exam1sol_s07

exam1sol_s07 - MATH 2400 CALCULUS 3 MIDTERM 1 I have...

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Unformatted text preview: MATH 2400: CALCULUS 3 February 14, 2007 MIDTERM 1 I have neither given nor received aid on this exam. Name: 001 E. Kim . . . . . . . . . . . . . . . . (9am) 002 E. Angel . . . . . . . . . . . . .(10am) 003 I. Mishev . . . . . . . . . . . . (11am) 004 J. Boisvert . . . . . . . . . . (12am) 005 A. Gorokhovsky . . . . . . (1pm) If you have a question raise your hand and remain seated. In order to receive full credit your answer must be complete , legible and correct . Show all of your work, and give adequate explanations. DO NOT WRITE IN THIS BOX! Problem Points Score 1 20 pts 2 20 pts 3 20 pts 4 20 pts 5 20 pts 6 20 pts 7 20 pts 8 20 pts TOTAL 160 pts 1. (9 pt.) Let v = h 1 , 1 , 1 i , b = h 1 , , 1 i . Find the orthogonal projection of v onto b . proj b v = v · b || b || 2 b = h 1 , 1 , 1 i · h 1 , , 1 i ||h 1 , , 1 i|| 2 h 1 , , 1 i = 2 ( √ 2 ) 2 h 1 , , 1 i = h 1 , , 1 i 2. (9 pt.) Find an equation of the plane passing through (- 1 , , 1) and perpendicular to the line h x, y, z i = h 5 t, 1 + t,- t i . Let n = h 5 , 1 ,- 1 i (a direction vector for the line). Then n will serve as a normal vector for our plane, since our plane is to be perpendicular to the line. This gives an equation for the plane to be 5( x + 1) + 1( y- 0)- 1( z- 1) = 0, which simplifies to 5 x + y- z + 6 = 0....
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exam1sol_s07 - MATH 2400 CALCULUS 3 MIDTERM 1 I have...

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