Prelim1_fall05soluti - Math 192 Prelim 1 Solution 1 a In order for the lines to intersect there must be a point of intersection at which x1 = x2 so

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Math 192, Prelim 1. Solution September 29, 2005 1) a) In order for the lines to intersect, there must be a point of intersection at which x 1 = x 2 , so 3 + 2 t =15 7 s , t =6 7 s/ 2. The point of intersection also must have y 1 = y 2 ,so 3+3 t = 2+ s . Substituting t =6 7 t/ 2 and solving yields s =2 , t = 1. With these values of t and s Fnd z 1 = 1= z 2 . So there is a point of intersection, and its coordinates are (1 , 0 , 1). b) Let the plane be given by A ( x x o )+ B ( y y o )+ C ( z z 0 ) = 0. The point of intersection is a point in the plane, so can take ( x o ,y o ,z o )=(1 , 0 , 1). The direction of L 1 is given by ~v 1 =2 ~ i +3 ~ j +2 ~ k ,thato f L 2 by ~v 2 = 7 ~ i + ~ j +3 ~ k . The normal to the plane is ~v 1 × ~v 2 = ± ± ± ± ± ± ~ i ~ j ~ k 232 713 ± ± ± ± ± ± =7 ~ i 20 ~ j +23 ~ k .So A =7 , B = 20 and C = 23. The equation for the plane is 7( x 1) 20( y 0) + 23( z + 1) = 0, or 7 x 20 y +23 z = 16
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This note was uploaded on 06/01/2008 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).

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Prelim1_fall05soluti - Math 192 Prelim 1 Solution 1 a In order for the lines to intersect there must be a point of intersection at which x1 = x2 so

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