pset2sol[1]

pset2sol[1] - 18.02 Problem Set 2 Solutions 1. The...

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Unformatted text preview: 18.02 Problem Set 2 Solutions 1. The directional vector of the line ( AB ) is ~u =- + 2 k . The directional vector of the line ( CD ) is ~v =- 3 + + 3 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 + b + c k must be perpendicular to both ~u and ~v . Therefore we may take ~n = ~u ~v : ~n = k- 1 0 2- 3 1 3 =- 2 - 3 - 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 ) + (1) + (- 1 + 2 t 1 ) k (5) ~ r 2 ( t 2 ) =-- OC + t 2 ~v = (1- 3 t 2 ) + (- 1 + t 2 ) + (3 t 2 ) 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 ) + ( t 2- 2) + (3 t 2- 2 t 1 + 1) k (7) The condition that P...
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pset2sol[1] - 18.02 Problem Set 2 Solutions 1. The...

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