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Integrals

Question
CBSEENMA12036132

The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal perpendicular to both the linesfraction numerator straight x minus 1 over denominator 1 end fraction space equals space fraction numerator straight y plus 2 over denominator negative 2 end fraction space equals space fraction numerator straight z minus 4 over denominator 3 end fraction space and space fraction numerator straight x minus 2 over denominator 2 end fraction space equals space fraction numerator straight y plus 1 over denominator negative 1 end fraction space equals space fraction numerator straight z plus 7 over denominator negative 1 end fraction is

  • fraction numerator 10 over denominator square root of 74 end fraction
  • fraction numerator 20 over denominator square root of 74 end fraction
  • fraction numerator 10 over denominator square root of 83 end fraction
  • fraction numerator 5 over denominator square root of 83 end fraction

Solution

C.

fraction numerator 10 over denominator square root of 83 end fraction

Normal vector
open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 1 cell negative 2 end cell 3 row 2 cell negative 1 end cell cell negative 1 end cell end table close vertical bar space equals space 5 straight i with hat on top space plus space 7 straight j with hat on top space plus 3 straight k with hat on top
So space plane space is space
5 left parenthesis straight x minus 1 right parenthesis space plus space 7 left parenthesis straight y plus 1 right parenthesis space plus space 3 left parenthesis straight z plus 1 right parenthesis space equals space 0 space
rightwards double arrow space 5 space straight x space plus 7 straight y space plus space 3 straight z space plus space 5 space equals space 0
Distance space fraction numerator 5 plus 21 minus 21 plus 5 over denominator square root of 25 plus 49 plus 9 end root end fraction space equals space fraction numerator 10 over denominator square root of 83 end fraction

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