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Exam_solutions_1_(2)

Exam_solutions_1_(2) - Mathematics 241 First Exam Solutions...

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Mathematics 241 First Exam Solutions Dr. Rosenberg Friday, February 28, 2003 1. a) Show that the points P 1 = ( - 1 , - 2 , 6) , P 2 = (0 , 4 , 1) , P 3 = (1 , 0 , 1) determine a unique plane P , and determine the equation of P . Solution: To show they determine a unique plane, it’s enough to show that the line segments P 1 P 2 and P 1 P 3 are not parallel. But ( P 2 - P 1 ) × ( P 3 - P 1 ) = P 2 × P 3 - P 1 × P 3 + P 2 × P 1 = (4 , 1 , - 4) + ( - 26 , 2 , - 4) - (2 , 7 , 2) = ( - 20 , - 5 , - 10) = - 5(4 , 1 , 2) 6 = (0 , 0 , 0) . So the points are not collinear and (4 , 1 , 2) is perpendicular to P . Then the equation of P is (4 , 1 , 2) · ( x, y, z ) = (4 , 1 , 2) · ( - 1 , - 2 , 6) or 4 x + y + 2 z = 6 . b) Find the area of the triangle with vertices P 1 , P 2 , and P 3 . Solution: 1 2 ( P 2 - P 1 ) × ( P 3 - P 1 ) = 5 2 (4 , 1 , 2) = 5 2 21 . 2. (10 points) Let P 1 and P 2 be the planes with equations 3 x - 4 y + z = 2 , 3 x + 2 y - z = 7 . Find symmetric equations of the line L 1 where P 1 and P 2 intersect. You may use the fact that (3 , - 4 , 1) × (3 , 2 , - 1) = (2 , 6 , 18). Solution: L 1 is perpendicular to the normals to both P 1 and P 2 , hence is parallel to (3 , - 4 , 1) × (3 , 2 , - 1) = (2 , 6 , 18) = 2(1 , 3 , 9) .

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