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Unformatted text preview: MATH 241, 1st Examination Solutions and Grading Key Prof. Jonathan Rosenberg Friday, September 30, 2005 1. (15 points, 5 per part) Consider the lines L 1 : x 1 7 = y 2 3 = z, L 2 : x + 6 7 = y + 1 3 = z + 1 , L 3 : x 1 1 = y 2 2 = z, L 4 : x 1 = y 2 2 = z, (a) Are L 1 and L 2 the same line? If not, are they parallel? If not, do they intersect? If they do intersect, do they intersect perpendicularly? Explain. Solution. They are the same line, since they both point in the direction of (7 , 3 , 1) , and they both contain the point (1 , 2 , 0) , since 1 1 7 = 2 2 3 = 0 , 1 + 6 7 = 2 + 1 3 = 0 + 1 = 1 . (b) Are L 1 and L 3 the same line? If not, are they parallel? If not, do they intersect? If they do intersect, do they intersect perpendicularly? Explain. Solution. They intersect, since they both contain the point (1 , 2 , 0) , since 1 1 7 = 2 2 3 = 0 , 1 1 1 = 2 2 2 = 0 . But L 1 points in the directionof (7 , 3 , 1) , and L 3 points in the directionof ( 1 , 2 , 1) . Since (7 , 3 , 1) · ( 1 , 2 , 1) = 7 + 6 + 1 = 0 , the lines intersect perpendicularly. (c) Are L 1 and L 4 the same line? If not, are they parallel? If not, do they intersect? If they do intersect, do they intersect perpendicularly? Explain. Solution. These two lines do not intersect, since if they did, one would have to have z = x 1 7 = y 2 3 = x 1 = y 2 2 1 for some ( x, y, z ) . But x 1 7 = x 1 implies 1 x = 7 x or x = 1 / 8 , z = x 1 7 = 1 / 8 , while y 2 3 = y 2 2 implies y 2 = 0 or y = 2 , hence z = y 2 3 = 0 ....
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This note was uploaded on 09/09/2009 for the course MATH 241 taught by Professor Wolfe during the Spring '08 term at Maryland.
 Spring '08
 Wolfe
 Math, Calculus

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