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)

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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