Find the shortest distance between the given lines: r ?=(i ?+2j ?-4k ? )+?(2i ?+3j ?+6k ? ), r ?=(3i ?+3j ?-5k ?)+?(-2i ?+3j ?+3k ?) The answer given in the book is:14/?241 units.

Asked by Manoj | 18th May, 2013, 04:40: AM

Expert Answer:

 r1 = (1,2,-4) + ? (2,3,6)   and   r = (3,3,-5) + ?(-2,3,3).

The point P with position vector p = (1,2,-4) lies on r1. 
The point Q with position vector q = (3,3,-5) lies on r2 

The cross product of the two direction vectors, (2,3,6) and (-2,3,3) gives a common normal to both planes.

Working this out gives the vector (-9,-18,12).
So a unit normal vector to both planes is given by  1/sqrt(61) (-3,-6,4). 
The equation of P(1) is  1/sqrt(13) (3, 0, 2) . r = 1/sqrt(61) (-3,-6,4) . (1,2,-4) = -31/sqrt(61).
The equation of P(2) is  1/sqrt(13) (3, 0, 2) . r = 1/sqrt(61) (-3,-6,4). (3,3,-5) = -47/sqrt(61).
So the shortest distance between the 2 skew lines is | -31/sqrt(61)) + 47/sqrt(61) | = 16/sqrt(61) units.

The answer in the book seems wrong. 

Answered by  | 21st May, 2013, 06:18: AM

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