Question

please solve ASAP. :(

Find the equation of the plane which passes through the line of intersection of the planes x+y−2z = 1 and 2x+y−4z = 3 and which is perpendicular to the plane x+y+z = 1

Answer 1

$x+y-2z-1=0\ \ and\ \ 2x+y-4z-3=0$

Equation of plane passing through the line of intersection of the above two planes is

$(x+y-2z-1)+\lambda(2x+y-4z-3)=0\\\therefore x(1+2\lambda)+y(1+\lambda)+z(-2-4\lambda)-1-3\lambda=0$

This plane is perpendicular to $x+y+z-1=0$ $\implies$ Normals are also perpendicular

Direction ratios of normals of new plane $a_1,b_1,c_1$ are $(1+2\lambda),(1+\lambda),(-2-4\lambda)$ respectively

Direction ratios of normals of plane x+y+z-1=0 $a_2,b_2,c_2$ $are\ 1,1,1$ respectively

Since the planes are perpendicular,

So, $a_1a_2+b_1b_2+c_1c_2=0$

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

$\therefore -\lambda=$ 0 $\implies \lambda=0$

Substituting \lambda=0\ in equation of line , we get

$\implies x+y-2z-1=0$

This is the required equation of line.