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AB = 3 i ^ -   j ^ + k ^ and   C D = -3 i ^ + 2 j ^ + 4 k ^ are two vectors. The position vectors of the points A and C are 6 i ^ + 7 j ^  + 4 k ^  and -9 i ^  + 2 j ^  + k , ^ respectively. Find the position vector of a point P on the line AB and a point Q on the line CD such that P Q →  is perpendicular to both   A B → and   C D → both.

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

    P o s i t i o n     v e c t o r     o f     A     i s     6 i ^ + 7 j ^ + 4 k ^     a n d     A B → = 3 i ^ − j ^ + k ^ So,  equation  of  any  line  passing  through  A  and  parallel  to  AB→                                         r → = ( 6 i ^ + 7 j ^ + 4 k ^ ) + λ ( 3 i ^ − j ^ + k ^ )                                                                                     … ( i ) Now  any  point  P  on  AB→=(6+3λ, 7−λ, 4+λ) S i m i l a r l y ,     p o s i t i o n     v e c t o r     o f     C     i s     − 9 j ^ + 2 k ^     a n d     C D → = − 3 i ^ + 2 j ^ + 4 k ^ So,  equation  of  any  line  passing  through  C  and  parallel  to  CD→  is                                         r → = ( − 9 j ^ + 2 k ^ ) + μ ( − 3 i ^ + 2 j ^ + 4 k ^ )                                                                                     … ( i i ) Any  point  Q  on  CD→=(−3μ, −9+2μ, 2+4μ) d ' r a t i o s     o f     P Q →     a r e                                   ( 6 + 3 λ + 3 μ ,   7 − λ + 9 − 2 μ ,   4 + λ − 2 − 4 μ ) ⇒                           ( 6 + 3 λ + 3 μ ) ,   ( 1 6 − λ − 2 μ ) , ( 2 + λ − 4 μ ) N o w     P Q →     i s     ⊥     t o     e q . ( i ) ,     t h e n           3 ( 6 + 3 λ + 3 μ ) − 1 ( 1 6 − λ − 2 μ ) + 1 ( 2 + λ − 4 μ ) = 0 ⇒                                   1 8 + 9 λ + 9 μ − 1 6 + λ + 2 μ + 2 + λ − 4 μ = 0 ⇒                                                                                                                                                     1 1 λ + 7 μ + 4 = 0                                                               … ( i i i ) P Q →     i s     ⊥     t o     e q . ( i i ) ,     t h e n           − 3 ( 6 + 3 λ + 3 μ ) + 2 ( 1 6 − λ − 2 μ ) + 4 ( 2 + λ − 4 μ ) = 0 ⇒                       − 1 8 − 9 λ − 9 μ + 3 2 − 2 λ − 4 μ + 8 + 4 λ − 1 6 μ = 0 ⇒                                                                                                                                                 − 7 λ − 2 9 μ + 2 2 = 0 ⇒                                                                                                                                                           7 λ + 2 9 μ − 2 2 = 0                                                             … ( i v ) S o l v i n g     e q n . ( i i i )     a n d ( i v )     w e     g e t ⇒                   7 7 λ           + 4 9 μ         +         2 8       = 0                             7 7 λ           + 3 1 9 μ − 2 4 2       = 0                   ( − )                   ( − )                             ( + )                                     _                                                           − 2 7 0 μ     + 2 7 0     = 0                     ∴ μ = 1 P u t t i n g     t h e     v a l u e     o f     μ     i n     e q n . ( i v )     w e     g e t ,                                               7 λ + 2 9 − 2 2 = 0                   ⇒ λ = − 1 ∴               P o s i t i o n     v e c t o r     o f     P = [ 6 + 3 ( − 1 ) ,   7 + 1 ,   4 − 1 ] = ( 3 ,   8 ,   3 ) a n d     P o s i t i o n     v e c t o r     o f     Q = [ − 3 ( 1 ) ,   − 9 + 2 ( 1 ) ,   2 + 4 ( 1 ) ] = ( − 3 ,   − 7 ,   6 ) H e n c e ,     t h e     p o s i t i o n     v e c t o r     o f                 P = 3 i ^ + 8 j ^ + 3 k ^     a n d     Q = − 3 i ^ − 7 j ^ + 6 k ^

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