The equation of a plane containing the line of intersection of the planes 2x-y-4=0 and y+2z-4=0 and passing through the point (1 , 1 , 0) is :

Question

The equation of a plane containing the line of intersection of the planes 2x-y-4=0 and y+2z-4=0 and passing through the point (1 , 1 , 0) is : (A) x+3y+z=4 (B) x-y-z=0 (C) x-3y-2z=-2 (D) 2x-z=2

Tardigrade Admin · Accepted Answer

The required plane is (2 x - y - 4)+λ (y + 2 z - 4)=0 it passes through (1 , 1 , 0) ⇒ (2 - 1 - 4)+λ (1 - 4)=0 ⇒ -3-3λ =0⇒ λ =-1 ⇒ x-y-z=0

Q. 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 :

$x + 3 y + z = 4$

$x - y - z = 0$

$x - 3 y - 2 z = - 2$

$2 x - z = 2$

The required plane is
$(2 x - y - 4) + λ (y + 2 z - 4) = 0$
it passes through $(1, 1, 0)$
$\Rightarrow (2 - 1 - 4) + λ (1 - 4) = 0$
$\Rightarrow - 3 - 3 λ = 0 \Rightarrow λ = - 1$
$\Rightarrow x - y - z = 0$

Q. The equation of a plane containing the line of intersection of the planes 2x−y−4=0 and y+2z−4=0 and passing through the point (1,1,0) is :

x+3y+z=4

x−y−z=0

x−3y−2z=−2

2x−z=2