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Three Dimensional Geometry

Question
CBSEENMA12033428

Find the equation of the plane passing through the points (0, –1, –1), (4, 5, 1) and  (3, 9, 4).

Solution

The equation of plane passing through (0, – 1,–1) is
A (x – 0) + B(y + 1) + C(z + 1) = 0    ...(1)
∴ it passes through (4, 5, 1)
∴ A(4 – 0) + B(5 + 1) + C( 1 + 1) = 0
∴ 4A + 6B + 2C = 0 ⇒ 2A + 3B + C = 0    ...(2)
Again plane (1) passes through (3, 9, 4)
∴ A(3 – 0) + B(9 + 1) + C(4 + 1) = 0
∴ 3A + 10B + 5C = 0    ....(3)
Solving (2) and (3), we get,
                      fraction numerator straight A over denominator 15 minus 10 end fraction space equals space fraction numerator straight B over denominator 3 minus 10 end fraction equals fraction numerator straight C over denominator 20 minus 9 end fraction
therefore space space space space space space space straight A over 5 space equals space fraction numerator straight B over denominator negative 7 end fraction space equals straight C over 11 space equals space straight k space left parenthesis say right parenthesis
therefore space space space space space space straight A space equals space 5 space straight k comma space space space space straight B space equals space minus 7 space straight k comma space space space straight C space equals space 11 space straight k

Putting values of A, B, C in (1), we get,
5 k (x – 0) – 7 k (y + 1) + 11 k (z + 1) = 0
or  5 x – 7 y – 7 + 11 z + 11 = 0
or  5 x – 7 y + 11 z + 4 = 0
which is required equation of plane.