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Math241FinalAnswSp01 - → n 2 = | → n 1 || → n 2 | cos...

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Answers to Math 241 Final, Spring 2001 Part I: (1) (a) 1 , - 4 , 5 (b) - 1 (c) 2 , 3 , 2 (2) (a) - 57 / 5 (b) 3 , 12 or 1 , 4 or 1 / 17 , 4 / 17 (3) (0 , 1 , 0) (0 , - 1 , 0) (4) 0 (5) Maximum 1 / 4 Minimum - 1 / 4 (6) 36 π 3 / 5 (7) (a) saddle (b) not a critical point (c) local mininimum Part II: (1) - 1 / 2 (2) (a) (2 - t - s ) 2 + ( - 4 - t - s ) 2 + (3 - 3 s ) 2 (b) t = - 2 and s = 1 (c) (3 , 1 , 2) and (0 , 4 , 2) (3) (a) 61 π/ 3 (b) 32 / 5 (4) (a) Let n 1 = 1 , - 2 , - 1 and n 2 = a, b, c . Then n 1 is perpendicular to P and n 2 is perpendicular to ax + by + cz = d . The angle θ between n 1 and n 2 is either 60 or 120 . Deduce the given equation from n 1 ·
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Unformatted text preview: → n 2 = | → n 1 || → n 2 | cos θ . (b) Use that a 2 + b 2 + c 2 = 6 in the equation in part (a). (c) Suppose ax + by + cz = d is P or P 00 so that Q and R are on ax + by + cz = d . Then → QR = h 1 , 1 ,-1 i is perpendicular to → n 2 in part (a). From → n 2 · → QR = 0, deduce a + b-c = 0. (d) (-1 ,-1 ,-2) and (2 ,-1 , 1) (e) x + y + 2 z = 9 and 2 x-y + z = 12...
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