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Fall calc test

Fall calc test - MATH 162 – FALL 2006 – FIRST EXAM...

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Unformatted text preview: MATH 162 – FALL 2006 – FIRST EXAM SEPTEMBER 18, 2006 SOLUTIONS 1) (4 points) The center and the radius of the sphere given by x 2 + y 2 + z 2 = 4 x + 3 y are A) Center (0 , 3 / 2 , 2) and radius 3 / 2 B) Center (2 , 3 / 2 , 0) and radius 3 / 2 C) Center (2 , 3 / 2 , 0) and radius 5 / 2 D) Center (1 , 2 , 3) and radius 2 / 3 E) Center (2 , 2 / 3 , 1) and radius 5 / 2 Solution: Complete the squares and write x 2 + y 2 + z 2 = 4 x +3 y as ( x − 2) 2 +( y − 3 2 ) 2 + z 2 = 25 4 . So the center of the sphere is (2 , 3 2 , 0) and its radius is 5 2 . Correct answer C. 2) ( 8 points) The point 1 / 4 of the way from (1 , − 3 , 1) and (7 , 9 , − 9) is A) (4 , 3 , − 4) B) (5 / 2 , , − 3 / 2) C) (3 / 2 , 3 , − 3 / 2) D) (3 / 2 , 6 , − 5) E) (11 / 4 , 6 , − 13 / 2) Solution: The segment of line joining the points P 1 (1 , − 3 , 1) and P 2 (7 , 9 , − 9) is given by (1 , − 3 , 1) + t vector P 1 P 2 . But vector P 1 P 2 = < 6 , 12 , − 10 > = (7 , 9 , − 9) − (1 , − 3 , 1) . Taking t = 1 4 , we find the point which is 1 / 4 of the way between the two points. This point is ( 5 2 , , − 3 2 ) . Correct answer B. 3) (8 points) The area of the triangle with vertices ( − 1 , 1 , 1) , (2 , , 2) and (3 , 2 , 2) is A) 3 √ 6 2 B) 5 √ 6 3 1 2 C) 2 √ 3 D) √ 6 E) √ 3 2 Solution: Let P 1 ( − 1 , 1 , 1) , P 2 (2 , , 2) and P 3 (3 , 2 , 2) . These three points give two vectors: vector v 1 = vector P 1 P 2 = ( 3 , − 1 , 1 ) and vector v 2 = vector P 1 P 3 = ( 4 , 1 , 1 ) . The area of the triangle is equal to one half of the area of the parallelogram formed by the vectors. So the area of the triangle is equal to 1 2 | vector v 1 × vector v 2 | . We find that vector v 1 × vector v 2 = (− 2 , 1 , 7 ) . So 1 2 | vector v 1 × vector v 2 | = √ 542 = 3 √ 6 2 . Correct answer A. 4)(8 points) Let vectora = ( − 5 , 4 , 3) and vector b = ( − 1 , − 1 , − 2) . Which one of the following is true?...
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Fall calc test - MATH 162 – FALL 2006 – FIRST EXAM...

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