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Find the equation of the plane which is perpendicular to the plane   5 x + 3 y + 6 z + 8 = 0 and contains the line of intersection of the planes   x+2y+3z–4=0 and 2x+y–z+5=0 .

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

    T h e     g i v e n     p l a n e s     a r e         P 1 :           5 x + 3 y + 6 z + 8 = 0         P 2 :             x + 2 y + 3 z − 4 = 0         P 3 :                   2 x + y − z + 5 = 0 Equations  of  the  plane  passing  through  the  line  of  intersection  of  P2  and  P3  is                               ( x + 2 y + 3 z − 4 ) + λ ( 2 x + y − z + 5 ) = 0 ⇒ ( 1 + 2 λ ) x + ( 2 + λ ) y + ( 3 − λ ) z − 4 + 5 λ = 0                                                                               … ( i ) P l a n e ( i )     i s     p e r p e n d i c u l a r     t o     P 1 , t h e n                   5 ( 1 + 2 λ ) + 3 ( 2 + λ ) + 6 ( 3 − λ ) = 0 ⇒                               5 + 1 0 λ + 6 + 3 λ + 1 8 − 6 λ = 0 ⇒                               7 λ + 2 9 = 0         ⇒ λ = − 2 9 7 P u t t i n g     t h e     v a l u e     o f     λ     i n     e q . ( i ) ,     w e     g e t                     [ 1 + 2 ( − 2 9 7 ) ] x + [ 2 − 2 9 7 ] y + [ 3 + 2 9 7 ] z − 4 + 5 ( − 2 9 7 ) = 0 ⇒         − 1 5 7 x − 1 5 7 y + 5 0 7 z − 4 − 1 4 5 7 = 0 ⇒               − 1 5 x − 1 5 y + 5 0 z − 2 8 − 1 4 5 = 0 ⇒               − 1 5 x − 1 5 y + 5 0 z − 1 7 3 = 0           ⇒ 1 5 x + 1 5 y − 5 0 z + 1 7 3 = 0

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