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CBSE Class 12-science Answered

find the equation of the plane passing through the point (-1,-3,2) and perpendicular to each of the planes: x+2y+3z=5 and 3x+3y+z=0.

Asked by Pragyashree Jain | 15 Mar, 2015, 09:22: PM
answered-by-expert Expert Answer
L e t space a comma space b space a n d space c space b e space t h e space d i r e c t i o n space r a t i o s space o f space t h e space n o r m a l space t o space t h e space p l a n e. 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 open parentheses x subscript 1 comma y subscript 1 comma z subscript blank close parentheses space i s a open parentheses x minus x subscript 1 close parentheses plus b open parentheses y minus y subscript 1 close parentheses plus c open parentheses z minus z subscript 1 close parentheses equals 0 T h e r e f o r e 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 open parentheses negative 1 comma negative 3 comma 2 close parentheses space i s a open parentheses x plus 1 close parentheses plus b open parentheses y plus 3 close parentheses plus c open parentheses z minus 2 close parentheses equals 0... left parenthesis 1 right parenthesis  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 i s space p e r p e n d i c u l a r space t o space t h e space p l a n e s x plus 2 y plus 3 z equals 5 space a n d space 3 x plus 3 y plus z equals 0 T h u s comma space w e space h a v e a plus 2 b plus 3 c equals 0 a n d 3 a plus 3 b plus c equals 0 S o l v i n g space b y space c r o s s space m u l t i p l i c a t i o n comma space w e space h a v e comma fraction numerator a over denominator 2 cross times 1 minus 3 cross times 3 end fraction equals fraction numerator b over denominator 3 cross times 3 minus 1 cross times 1 end fraction equals fraction numerator c over denominator 1 cross times 3 minus 3 cross times 2 end fraction rightwards double arrow fraction numerator a over denominator 2 minus 9 end fraction equals fraction numerator b over denominator 9 minus 1 end fraction equals fraction numerator c over denominator 3 minus 6 end fraction rightwards double arrow fraction numerator a over denominator negative 7 end fraction equals b over 8 equals fraction numerator c over denominator negative 3 end fraction equals k rightwards double arrow a equals negative 7 k semicolon b equals 8 k semicolon c equals negative 3 k s u b s t i t u t i n g space t h e space v a l u e s space o f space a comma space b space a n d space c space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e comma minus 7 k open parentheses x plus 1 close parentheses plus 8 k open parentheses y plus 3 close parentheses minus 3 k open parentheses z minus 2 close parentheses equals 0 rightwards double arrow negative 7 open parentheses x plus 1 close parentheses plus 8 open parentheses y plus 3 close parentheses minus 3 open parentheses z minus 2 close parentheses equals 0 rightwards double arrow negative 7 x minus 7 plus 8 y plus 24 minus 3 z plus 6 equals 0 rightwards double arrow negative 7 x plus 8 y minus 3 z plus 23 equals 0 rightwards double arrow 7 x minus 8 y plus 3 z minus 23 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 r e q u i r e d space p l a n e space i s space 7 x minus 8 y plus 3 z minus 23 equals 0
Answered by Vimala Ramamurthy | 16 Mar, 2015, 10:51: AM
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