JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Three Dimensional Geometry

The equation of a plane containing the line of intersection of the planes

2 x - y - 4 = 0

and

y + 2 z - 4 = 0

and passing through the point

(1, 1, 0)

is:

(A) $x + 3 y + z = 4$
(B) $2 x - z = 2$
(C) $x - 3 y - 2 z = - 2$
(D) $x - y - z = 0$ ✓ Correct

Answer: (D)

Solution

Any plane through the common line of the two given planes can be written as

(2 x - y - 4) + λ (y + 2 z - 4) = 0

. Forcing it through

(1, 1, 0)

gives

(2 - 1 - 4) + λ (1 - 4) = 0

, so

- 3 - 3 λ = 0

and

λ = - 1

. Substituting back,

(2 x - y - 4) - (y + 2 z - 4) = 0

simplifies to

2 x - 2 y - 2 z = 0

, that is

x - y - z = 0

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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