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Unformatted text preview: M341, F10, Homework 1 Section 1.1 1. b) [ − 5 , 1 , − 2], d = √ 25 + 1 + 4 = √ 30 2. b) (1 , 1 , 1) + [ − 1 , 4 , 2] = (0 , 5 , 3), 5. b) √ 4 2 + 1 2 + 0 2 + ( − 2) 2 = √ 21 , u = 1 √ 21 (4 , 1 , , − 2), the normalized (unit) vector is shorter since 1 < √ 21. 6. a) [12 , − 16] = 4 / 3 [9 , − 12] , b) [4 , − 14] = − 2 [ − 2 , 7] 7. [b) [2 , , − 6] , d) [ − 5 , 1 , 1]] , f) [ − 23 , 12 , 11]. 19. The acceleration is 6 times a unit vector in the direction of the vector [ − 2 , 3 , 1], say: a = 6 / √ 14[ − 2 , 3 , 1]. So by the Newton’s law of mechanics: f = 30 a = 180 √ 14[ − 2 , 3 , 1] = [ − 96 . 21 , 144 . 32 , 48 . 10] . Section 1.2 1. b) cos θ = 13 / √ 13 √ 66 = 0 . 44 . θ ≃ 64 ◦ , d) 180 ◦ ( π ). 3. a) − ab + ba = 0 and also ab − ba = 0. b) if a or b is 0, the lines are orthogonal since they are parallel to the coordinate axes. For instance if a = 0, the first one, by + c = 0 is parallel to the xaxis, while the second one, bx + d = 0 is parallel to the yaxis. If neither a nor b is 0, we first replace the lines by the parallel lines through the origin without changing the orthogonality question. Say we consider ax + by = 0, and bx − ay = 0. The first line contains the vector [ − b, a ] while the second contains [ a, b ]. From a) these vectors are orthogonal and so are the lines through the origin and as a consequence the given lines that are parallel to them....
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 Spring '08
 HIETMANN
 Contrapositive, normed vector space, Lp space, y∥2

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