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The equation of the plane containing the line 2x-5y+z=3;x+y+4z=5,

and parallel to the plane, x+3y+6z=1,\, is:

  • Option 1)

    2x+6y+12z=13

  • Option 2)

    x+3y+6z=-7

  • Option 3)

    x+3y+6z=7

  • Option 4)

    2x+6y+12z=-13

 

Answers (1)

best_answer

As we learnt in

Equation of any plane passing through the line of intersection of two planes (Cartesian form ) -

  The equation of any plane passing through the line of intersection of two planes

ax+by+cz+d= 0 and

a_{1}x+b_{1}y+c_{1}z+d_{1}= 0 is given by 

\left ( ax+by+cz+d \right )+\lambda \left ( a_{1}x+b_{1}y+c_{1}z+d _{1}\right )= 0

 

 

-

 

 Equation of a plane be \left ( 2x-5y+z-3 \right ) +\lambda \left ( x+y+4z-5 \right ) = 0            ..........(i)

Since a plane is parallel to  x+3y+6z-1

\frac{2+\lambda }{1} = \frac{\lambda - 5}{3}= \frac{1+4\lambda }{6}\Rightarrow \lambda = \frac{-11}{2}

After substituting \lambda = \frac{-11}{2} in (i)

Hence requred plane will be   

x+3y+6z=7


Option 1)

2x+6y+12z=13

This option is incorrect

Option 2)

x+3y+6z=-7

This option is incorrect

Option 3)

x+3y+6z=7

This option is correct

Option 4)

2x+6y+12z=-13

This option is incorrect

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