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HW-1-solution

# HW-1-solution - HW-1 solution 1 For the two vectors A x 2 y...

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HW-1 solution 1. For the two vectors z y x A 2 and z y x B 3 2 , find (a) B A and B A , (b) component of B along A , (c) angle between A and B , (d) B A , (e) B A B A . Solution: (a) z y x z y x z y x B A 2 3 ) 3 2 ( ) 2 ( 74 . 3 ) 2 ( ) 1 ( 3 | | 2 2 2 B A (b) 22 . 1 ) 1 ( 2 1 1 ) 1 ( 3 2 ) 2 ( 1 | | 2 2 2 A A B (c) 327 . 0 1 3 ) 2 ( ) 1 ( 2 1 3 | | | | cos 2 2 2 2 2 2 B A B A , =cos -1 (0.327)=70.89 o (d) z y x 7 5 1 3 2 - 1 - 2 1 z y x B A (e) z y x B A A B B A B A B B A A B A B A 14 2 10 2 ) ( ) ( ) ( ) ( . 2. A , B , and C are three vectors pointing from the origin of a coordinate to three points A, B, and C, respectively. Find the distance between the origin to the ABC plane and the area of the ABC triangle. Express your result in terms of A , B , and C vectors. Solution: The ABC plane can be defined by the two vectors C A and C B . Thus, the surface normal direction is along vector A C C B B A B C C A B A C B C A

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