Question

Find the equation of planes passing through the intersection of the planes $x+3y+6=0$ and $3x-y+4z=0$ whose distance from origin is $1$.

Answer 1

Equation of planes which intersect two planes
$p_1+\lambda p_2=0$
$(x+3y+6)+\lambda (3x-y+4z)=0$ ..........(i)
$x+3y+6+3\lambda x-\lambda y+4\lambda z=0$
$x(1+3\lambda )+y(3-\lambda )+4\lambda z+6=0$
Length of perpendicular draw from $(0,0,0)$ is equal to $1$ is given
$\therefore 1=\frac{6}{\sqrt{(1+3\lambda {)}^2+(3-\lambda {)}^2+(4\lambda {)}^2}}$
$(1+3\lambda {)}^2+(3-\lambda {)}^2+(4\lambda {)}^2=36$ by squaring
$1+9{\lambda }^2+6\lambda +9+{\lambda }^2-6\lambda +16{\lambda }^2=36$
$26{\lambda }^2=36-10$
$26{\lambda }^2=26$
${\lambda }^2=\frac{26}{26}=1$
$\lambda =\pm 1$
Putting the value of $\lambda$ in (i) $\lambda =1$
We get
$(x+3y+6)+\lambda (3x-y+4z)=0$
$x+3y+6+1(3x-y+4z)=0$
$x+3y+6+3x-y+4z=0$
$4x+2y+4z+6=0$
$2x+y+2z+3=0$
When $\lambda =-1$
$(x+3y+6)-1(3x-y+4z)=0$
$x+3y+6-3x+y-4z=0$
$-2x+4y-4z=0$
$x-2y+2z-3=0$.