exam1s4soln

exam1s4soln - Math 241 Sections 01* Exam 1 Sample 4...

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Math 241 Sections 01** Exam 1 Sample 4 Solutions Dr. Justin O. Wyss-Gallifent 1. Given the following data: P = ( - 1 , 0 , 3) Q = (2 , 5 , 5) ¯ a = 1ˆ ı + 2 ˆ + 0 ˆ k ¯ b = 3ˆ ı + 2 ˆ + 1 ˆ k (a) We use: 2 −→ PQ || −→ PQ || = 2 p ı + 5 ˆ + 2 ˆ k 9 + 25 + 4 P (b) We have: ¯ a · ¯ b = || ¯ a |||| ¯ b || cos θ 7 = 1 + 4 + 0 9 + 4 + 1 cos θ cos θ = 7 5 14 (c) We have: Pr ¯ a ¯ b = ± ¯ a · ¯ b ¯ a · ¯ a ² ¯ a = ± 7 5 ² (1ˆ ı + 2 ˆ + 0 ˆ k )
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2. (a) The line has ¯ L = 3ˆ ı + 0 ˆ + 1 ˆ k and P = ( - 2 , - 2 , 3). We have Q = (3 , 2 , 1) of the line. ThereFore −→ PQ = 5ˆ ı + 4 ˆ - 2 ˆ k and so the distance is: Dist = || ¯ L × −→ PQ || || ¯ L || = || - ı + 11 ˆ + 12 ˆ k || 9 + 1 + 0 = 16 + 121 + 144 9 + 1 + 0 (b) We have ¯ r ( t ) = - sin( t ı + 3 cos( t ) ˆ ¯ T ( π/
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exam1s4soln - Math 241 Sections 01* Exam 1 Sample 4...

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