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Find the equation of a plane which is at a distance of   3 3 units from the origin and the normal to which is equally inclined to coordinate axes.

0 3 Views | Posted 3 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

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

    Since,  the  normal  to  the  plane  is  equally  inclined  to  the  axis ∴                       c o s α = c o s β = c o s γ ⇒                 c o s 2 α + c o s 2 α + c o s 2 α = 1 ⇒                 3 c o s 2 α = 1                   ⇒ c o s α = 1 3 ⇒                 c o s α = c o s β = c o s γ = 1 3 S o ,     t h e     n o r m a l     i s     N → = 1 3 i ^ + 1 3 j ^ + 1 3 k ^ ∴     E q u a t i o n     o f     t h e     p l a n e     i s     r → . N → = d ⇒                                                                                   r → . N → | N → | = d ⇒                                                                                   r → . ( 1 3 i ^ + 1 3 j ^ + 1 3 k ^ ) 1 = 3 3 ⇒                                                                                       r → . ( 1 3 i ^ + 1 3 j ^ + 1 3 k ^ ) = 3 3 ⇒                                                                       ( x i ^ + y j ^ + z k ^ ) . 1 3 ( i ^ + j ^ + k ^ ) = 3 3 ⇒                                                                       x + y + z = 3 3 . 3 = 9 H e n c e ,     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     i s     x + y + z = 9 .

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