ch03-p067 - 67. The three vectors given are a = 5.0 b = 2.0...

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67. The three vectors given are ˆˆ ˆ 5.0 i 4.0 j 6.0 k ˆ 2.0 i 2.0 j 3.0 k ˆ 4.0 i 3.0 j 2.0 k a b c =+ =− + + + G G G (a) The vector equation rabc =−+ G GG G is ˆ [5.0 ( 2.0) 4.0]i (4.0 2.0 3.0)j ( 6.0 3.0 2.0)k ˆ =11i+5.0j 7.0k. r −+ +− + + + G (b) We find the angle from + z by “dotting” (taking the scalar product) G r with # k. Noting that 22 2 = | | = (11.0) + (5.0) + ( 7.0) = 14, rr G Eq. 3-20 with Eq. 3-23 leads to () ( ) k 7.0 14 1 cos 120 . r φφ ⋅= − = ¡ G G (c) To find the component of a vector in a certain direction, it is efficient to “dot” it (take the scalar product of it) with a unit-vector in that direction. In this case, we make the desired unit-vector by 2 ˆ 2.0i+2.0j+3.0k ˆ . || 2.0 (2.0) (3.0) b b b == + G G We therefore obtain ( ) ( ) 2 5.0 2.0 4.0 2.0 6.0 3.0 ˆ 4.9 . 2.0 (2.0) (3.0) b aa b + =⋅= = + G (d) One approach (if all we require is the magnitude) is to use the vector cross product, as the problem suggests; another (which supplies more information) is to subtract the result
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ch03-p067 - 67. The three vectors given are a = 5.0 b = 2.0...

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