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exam1s3soln

# exam1s3soln - Math 241 Exam 1 Sample 3 Solutions ^ 1(a We...

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Math 241 Exam 1 Sample 3 Solutions 1. (a) We have ¯ PQ = 2ˆ ı - 2 ˆ - 1 ˆ k and to make it length 1 we: ¯ PQ || ¯ PQ || = ı - 2 ˆ - 1 ˆ k 4 + 4 + 1 (b) We need ( α ˆ ı - 2 ˆ + α ˆ k ) · (2ˆ ı + 5 ˆ ) = 0 2 α - 10 = 0 α = 5 (c) We have Pr ¯ b ¯ a = ¯ b · ¯ a ¯ b · ¯ b ¯ b = 2 + 10 1 + 4 + 9 (1ˆ ı + 2 ˆ + 3 ˆ k )

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2. (a) The plane has ¯ N = 2ˆ ı + 3 ˆ - 1 ˆ k and a point is P = (2 , 0 , 0) (any points satisfying the equation). Then with Q = (3 , 2 , 1) we have ¯ PQ = 1ˆ ı + 2 ˆ + 1 ˆ k and so d = | ¯ N · ¯ PQ | || ¯ N || = | 2 + 6 - 1 | 4 + 9 + 1 (b) First: ¯ r ( t ) = t ˆ ı + sin t ˆ ¯ v ( t ) = 1ˆ ı + cos t ˆ ¯ a ( t ) = 0ˆ ı - sin t ˆ Then || ¯ a × ¯ v || = sin t ˆ k and so κ ( t ) = || ¯ a × ¯ v || || ¯ v || 3 κ ( t ) = sin 2 t (1 + cos 2 t ) 3 / 2 κ ( π/ 2) = 1 (1 + 0) 3 / 2
3. (a) The graph is: Start=(3,-2,0) End=(-3,-2,0) (b) The parabolic part is ¯ r ( t ) = t ˆ ı + t 2 ˆ for - 1 t 2. The straight part is ¯ r ( t ) = (2 - 3 t ı + (4 - 3 t ) ˆ for 0 t 1.

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4. (a) The vector is ¯ L = 3ˆ ı - 4 ˆ - 2 ˆ k and so using the first point we have x - 1 3 = y - 2 - 4 = z - 3 - 2 (b) Start with: ¯ a ( t ) = 2ˆ ı ¯ v ( t ) = integraldisplay ¯ a ( t ) dt = 2 t ˆ ı + ¯ C ¯ v (1) = 2ˆ ı + ¯ C = 2ˆ ı + ˆ + ˆ k ¯ C = ˆ + ˆ
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