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Three Dimensional Geometry

Question
CBSEENMA12033383
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In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
x + y + z = 1

Solution
The equation of plane is
x + y + z = 1     ...(1)
Dividing both sides by square root of left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared end root space equals space square root of 3 comma space space space we space get comma
fraction numerator 1 over denominator square root of 3 end fraction straight x plus fraction numerator 1 over denominator square root of 3 end fraction straight y plus fraction numerator 1 over denominator square root of 3 end fraction straight z space equals space fraction numerator 1 over denominator square root of 3 end fraction
∴ direction ratios of the normal OP to the plane are fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma where O is origin and P (x1, y1, z1) is foot of perpendicular.
Direction ratios of OP are x1 – 0, y1 – 0, z1– 0 i.e. x1, y1, z1.
Since direction cosines and direction ratios of a line are proportional.
therefore space space space space space space space space fraction numerator straight x subscript 1 over denominator begin display style fraction numerator 1 over denominator square root of 3 end fraction end style end fraction space equals space fraction numerator straight y subscript 1 over denominator begin display style fraction numerator 1 over denominator square root of 3 end fraction end style end fraction space equals fraction numerator straight z subscript 1 over denominator begin display style fraction numerator 1 over denominator square root of 3 end fraction end style end fraction space equals space straight k comma space space say therefore space space space space space space space space straight x subscript 1 space equals space fraction numerator 1 over denominator square root of 3 end fraction straight k. space space space space space straight y subscript 1 space equals space fraction numerator 1 over denominator square root of 3 end fraction straight k comma space space space space straight z subscript 1 space equals space fraction numerator 1 over denominator square root of 3 end fraction straight k therefore space space space space straight P space is space open parentheses fraction numerator 1 over denominator square root of 3 end fraction straight k comma space fraction numerator 1 over denominator square root of 3 end fraction straight k comma space fraction numerator 1 over denominator square root of 3 end fraction straight k close parentheses
Since P lies on plane (1)
therefore space space space space fraction numerator straight k over denominator square root of 3 end fraction plus fraction numerator straight k over denominator square root of 3 end fraction plus fraction numerator straight k over denominator square root of 3 end fraction space equals space 1 space space space or space space straight k plus straight k plus straight k space equals space square root of 3 space space space or space space 3 space straight k space equals space square root of 3 space space rightwards double arrow space space straight k space equals space fraction numerator 1 over denominator square root of 3 end fraction therefore space space space straight P space is space open parentheses 1 third comma space 1 third comma space 1 third close parentheses comma space which space is space foot space of space perpendicular. space

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