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Find the shortest distance between the lines r → = ( 8 + 3 λ ) i ^ − ( 9 + 1 6 λ ) j ^ + ( 1 0 + 7 λ ) k ^ and r → = 1 5 i ^ − 2 9 j ^ + 5 k ^ + μ ( 3 i ^ + 8 j ^ − 5 k ^ )    

0 2 Views | Posted 3 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is a long answer type question as classified in NCERT Exemplar

    G i v e n     e q u a t i o n s     o f     l i n e s     a r e                     r → = ( 8 + 3 λ ) i ^ − ( 9 + 1 6 λ ) j ^ + ( 1 0 + 7 λ ) k ^                                                   … ( i ) a n d     r → = 1 5 i ^ − 2 9 j ^ + 5 k ^ + μ ( 3 i ^ + 8 j ^ − 5 k ^ )                                                               … ( i i ) E q u a t i o n ( i )     c a n     b e     r e − w r i t t e n     a s                     r → = 8 i ^ − 9 j ^ + 1 0 k ^ + λ ( 3 i ^ − 1 6 j ^ + 7 k ^ )                                                               … ( i i i ) H e r e ,     a 1 → = 8 i ^ − 9 j ^ + 1 0 k ^     a n d     a 2 → = 1 5 i ^ − 2 9 j ^ + 5 k ^                             b 1 → = 3 i ^ − 1 6 j ^ + 7 k ^     a n d     b 2 → = 3 i ^ + 8 j ^ − 5 k ^                   a 2 → − a 1 → = 7 i ^ + 3 8 j ^ − 5 k ^                       b 1 → × b 2 → = | i ^ j ^ k ^ 3 − 1 6 7 3 8 − 5 |                                                   = i ^ ( 8 0 − 5 6 ) − j ^ ( − 1 5 − 2 1 ) + k ^ ( 2 4 + 4 8 )                                                   = 2 4 i ^ + 3 6 j ^ + 7 2 k ^ ∴         Shortest  distance,  SD=|(a2→−a1→).(b1→×b2→)|b1→×b2→||=|(7i^+38j^−5k^).(24i^+36j^+72k^)(24)2+(36)2+(72)2|                                                                           = | 1 6 8 + 1 3 6 8 − 3 6 0 5 7 6 + 1 2 9 6 + 5 1 8 4 | = | 1 6 8 + 1 0 0 8 7 0 5 6 | = 1 1 7 6 8 4 = 1 4   u n i t s Hence,  the  required  distance  is  14 units.

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