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Unformatted text preview: Math 20C  Final (Lecture D, Winter 2007) Duration: 3 hours Please close your books and turn off your phones. You can use one page of handwritten notes. To get full credit, you should support your answers. 1. Let A = (1 , 1 , 2) , B = (3 , 3 , 2) and C = (1 , 3 , 0). a) (3 points) Find the angle θ between the displacement vectors vector AB and vector AC and the area of the triangle ABC . Solution. It is easy to see that vector AB = ( 2 , 2 , ) and vector AC = ( , 2 , − 2 ) . From the properties of the dot product, we get that cos θ = vector AB · vector AC  vector AB  vector AC  = 4 √ 8 √ 8 = 1 2 Thus, cos θ = 1 2 which implies θ = π 3 or 60 degrees. The area of the triangle ABC equals half of the length of the cross product vector AB × vector AC . This cross product is vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vector i vector j vector k 2 2 0 2 − 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = − 4 vector i + 4 vector j − 4 vector k The length of vector AB × vector AC is radicalbig ( − 4) 2 + 4 2 + ( − 4) 2 = 4 √ 3. Thus, the area is 2 √ 3. b) (2 points) Find an unit vector that is normal to the plane formed by the points A, B and C . Find the equation of the plane determined by the points A, B and C . Solution. A vector perpendicular to the plane formed by A, B and C is the vector vector AB × vector AC = (− 4 , 4 , 4 ) . A unit vector perpendicular to the plane formed by A, B and C is obtained by dividing the previous vector by its length. Thus, the answer is (− 1 √ 3 , 1 √ 3 , − 1 √ 3 ) . Also, a correct answer is the vector ( 1 √ 3 , − 1 √ 3 , 1 √ 3 ) . The plane formed by A, B and C is the plane passing through A = (1 , 1 , 2) which is perpendicular on the vector vector AB × vector AC = (− 4 , 4 , − 4 ) . An equation of this plane is − 4( x − 1) + 4( y − 1) − 4( z − 2) = 0 which simplifies to x − y + z = 2. 2. Let f ( x, y ) = radicalbig 9 − ( x − 1) 2 − ( y + 2) 2 . a) (3 points) Find an equation of the tangent plane to the graph of f ( x, y ) at the point (2 , − 4 , 2). Use the tangent plane at (2 , − 4 , 2) to find an approximation for f (1 . 99 , − 3 . 99). Is f (1 . 99 , − 3 . 99) greater or less than your approximation ? Explain your answer. Hint: Use the graph of z = f ( x, y ). Solution. An equation of the tangent plane is z = f (2 , − 4) + f x (2 , − 4)( x − 2) + f y (2 , − 4)( y + 4) Since f x ( x, y ) = − 2( x − 1) 2 √ 9 − ( x − 1) 2 − ( y +2) 2 = − x − 1 √ 9 − ( x − 1) 2 − ( y +2) 2 and f y ( x, y ) = − 2( y +2) 2 √ 9 − ( x − 1) 2 − ( y +2) 2 = − y +2 √ 9 − ( x − 1) 2 − ( y +2) 2 , it follows that f x (2 , − 4) = − 1 2 and...
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This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Fall '08 term at UCSD.
 Fall '08
 Helton
 Math

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