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Unformatted text preview: CHAPTER 1 1.1. Given the vectors M = − 10 a x + 4 a y − 8 a z and N = 8 a x + 7 a y − 2 a z , find: a) a unit vector in the direction of − M + 2 N . − M + 2 N = 10 a x − 4 a y + 8 a z + 16 a x + 14 a y − 4 a z = (26 , 10 , 4) Thus a = (26 , 10 , 4)  (26 , 10 , 4)  = (0 . 92 , . 36 , . 14) b) the magnitude of 5 a x + N − 3 M : (5 , , 0) + (8 , 7 , − 2) − ( − 30 , 12 , − 24) = (43 , − 5 , 22), and  (43 , − 5 , 22)  = 48 . 6 . c)  M  2 N  ( M + N ):  ( − 10 , 4 , − 8)  (16 , 14 , − 4)  ( − 2 , 11 , − 10) = (13 . 4)(21 . 6)( − 2 , 11 , − 10) = ( − 580 . 5 , 3193 , − 2902) 1.2. The three vertices of a triangle are located at A ( − 1 , 2 , 5), B ( − 4 , − 2 , − 3), and C (1 , 3 , − 2). a) Find the length of the perimeter of the triangle: Begin with AB = ( − 3 , − 4 , − 8), BC = (5 , 5 , 1), and CA = ( − 2 , − 1 , 7). Then the perimeter will be ` =  AB  +  BC  +  CA  = √ 9 + 16 + 64+ √ 25 + 25 + 1 + √ 4 + 1 + 49 = 23 . 9 . b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side BC : The vector from the origin to the midpoint of AB is M AB = 1 2 ( A + B ) = 1 2 ( − 5 a x +2 a z ). The vector from the origin to the midpoint of BC is M BC = 1 2 ( B + C ) = 1 2 ( − 3 a x + a y − 5 a z ). The vector from midpoint to midpoint is now M AB − M BC = 1 2 ( − 2 a x − a y + 7 a z ). The unit vector is therefore a MM = M AB − M BC  M AB − M BC  = ( − 2 a x − a y + 7 a z ) 7 . 35 = − . 27 a x − . 14 a y + 0 . 95 a z where factors of 1 / 2 have cancelled....
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This test prep was uploaded on 04/01/2008 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.
 Fall '08
 Joshi
 Electrical Engineering

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