Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:

Options:

x – 2y + 11 = 0

x + 2y + 11 = 0

x + 2y – 11 = 0

x – 2y – 11 = 0

Correct Answer:

x + 2y + 11 = 0

Explanation:

Equation of the required plane is (x + y + z – 6) + λ(2x + 3y + z + 5) = 0

i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0

This plane is perpendicular to xy plane whose equation is z = 0

i.e. 0 . x + 0 . y + z = 0

∴ By condition of perpendicularity

0.(1 + 2λ) + 0. (1 + 3λ) + (1 + λ) .1 = 0 i.e. λ = –1

∴ Equation of required plane is

(1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) = 0 or x + 2y + 11 = 0