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The plane   a x + b y = 0 is rotated about its line of intersection with the plane z = 0 through an angle α . Prove that the equation of the plane in its new position is a x + b y ± a 2 + b 2 t a n ( α ) z = 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                                       a x + b y = 0                                                                                                                                                   … ( i )                                                               z = 0                                                                                                                                                   … ( i i ) Equation  of  any  plane  passing  through  the  line  of  intersection  of  plane(i)  and  (ii)  is             ( a x + b y ) + k z = 0         ⇒ a x + b y + k z = 0                                                               … ( i i i ) D i v i d i n g     b o t h     s i d e s     b y     a 2 + b 2 + k 2 ,     w e     g e t                   a a 2 + b 2 + k 2 x + b a 2 + b 2 + k 2 y + k a 2 + b 2 + k 2 z = 0 ∴  Direction  cosines  of  the  normal  to  the  plane  are                       a a 2 + b 2 + k 2 ,     b a 2 + b 2 + k 2 ,     k a 2 + b 2 + k 2   and  the  direction  cosines  of  the  plane(i)  are                       a a 2 + b 2 ,     b a 2 + b 2 ,     0 Since,  α  is  the  angle  between  the  planes(i)  and  (iii),  we  get                           c o s α = a . a + b . b + k . 0 a 2 + b 2 + k 2 . a 2 + b 2 ⇒                 c o s α = a 2 + b 2 a 2 + b 2 + k 2 . a 2 + b 2 ⇒                 c o s α = a 2 + b 2 a 2 + b 2 + k 2                   ⇒ c o s 2 α = a 2 + b 2 a 2 + b 2 + k 2 ⇒                   ( a 2 + b 2 + k 2 ) c o s 2 α = a 2 + b 2 ⇒                     a 2 c o s 2 α + b 2 c o s 2 α + k 2 c o s 2 α = a 2 + b 2 ⇒                     k 2 c o s 2 α = a 2 − a 2 c o s 2 α + b 2 − b 2 c o s 2 α ⇒                     k 2 c o s 2 α = a 2 ( 1 − c o s 2 α ) + b 2 ( 1 − c o s 2 α ) ⇒                     k 2 c o s 2 α = a 2 s i n 2 α + b 2 s i n 2 α ⇒                     k 2 c o s 2 α = ( a 2 + b 2 ) s i n 2 α ⇒                     k 2 = ( a 2 + b 2 ) s i n 2 α c o s 2 α             ⇒ k = ± a 2 + b 2 . t a n α P u t t i n g     t h e     v a l u e     o f     k     i n     e q . ( i i i )     w e     g e t                           a x + b y ± ( a 2 + b 2 . t a n α ) z = 0     w h i c h     i s     t h e     r e q u i r e d     E q u a t i o n     o f     p l a n e . H e n c e     p r o v e d .

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