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Winter 2007 - Takeda's Class - Exam 1 (Version 1)

# Winter 2007 - Takeda's Class - Exam 1 (Version 1) - Math...

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Math 20C, Midterm 1 Solutions. January 29, 2007 Name : PID : TA : Sec. No : Sec. Time : This exam consists of 6 pages including this front page. Ground Rules 1. No calculator is allowed. 2. Show your work for every problem. A correct answer without any justification will receive no credit. 3. You may use one 4-by-6 index card, both sides. 4. You are not allowed to leave your seat if the remaining time is less than 5 minutes. Score 1 10 2 10 3 10 4 10 5 10 Total 50 1

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1. (a) Find a symmetric equation of the line passing through the point (1 , 1 , 1) and parallel to the line whose symmetric equation is x - 2 - 1 = y - 6 5 = z +10. x - 1 - 1 = y - 1 5 = z - 1 . (b) Find the number a so that the vector a, 1 , 2 is orthogonal to the normal vector of the plane given by - x + 3 y + 2 z = 10. We must have - a + 3 + 4 = 0 . So a = 7 . 2
2. Let a and b be 3 dimensional vectors. Show that a × b = - b × a . Let a = a 1 , a 2 , a 3 and b = b 1 , b 2 , b 3 . Then a × b = a 2 b 3 - a 3 b 2 , - ( a 1 b 3 - a 3 b 1 ) , a 1 b 2 - a 2 b 1 - b × a = - b 2 a 3 - b 3 a 2 , - ( b 1 a 3 - b 3 a 1 ) , b 1 a 2 - b 2 a 1 = - b 2 a 3 + b 3 a 2 , b 1 a 3 - b 3 a 1 , - b 1 a 2 + b 2 a 1 = a 2 b 3 - a 3 b 2 , - ( a 1 b 3 - a 3 b 1 ) , a 1 b 2 - a 2 b 1 . Thus a × b = - b × a . 3

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3. Find a vector with magnitude 5 which is orthogonal to the plane x + 3 y - 2 z + 11 = 0 Let a = 1 , 3 , - 2 . Then a is orthogonal to the plane, because it is a normal

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