Sponsor Area

Three Dimensional Geometry

Question
CBSEENMA12033449

Find the image of the point (1, 3, 4) in the plane x – y + z = 5.

Solution
The equation of plane is x – y + z = 5          ...(1)

From P (1, 3, 4) , draw PM ⊥ plane and produce it to P'such that M is mid-point of PP’ Then P' (α,β, γ) is image of P.
Direction ratios of PM are 1, – 1, 1
The equations of PM are
fraction numerator straight x minus 1 over denominator 1 end fraction space equals space fraction numerator straight y minus 3 over denominator negative 1 end fraction space equals space fraction numerator straight z minus 4 over denominator 1 end fraction
Any point on it is M (r + 1, – r + 3, r + 4)
∵ M lies on plane (1)
∴  (r + 1)–(–r + 3)+.(r + 4) = 5,    ∴ r + 1 + r – 3 + r + 4 = 5
∴ 3r = 3    ⇒ r = 1
∴ M is (2, 2, 5)
Since M is mid-point of PP'
therefore space space space fraction numerator straight alpha plus 1 over denominator 2 end fraction space equals space 2 comma space space space space fraction numerator straight beta plus 3 over denominator 2 end fraction space equals space 2 comma space space space space fraction numerator straight gamma plus 4 over denominator 2 end fraction space equals space 5
therefore space space space space space straight alpha plus 1 space equals space 4 comma space space space space straight beta plus 3 space equals space 4 comma space space space space space straight gamma plus 4 space equals space space 10
therefore space space space space space space space straight alpha space equals space 3 comma space space space space space space straight beta space equals space 1 comma space space space space space straight gamma space equals space 6
therefore space space space space space straight P apostrophe space is space left parenthesis 3 comma space 1 comma space 6 right parenthesis comma space which space is space image space of space straight P space in space the space plane. space