solution1 - M341 Homework Solution 1 Section 1.1 - 7(f) : 2...

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Unformatted text preview: M341 Homework Solution 1 Section 1.1 - 7(f) : 2 x + 3 y 4 z = [ 4 , 8 , 10] + [ 3 , , 9] [16 , 4 , 8] = [ 23 , 12 , 11] Section 1.1 - 9 : Let A , B , C denote the points (7 , 3 , 6) , (11 , 5 , 3) and (10 , 7 , 8) respectively. Then by distance formula we have AB = 16 + 4 + 9 = 29 , BC = 1 + 4 + 25 = 30 , AC = 9 + 16 + 4 = 29 Thus AB = AC = BC and triangle ABC is isosceles but not equilateral. Section 1.1 - 19 : First notice that f 1 = 22 [9 , 6 , 2] 81 + 36 + 4 = [18 , 12 , 4] . f 2 = 27 [7 , 4 , 4] 49 + 16 + 16 = [21 , 12 , 12] . Hence the total force f is f = f 1 + f 2 = [39 , , 8] and the acceleration vector a is a = f m = [39 , , 8] 6 = [ 13 2 , , 4 3 ] ( m/sec 2 ) Section 1.1 - 22(a) : Let x = [ x 1 , x 2 , . . . , x n ] be any vector in R n , then x = x 2 1 + x 2 2 + . . . + x 2 n since a for any nonnegative number a ....
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solution1 - M341 Homework Solution 1 Section 1.1 - 7(f) : 2...

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