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find the equation of the plane through the intersection of the planes 3x-y+2z=4 and x+y+z-2=0 and the point (2,2,1)

Asked by vidyuti.varma | 15th Nov, 2015, 12:04: AM

Expert Answer:

$C o n s i d e r space t h e space g i v e n space p l a n e s comma space 3 x minus y plus 2 z minus 4 equals 0 space a n d space x plus y plus z minus 2 equals 0 T h u s space t h e space e q u a t i o n space o f space t h e space p l a n e space p a s sin g space t h r o u g h space t h e space i n t e r s e c t i o n space o f space t h e space g i v e n space p l a n e s space i s 3 x minus y plus 2 z minus 4 plus lambda open parentheses space x plus y plus z minus 2 close parentheses equals 0 rightwards double arrow open parentheses 3 plus lambda close parentheses x minus open parentheses 1 minus lambda close parentheses y plus open parentheses 2 plus lambda close parentheses z minus open parentheses 4 plus 2 lambda close parentheses equals 0 G i v e n space t h a t space t h e space r e q u i r e d space p l a n e space p a s sin g space t h r o u g h space t h e space p o i n t space left parenthesis 2 comma 2 comma 1 right parenthesis rightwards double arrow open parentheses 3 plus lambda close parentheses cross times 2 minus open parentheses 1 minus lambda close parentheses cross times 2 plus open parentheses 2 plus lambda close parentheses cross times 1 minus open parentheses 4 plus 2 lambda close parentheses equals 0 rightwards double arrow 6 plus 2 lambda minus 2 plus 2 lambda plus 2 plus lambda minus 4 minus 2 lambda equals 0 rightwards double arrow 2 plus 3 lambda equals 0 rightwards double arrow lambda equals negative 2 over 3 T h u s comma space t h e space e q u a t i o n space o f space t h e space r e q u i r e d space p l a n e space i s open parentheses 3 plus lambda close parentheses x minus open parentheses 1 minus lambda close parentheses y plus open parentheses 2 plus lambda close parentheses z minus open parentheses 4 plus 2 lambda close parentheses equals 0 S u b s t i u t i n g space t h e space v a l u e space o f space lambda space i n space t h e space a b o v e space e q u a t i o n space w e space h a v e comma open parentheses 3 minus 2 over 3 close parentheses x minus open parentheses 1 plus 2 over 3 close parentheses y plus open parentheses 2 minus 2 over 3 close parentheses z minus open parentheses 4 minus 4 over 3 close parentheses equals 0 rightwards double arrow 7 x minus 5 y plus 4 z minus 8 equals 0$

Answered by Vimala Ramamurthy | 15th Nov, 2015, 11:06: PM

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