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Unformatted text preview: 18.02 Fall 2008 – Problem Set 3, Part B Solutions 1. a) The line from E to P has the direction of EP = ( x − 1 , y , z ) and goes through ( 1 , , ) , so it can be parametrized as x = 1 + ( x − 1) t, y = y t, z = z t. It intersects the yzplane when x = 0, i.e. 1 + ( x − 1) t = 0, which gives t = 1 1 − x . So y = y 1 − x , z = z 1 − x . (We assume x < 1 because otherwise the point P would lie behind the observer). b) The image on the screen of a line segment in space is contained in the intersection of the plane containing E and the line segment with the yzplane. This intersection is a line, therefore the image is a line segment on the screen. c) Using question a), (0 , − 3 , 1) is displayed at ( − 3 , 1) in the yzplane, and ( − 1 , 2 , 5) is displayed at (1 , 5 2 ). So the image on the screen is the line segment from ( − 3 , 1) to (1 , 5 2 ). d) The trajectory will again be a line segment. The velocity of the bird is vectorv = −−−→ P P 1 = (− 1 , 5 , 4 ) , so at time t its position is ( − t, − 3 + 5 t, 1 + 4 t ). By the formula in a), this is displayed at y = 5 t − 3 t + 1 = 5 − 3 t 1 + 1 t , z = 4 t + 1 t + 1 = 4 + 1 t 1 + 1 t ....
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This note was uploaded on 04/28/2009 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.
 Fall '08
 Auroux

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