Question

Find the equations of angle bisectors between the planes
$x+2y+2z-3=0$
$3x+4y+12z+1=0$

Answer 1

Equations of bisector between planes

$x+2y+27-3=0$

$3x+4y+127+1=0$

are given by

$\frac{x+2y+2z-3}{\sqrt{1+2^2+2^2}}=\pm \frac{3x+4y+12z+1}{\sqrt{3^2+4^2+{12}^2}}$

$\Rightarrow \frac{x+2y+2z-3}{3}=\pm \frac{3x+4y+12z+1}{13}$

$\Rightarrow 13(x+2y+2z-3)=\pm 3(3x+4y+12z+1)$

$\Rightarrow 13x+26y+26z-39=\pm (9x+12y+36z+3)$

$\therefore 4x+14y-10z-42=0$

$\Rightarrow 2x+7y-5z-21=0$

or

$22x+38y+62z-36=0$

$11x+19y+31z=18-0$