20075ee1_1_HW1

# 20075ee1_1_HW1 - CHAPTER 1 1.1 Given the vectors M =-10ax...

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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 , 0 . 36 , 0 . 14) b) the magnitude of 5 a x + N 3 M : (5 , 0 , 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 = 0 . 27 a x 0 . 14 a y + 0 . 95 a z where factors of 1 / 2 have cancelled. c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the unit vector is therefore parallel to AC . First we find AC = 2 a x + a y 7 a z , which we recognize as 7 . 35 a MM . The vectors are thus parallel (but oppositely-directed).

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