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Unformatted text preview: 18.02 Problem Set 2 Solutions 1. The directional vector of the line ( AB ) is ~u = ˆ ı + 2 ˆ k . The directional vector of the line ( CD ) is ~v = 3ˆ ı + ˆ + 3 ˆ k . For part (a), we first observe that in order for a plane ax + by + cz = d to be parallel to both lines, its normal vector ~n = a ˆ ı + b ˆ + c ˆ k must be perpendicular to both ~u and ~v . Therefore we may take ~n = ~u × ~v : ~n = ˆ ı ˆ ˆ k 1 0 2 3 1 3 = 2ˆ ı 3ˆ  ˆ k (1) Now we are looking for planes of the form 2 x 3 y z = d . The value of d for which this plane contains A (and hence the line ( AB )) is d = 2(1) 3(1) ( 1) = 4. The value of d for which this plane contains C (and hence the line ( CD )) is d = 2(1) 3( 1) (0) = 1. We now want the distance between the two planes 2 x 3 y z = 4 (2) 2 x 3 y z = 1 (3) The distance between two planes with the same normal vector is the difference in the “ d ” values of the two equations, divided by the length of the normal vector: distance =   4 1   ~n  = 5 √ 2 2 + 3 2 + 1 2 = 5 √ 14 (4) (The last assertion could be demonstrated by computing the length of the line segment running from the first plane to the second along a direction parallel to ~n .) For part (b), we write the parametric equations of ( AB ) and ( CD ), using A and C as the initial points: ~ r 1 ( t 1 ) =→ OA + t 1 ~u = (1 t 1 )ˆ ı + (1)ˆ + ( 1 + 2 t 1 ) ˆ k (5) ~ r 2 ( t 2 ) =→ OC + t 2 ~v = (1 3 t 2 )ˆ ı + ( 1 + t 2 )ˆ + (3 t 2 ) ˆ k (6) Each point P 1 on the line ( AB ) corresponds to ~ r 1 ( t 1 ) for some particular value of t 1 , and each point P 1 on the line ( CD ) corresponds to ~ r 2 ( t 2 ) for some particular value of t 2 . The line segement P 1 P 2 from P 1 to P 2 can then be identified with the vector ~ r 2 ( t 2 ) ~ r 1 ( t 1 ). We compute: ~ r 2 ( t 2 ) ~ r 1 ( t 1 ) = ( t 1 3 t 2 )ˆ ı + ( t 2 2)ˆ + (3 t 2 2 t 1 + 1) ˆ k (7) The condition that P...
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 Spring '08
 Auroux

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