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102f05fin - B U Department of Mathematics Math 102 Calculus...

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B U Department of Mathematics Math 102 Calculus II Fall 2005 Final This archive is a property of Bo˘ gazi¸ ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1. (a) Find an equation for the plane through A (0 , 0 , 1) , B (2 , 0 , 0) , and C (0 , 3 , 0). (b) Find the cosine of the angle between this plane and xy-plane. (c) Find the area of the triangle ABC. Solution: (a) We find a vector normal to the plane and use it with one of the points (it doesn’t matter which) to write an equation for the plane. -→ AB = < 2 - 0 , 0 - 0 , 0 - 1 > = < 2 , 0 , - 1 > -→ AC = < 0 - 0 , 3 - 0 , 0 - 1 > = < 0 , 3 , - 1 > The cross product -→ AB × -→ AC = i j k 2 0 - 1 0 3 - 1 = 3 i + 2 j + 6 k is normal to the plane. By using components of this vector and coordinates of the point (0,0,1) we can write the equation of the plane as 3( x - 0) + 2( y - 0) + 6( z - 1) = 0 3 x + 3 y + 6 z = 6 (b) The angle between two intersecting planes is defined to be the (acute) angle determined by their normal vectors. The vectors n 1 = < 0 , 0 , 1 > and n 2 = < 3 , 2 , 6 > are normal vectors for xy-plane and the plane 3 x + 2 y + 6 z = 6 , respectively. To find the cosine of the angle
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