26_PartUniversity Physics Solution

26_PartUniversity Physics Solution - 1-26 1.87 1.88 Chapter...

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1-26 Chapter 1 1.87. I DENTIFY : Compare the magnitude of the cross product, sin AB φ , to the area of the parallelogram. S ET U P : The two sides of the parallelogram have lengths A and B . φ is the angle between A " and B " . E XECUTE : (a) The length of the base is B and the height of the parallelogram is sin A φ , so the area is sin AB φ . This equals the magnitude of the cross product. (b) The cross product A° B " " is perpendicular to the plane formed by A " and B " , so the angle is 90 ° . E VALUATE : It is useful to consider the special cases 0 φ = ° , where the area is zero, and 90 φ = ° , where the parallelogram becomes a rectangle and the area is AB . 1.88. I DENTIFY : Use Eq.(1.27) for the components of the vector product. S ET U P : Use coordinates with the -axis x + to the right, -axis y + toward the top of the page, and -axis z + out of the page. 0 x A = , 0 y A = and 3.50 cm z A = − . The page is 20 cm by 35 cm, so 20 cm x B = and 35 cm y B = . E XECUTE : ( ) ( ) ( ) 2 2 122 cm , 70 cm , 0. x y z = = − = A ° B A ° B A ° B " " " " " " E VALUATE : From the components we calculated the magnitude of the vector product is 2 141 cm . 40.3 cm B = and 90 φ = ° , so 2 sin 141 cm AB φ = , which agrees. 1.89. I DENTIFY : A " and B " are given in unit vector form. Find A , B and the vector difference . A B " " S ET U P : 2.00 3.00 4.00 , = − + + A i j k " " " " 3.00 1.00 3.00 = + B i j k " " " " Use Eq.(1.8) to find the magnitudes of the vectors. E XECUTE : (a) 2 2 2 2 2 2 ( 2.00) (3.00) (4.00) 5.38 x y z A A A A = + + = + + = 2 2 2 2 2 2 (3.00) (1.00) ( 3.00) 4.36 x y z B B B B = + + = + + − = (b) ° ° ° ° ° ° ( 2.00 3.00 4.00 ) (3.00 1.00 3.00 ) = − + + + A B i j k i j k " " ° ° ° ° ° ° ( 2.00 3.00) (3.00 1.00) (4.00 ( 3.00)) 5.00 2.00 7.00 . = − + + − − = − + + A B i j k i j k " " (c) Let , = C A B " " " so 5.00, x C = − 2.00, y C = + 7.00 z C = + 2 2 2 2 2 2 ( 5.00) (2.00) (7.00) 8.83 x y z C C C C = + + = + + = ( ), = − B A A B " " " " so A B " " and B A " " have the same magnitude but opposite directions. E VALUATE : A , B and C are each larger than any of their components. 1.90. I DENTIFY : Calculate the scalar product and use Eq.(1.18) to determine φ . S ET U P : The unit vectors are perpendicular to each other. E XECUTE : The direction vectors each have magnitude 3 , and their scalar product is ( )( ) ( )( ) ( )( ) 1 1 1 1 1 1 1 , + + = 2 so from Eq. (1.18) the angle between the bonds is 1 1 arccos arccos 109 . 3 3 3 = = ° E VALUATE : The angle between the two vectors in the bond directions is greater than 90 ° . 1.91. I DENTIFY : Use the relation derived in part (a) of Problem 1.92: 2 2 2 2 cos , C A B AB φ = + + where φ is the angle between A " and B " . S ET U P : cos 0 φ = for 90 φ = ° . cos 0 φ < for 90 180 φ < < ° ° and cos 0 φ > for 0 90 φ < < ° ° . E XECUTE : (a) If 2 2 2 , cos 0, C A B φ = + = and the angle between A " and B " is 90 ° (the vectors are perpendicular).
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