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Unformatted text preview: Math 53, Solutions to the Review Problems for the First Midterm Sections 210, 213; GSI: Ivan Mati c 1. n 1 = ( 1 , 1 , 1 ) , n 2 = ( 2 , 1 , 1 ) . n 1 n 2 = ( , 3 , 3 ) and the point (2 , 4 , 0) belongs to both planes. Hence the intersection line is given by the parametric equations: x = 2 y = 4 + 3 t z = 3 t Since the dot product of these two vectors is 0, the angle is 90 . 2. v (0) = ( , , 1 ) and the line is x = 1, y = 0, z = t . Problem 2 Problem 3 Problem 4 3. Let p ( x, y, z ) be the equation of our point. Its distance from the point (0 , , 1) is equal to radicalbig x 2 + y 2 + ( z + 1) 2 and its distance from the plane z = 1 is equal to  z 1  . Hence the equation is  z 1  = radicalbig x 2 + y 2 + ( z + 1) 2 . After simplifying we get 4 z = x 2 + y 2 . That is elliptic paraboloid. 4. First we draw the picture as shown above, and then we see that the required area is integraldisplay 5 / 4 / 4 1 2 r 2 d = integraldisplay 5 / 4...
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This note was uploaded on 03/09/2010 for the course PHYSICS 7A taught by Professor Lanzara during the Fall '08 term at University of California, Berkeley.
 Fall '08
 Lanzara
 Physics

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