Question

Find the equation of the plane passing through the line of intersection of planes $2x-y+z=3$ , $4x-3y+5z+9=0$ and parallel to the line $\frac{x+1}{2}=-\frac{y+3}{4}=\frac{z-3}{5}$.

• A. $19x+3y+10z=126$
  
• B. $19x-3y-10z=126$
  
• C. $19x+3y-10z=126$
  
• D. $19x-3y+10z=126$
  

Answer 1

Answer: (B)

$19x-3y-10z=126$

Let the plane passing through line of intersection of two plane be : $2x-y+z-3+\lambda (4x-3y+5z+9)=0$

$\Rightarrow (2+4\lambda )x-(1+3\lambda )y+(1+5\lambda )z-(3-9\lambda )=0$ equation-(1)

For the above plane to be parallel to given line $\frac{x+1}{2}=\frac{y+3}{-4}=\frac{z-3}{5}$ ;the dot product of normal vector of plane and direction cosines of the plane should be zero

$\Rightarrow 2(2+4\lambda )+4(1+3\lambda )+5(1+5\lambda )=0$

$\Rightarrow \lambda =\frac{-13}{45}$

On substituting $\lambda$ in equation -(1) we get

the final equation of the plane :$19x-3y-10z=126$