ps3sol - 18.02 Fall 2008 Problem Set 3 Part B Solutions 1 a...

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18.02 Fall 2008 – Problem Set 3, Part B Solutions 1. a) The line from E to P has the direction of EP = ( x 0 1 ,y 0 ,z 0 ) and goes through ( 1 , 0 , 0 ) , so it can be parametrized as x = 1 + ( x 0 1) t, y = y 0 t, z = z 0 t. It intersects the yz -plane when x = 0, i.e. 1 + ( x 0 1) t = 0, which gives t = 1 1 x 0 . So y = y 0 1 x 0 , z = z 0 1 x 0 . (We assume x 0 < 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 yz -plane. 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 yz -plane, 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 0 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 . Taking the limit as t → ∞ , the position on the screen approaches (5 , 4) (even though the actual position of the bird in 3D space is further and further away).
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