Question

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

Solution

The equation of plane through (3, 2, 2) is
A (x – 3) + B (y – 2) + C (z – 2) = 0    ...(1)
Since it passes through (1, 0, –1)
∴ A (1 – 3) + B (0 – 2) + C (–1 – 2) = 0
or –2A – 2B – 3C = 0
∴ 2A + 2B + 3C = 0    .,.(2)
The equation of line is
$fraction numerator straight x minus 1 over denominator 2 end fraction space equals fraction numerator straight y minus 1 over denominator negative 2 end fraction space equals fraction numerator straight z minus 2 over denominator 3 end fraction$
Its direction ratios are 2, –2, 3
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 – 2B + 3C = 0    ...(3)
From (2) and (3), we get,
$fraction numerator straight A over denominator 6 plus 6 end fraction space equals fraction numerator straight B over denominator 6 minus 6 end fraction space equals fraction numerator straight C over denominator negative 4 minus 4 end fraction$
$therefore space space space space space space straight A over 12 space equals space straight B over 0 space equals space fraction numerator straight C over denominator negative 8 end fraction therefore space space space space space space space straight A over 3 space equals space straight B over 0 space equals space fraction numerator straight C over denominator negative 2 end fraction space equals space straight k space left parenthesis say right parenthesis therefore space space space space space space space space straight A space equals space 3 straight k comma space space straight B space equals space 0 comma space space straight C space equals space minus 2 straight k$
Putting these values of A, B, C in (1), we get,
3k (x – 3) + 0 (y – 2) – 2k (z –2) = 0
or 3 (x – 3) – 2 (z – 2 ) = 0
or 3x – 9 –2z + 4 = 0 or 3x –2z – 5 = 0

For Daily Free Study Material Join wiredfaculty

Three Dimensional Geometry

Some More Questions From Three Dimensional Geometry Chapter

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

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location