M341F10HW1 - M341, F10, Homework 1 Section 1.1 1. b) [ 5 ,...

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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 Newtons law of mechanics: f = 30 a = 180 14[ 2 , 3 , 1] = [ 96 . 21 , 144 . 32 , 48 . 10] . Section 1.2 1. b) cos = 13 / 13 66 = 0 . 44 . 64 , d) 180 ( ). 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 x-axis, while the second one, bx + d = 0 is parallel to the y-axis. 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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M341F10HW1 - M341, F10, Homework 1 Section 1.1 1. b) [ 5 ,...

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