Discussion Forum

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

Answer:

Explanation:

 Let equation of plane  containing the line 2x-5y+z=3  and x+y+4z=5 be

   $(2x-5y+z-3)+\lambda (x+y+4z-5)=0$

$\Rightarrow  (2+\lambda)x+(\lambda-5)y+(4\lambda+1)z-3-5\lambda=0$      ........(i)

 This plane is parallel to the plane

 x+3y+6z=1

$\therefore$     $\frac{2+\lambda}{1}=\frac{\lambda-5}{3}=\frac{4\lambda+1}{6}$

 On taking first wo equalities , we get

                   $6+3\lambda=\lambda-5$

$\Rightarrow$                 $2\lambda=-11$

$\Rightarrow$            $\lambda=-\frac{11}{2}$

On taking last two equalities , we get 

                          $6\lambda-30=3+12\lambda$

$\Rightarrow$   $-6\lambda=33$

$\Rightarrow$     $\lambda=-\frac{11}{2}$

 So, the equation of required  plane is

  $\left(2-\frac{11}{2}\right)x+\left(\frac{-11}{2}-5\right) y+\left(-\frac{44}{2}+1\right) z-3+5\times \frac{11}{2}=0$

   $\Rightarrow$             $-\frac{7}{2}x-\frac{21}{2}y-\frac{42}{2}z+\frac{49}{2}=0$

   $\Rightarrow$   x+3y+6z-7=0