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20esolns2

# 20esolns2 - 20E HW 2 Solutions 1.3.24 Answer The plane goes...

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20E HW 2 Solutions January 22, 2008 1.3.24) Answer The plane goes through the point (2 , - 1 , 3) and has normal vector 3 , - 2 , 4 . Thus the equation to the plane is 3( x - 2) - 2( y + 1) + 4( z - 3) = 0 . 1.3.33) Answer D = | 12(1) + 13(1) + 5( - 5) + 2 | 12 2 + 13 2 + 5 2 = 2 338 = 2 13 1.4.1(a)) Answer Cylindrical Standard Spherical (1 , π / 4 , 1) ( 2 / 2 , 2 / 2 , 1) ( 2 , π / 4 , π / 4) (2 , π / 2 , - 4) (0 , 2 , - 4) (2 5 , π / 2 , π - cos - 1 (2 5 / 2)) (0 , π / 4 , 10) (0 , 0 , 10) (10 , π / 4 , 0) (3 , π / 6 , 4) , (3 3 / 2 , 3 / 2 , 4) (5 , π / 6 , cos - 1 (4 / 5)) (1 , π / 6 , 0) , ( 3 / 2 , 1 / 2 , 0) (1 , π / 6 , π / 2) (2 , 3 π / 4 , - 2) ( - 2 , 2 , - 2) (2 2 , 3 π / 4 , 3 π / 4) 1.4.2) Answer a. Reflects across the x - y plane. b. Reflects through the origin. c. Rotates clockwise around the z by π / 4 , then reflects through the z -axis. (A perhaps easier answer is that it simply rotates counterclockwise by 3 π / 4 . 1

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1.4.3) Answer a. Rotation by π around the z axis. b. Reflection across the x - y plane. c. Rotation by π / 2 around the z axis; scale the radius by a factor of 2 . 1.4.4) a. When r = constant the surface is a cylinder of radius r about the origin. When θ = constant
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20esolns2 - 20E HW 2 Solutions 1.3.24 Answer The plane goes...

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