Sponsor Area

Three Dimensional Geometry

Question
CBSEENMA12033417

Find the equation of the plane passing through the point (– 1, – 1, 2) and perpendicular to each of the following planes:
2x + 3y – 3 = 2 and 5x – 4y + z = 6.

Solution

The equation of any plane through (– 1,–1, 2) is
a(x + 1) + b(y + 1) + c(z – 2) = 0    ...(1)
∴ it is perpendicular to the planes
2x + 3y – 3z = 2 and 5x – 4y + z = 6
∴ 2a + 3b – 3c = 0    ...(2)
and 5a – 4b + c = 0    ...(3)
Solving (2) and (3), we get,
                    fraction numerator straight a over denominator 3 minus 12 end fraction space equals space fraction numerator straight b over denominator negative 15 minus 2 end fraction space equals space fraction numerator straight c over denominator negative 8 minus 15 end fraction
therefore space space space space space space space space space space space space space fraction numerator straight a over denominator negative 9 end fraction space equals space fraction numerator straight b over denominator negative 17 end fraction equals space fraction numerator straight c over denominator negative 23 end fraction
therefore space space space space space space space space space space space space space space space space straight a over 9 space equals space straight b over 17 space equals space straight c over 28 space equals space straight k comma space space say.
therefore space space space straight a space equals space 9 space straight k comma space space space 6 space equals space 17 space straight k comma space space space straight c space equals space 23 space straight k

∴ a = 9 k, 6 = 17 k, c = 23 k
Putting values of a, b, c in (1), we get
9 k (x + 1)+ 17 k (y + 1) + 23 k (z – 2) = 0
or 9 (x + 1) + 17 (+ 1) + 23 (z – 2) = 0
or 9 x + 9 + 17 y + 17 + 23 z – 46 = 0
or 9 x + 17 y + 23 z – 20 = 0
which is required equation of plane.