ps2A-solns-2009

ps2A-solns-2009 - 18.02A pset 2A part II solutions fall...

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Unformatted text preview: 18.02A pset 2A, part II solutions, fall 2009 Problem 1 Direction vector for line =---→ P 1 P 2 = h 2 , 2 ,- 1 i ⇒ line is ( x,y,z ) = (2 t,- 1 + 2 t, 1- t ) or x = 2 t , y =- 1 + 2 t , z = 1- t . Distance to origin = p x 2 + y 2 + z 2 = p 4 t 2 + (- 1 + 2 t ) 2 + (1- t ) 2 . It’s easier to minimize the distance squared: f ( t ) = 4 t 2 + (- 1 + 2 t ) 2 + (1- t ) 2 . Look for critical points: f ( t ) = 8 t + 4(- 1 + 2 t )- 2(1- t ) = 18 t- 6 = 0 ⇒ t = 1 / 3. f 00 ( t ) = 18 > ⇒ the critical point is a minimum. (This is also obvious geometrically.) ⇒ the minimum distance is at (2 / 3 ,- 1 / 3 , 2 / 3) and the minimum distance is 1. Problem 2 a) Position of jet = h x,y,z i = h 1 , 1 , i + t h- 5 , , 1 i = h 1- 5 t, 1 ,t i . When the jet is at the point P = ( x,y,z ) the eye at point E will see it at Q on the yz-plane. ⇒ Q = intersection of the line-→ EP with the yz-plane. Line-→ EP is parameterized (using parameter u ) by E + u ( P- E ) = (1 , , 0) + u ( x- 1 ,y,z ). Q = point on-→ EP with x coordinate =0 x k k k k k k k k k k k k k y S S S S S S S S S S S S S S S S S S z S S S S S S S S S S S S S S S S S S § § § § § § § § §...
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ps2A-solns-2009 - 18.02A pset 2A part II solutions fall...

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