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Using vectors, find the value of k such that the points (k, –10, 3), (1, –1, 3), and (3, 5, 3) are collinear.

0 2 Views | Posted 3 months ago
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    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Short Answer type Questions as classified in NCERT Exemplar

    Sol:

    Let  the  given  points  are  A(k,−10,3)  ,B(1,−1,3)and  C(3,5,3)                           A B → = ( 1 − k ) i ^ + ( − 1 + 1 0 ) j ^ + ( 3 − 3 ) k ^ = ( 1 − k ) i ^ + 9 j ^ + 0 k ^ ∴                   | A B → | = ( 1 − k ) 2 + ( 9 ) 2 = ( 1 − k ) 2 + 8 1                             B C → = ( 3 − 1 ) i ^ + ( 5 + 1 ) j ^ + ( 3 − 3 ) k ^ = 2 i ^ + 6 j ^ + 0 k ^ ∴                     | B C → | = ( 2 ) 2 + ( 6 ) 2 = 4 + 3 6 = 4 0 = 2 1 0                               A C → = ( 3 − k ) i ^ + ( 5 + 1 0 ) j ^ + ( 3 − 3 ) k ^ = ( 3 − k ) i ^ + 6 j ^ + 0 k ^ ∴                       | A C → | = ( 3 − k ) 2 + ( 1 5 ) 2 = ( 3 − k ) 2 + 2 2 5 I f     A ,     B ,     a n d     C     a r e     c o l l i n e a r ,     t h e n                                                                   | A B → | + | B C → | = | A C → |                         ( 1 − k ) 2 + 8 1 + 2 1 0 = ( 3 − k ) 2 + 2 2 5 S q u a r i n g     b o t h     s i d e s ,     w e     h a v e                                                                                   [ ( 1 − k ) 2 + 8 1 + 4 0 ] 2 = [ ( 3 − k ) 2 + 2 2 5 ] 2 ⇒               ( 1 − k ) 2 + 8 1 + 4 0 + 2 4 0 ( 1 − k ) 2 + 8 1 = ( 3 − k ) 2 + 2 2 5 ⇒ 1 + k 2 − 2 k + 1 2 1 + 2 4 0 1 + k 2 − 2 k + 8 1 = 9 + k 2 − 6 k + 2 2 5 ⇒                                                 1 2 2 − 2 k + 2 4 0 k 2 − 2 k + 8 2 = 2 3 4 − 6 k D i v i d i n g     b y     2 ,     w e     g e t ⇒                                                 6 1 − k + 4 0 k 2 − 2 k + 8 2 = 1 1 7 − 3 k ⇒                                                                                     4 0 k 2 − 2 k + 8 2 = 1 1 7 − 6 1 − 3 k + k ⇒                                                                                     4 0 k 2 − 2 k + 8 2 = 5 6 − 2 k ⇒                                                                                     2 1 0 k 2 − 2 k + 8 2 = 5 6 − 2 k ⇒                                                                                           1 0 k 2 − 2 k + 8 2 = 2 8 − k                   ( D i v i d i n g     b y     2 ) S q u a r i n g     b o t h     s i d e s ,     w e     g e t ⇒                                                                                                 1 0 ( k 2 − 2 k + 8 2 ) = 7 8 4 + k 2 − 5 6 k ⇒                                                                                                 1 0 k 2 − 2 0 k + 8 2 0 = 7 8 4 + k 2 − 5 6 k ⇒                           1 0 k 2 − k 2 − 2 0 k + 5 6 k + 8 2 0 − 7 8 4 = 0 ⇒                       9 k 2 + 3 6 k + 3 6 = 0             ⇒ k 2 + 4 k + 4 = 0 ⇒                       ( k + 2 ) 2 = 0         ⇒ k + 2 = 0           ⇒ k = − 2 H e n c e ,     t h e     r e q u i r e d     v a l u e     i s     k = − 2 .

     

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