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Unformatted text preview: 126Chapter 1 1.87.IDENTIFY:Compare the magnitude of the cross product, sinABφ, to the area of the parallelogram. SET UP:The two sides of the parallelogram have lengths Aand B. φis the angle between A"and B". EXECUTE:(a) The length of the base is Band the height of the parallelogram is sinAφ, so the area is sinABφ. 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°. EVALUATE:It is useful to consider the special cases φ=°, where the area is zero, and 90φ=°, where the parallelogram becomes a rectangle and the area is AB. 1.88.IDENTIFY:Use Eq.(1.27) for the components of the vector product. SET UP:Use coordinates with the axisx+to the right, axisy+toward the top of the page, and axisz+out of the page. xA=, yA=and 3.50 cmzA= −. The page is 20 cm by 35 cm, so 20 cmxB=and 35 cmyB=. EXECUTE:( )( )( )22122 cm ,70 cm ,0.xyz==−=A & BA & BA & B""""""EVALUATE:From the components we calculated the magnitude of the vector product is 2141 cm . 40.3 cmB=and 90φ=°, so 2sin141 cmABφ=, which agrees. 1.89.IDENTIFY:A"and B"are given in unit vector form. Find A, Band the vector difference .−A B""SET UP:2.003.004.00 ,= −++Aijk""""3.001.003.00=+−Bijk""""Use Eq.(1.8) to find the magnitudes of the vectors. EXECUTE:(a)222222( 2.00)(3.00)(4.00)5.38xyzAAAA=++=−++=222222(3.00)(1.00)( 3.00)4.36xyzBBBB=++=++−=(b)&&&&&&( 2.003.004.00 )(3.001.003.00 )−= −++−+−A Bijkijk""&&&&&&( 2.00 3.00)(3.00 1.00)(4.00 ( 3.00))5.002.007.00 .−= −−+−+− −= −++A Bijkijk""(c)Let ,=−CA B"""so 5.00,xC= −2.00,yC= +7.00zC= +222222( 5.00)(2.00)(7.00)8.83xyzCCCC=++=−++=(),−= −−BAAB""""so −A B""and −BA""have the same magnitude but opposite directions. EVALUATE:A, Band Care each larger than any of their components. 1.90.IDENTIFY:Calculate the scalar product and use Eq.(1.18) to determine φ. SET UP:The unit vectors are perpendicular to each other. EXECUTE:The direction vectors each have magnitude 3 , and their scalar product is ( )( ) ( )( ) ( )( )1 111111,+−+−=2so from Eq. (1.18) the angle between the bonds is 11arccosarccos109 .33 3−⎛⎞⎛⎞=−=°⎜⎟⎜⎟⎝⎠⎝⎠EVALUATE:The angle between the two vectors in the bond directions is greater than 90°. 1.91.IDENTIFY:Use the relation derived in part (a) of Problem 1.92: 2222cos ,CABABφ=++where φis the angle between A"and B". SET UP:cosφ=for 90φ=°. cosφ<for 90180φ<<°°and cosφ>for 090φ<<°°. EXECUTE:(a) If 222, cos 0,CABφ=+=and the angle between A"and B"is 90°(the vectors are perpendicular). (b)If 222, cos0,CABφ<+<and the angle between A"and B"is greater than 90°. (c)If 222, cos0,CABφ>+>and the angle betweenA"and B"is less than 90 .°EVALUATE:It is easy to verify the expression from Problem 1.92 for the special cases φ=, where CA B=+, and for 180φ=°, where CA B=−....
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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