22 arccos = - = φ . b) , 136 , 34 , 60 = = = B A B A 28 136 34 60 arccos = = φ . c) . 90 , 0 = = φ B A 1.53: Use of the right-hand rule to find cross products gives (a) out of the page and b) into the page. 1.54: a) From Eq. (1.22), the magnitude of the cross product is ( 29 ( 29 ( 29 2 m 130 37 180 sin m 0 . 18 m 0 . 12 = - The right-hand rule gives the direction as being into the page, or the – z -direction.

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PHYS218_CHP01_51-60 - 1.51 a From Eq(1.21 A B b A B AB cos...

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