Sponsor Area

Three Dimensional Geometry

Question
CBSEENMA12033441

Find the equation of the plane passing through the points (3, 4, 1), (0, 1, 0) and parallel to the line fraction numerator straight x plus 3 over denominator 2 end fraction space equals fraction numerator straight y minus 3 over denominator 7 end fraction space equals space fraction numerator straight z minus 2 over denominator 5 end fraction.

Solution

The equation of plane through (3, 4, 1) is
A(x – 3) + B(y – 4) + C(z – 1) = 0    ...(1)
Since it passes through (0, 1, 0)
∴ A(0 – 3) + B(1 – 4) + C(0 – 1) = 0
or –3A –3B –C = 0
∴ 3A + 3B + C = 0    ..... (2)
The equation of line is
fraction numerator straight x plus 3 over denominator 2 end fraction space equals fraction numerator straight y minus 3 over denominator 7 end fraction space equals space fraction numerator straight z minus 2 over denominator 5 end fraction
Its direction ratios are 2, 7, 5
Since the line is parallel to plane (1) whose normal has direction ratios A, B, C.
∴ normal to plane (1) is perpendicular to line.
∴ 2A + 7B + 5C = 0    ...(3)
From (2) and (3), we get,
                 fraction numerator straight A over denominator 15 minus 7 end fraction space equals space fraction numerator straight B over denominator 2 minus 15 end fraction space equals fraction numerator straight C over denominator 21 minus 6 end fraction
therefore space space space space space space space space space space space space straight A over 8 space equals fraction numerator straight B over denominator negative 13 end fraction space equals space straight C over 15 space equals space straight k space left parenthesis say right parenthesis
therefore space space space space space space space space space space space space straight A space equals space 8 space straight k comma space space space space straight B space equals space minus 13 space straight k comma space space space straight C space equals space 15 space straight k

Putting these values of A, B, C in (1), we get,
8 k (x – 3) – 13 k (y – 4) + 15 k (z – 1) = 0
or 8 (x – 3) – 13(y – 4) + 15 (z – 1) = 0
or 8x – 24 – 13 y + 52 + 15 z – 15 = 0
or 8 x – 13 y + 15 z + 13 = 0,
which is required equation of plane.

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.