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:
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x – 2y + 11 = 0
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x + 2y + 11 = 0
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x + 2y – 11 = 0
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x – 2y – 11 = 0
Solution
B.
x + 2y + 11 = 0
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
