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The equation of a plane containg 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 :

Option 1)
$x+3y+z=4$

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

Option 3)
$x-y-z=0$

Option 4)
$2x-z=2$

Answers (1)

best_answer

Equation of plane passing through intersection of

$2x-y-4=0$ and $y+2z-4=0$

$=(2x-y-4)+\lambda (y+2z-4)=0$

$=2x-y+\lambda y-4+2\lambda z-4\lambda=0$

$=2x+y(\lambda-1) +2\lambda z-4-4\lambda =0$

$\because$ if passes through $(1,1,0)$

$=2+1(\lambda -1)+0-4-4\lambda =0$

$2+\lambda -1-4-4\lambda =0$

$-3=3\lambda$

$\lambda =-1$

$\because$ $2x+y(\lambda -1)+2\lambda z-4-4\lambda =0$

$2x+y(-1 -1)+2(-1) 2-4-4(-1) =0$

$2x-2y-2z=0$

$x-y-z=0$

Option 1)

$x+3y+z=4$

Option 2)

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

Option 3)

$x-y-z=0$

Option 4)

$2x-z=2$

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