Question

Find the equation of the planes bisecting the angles between the planes $2x-y+2z+3=0$ are $3x-2y+6z+8=0$

Answer 1

Equation of planes

$2x-y+2z+3=0......\left(1\right)$

$3x-2y+6z+8=0......\left(2\right)$

Equation of bisector planes are

$\left(\frac{2x-y+2z+3}{\sqrt{2^2+{(-1)}^2+2^2}}\right)=\pm \left(\frac{3x-2y+6z+8}{\sqrt{3^2+{(-2)}^2+6^2}}\right)$

Therefore,

$\left(\frac{2x-y+2z+3}{3}\right)=\pm \left(\frac{3x-2y+6z+8}{7}\right)$

$\text{Taking}\text{+ive}$

$\Rightarrow 7(2x-y+2z+3)=3(3x-2y+6z+8)$

$\Rightarrow 5x-y-4z-3=0$

$\text{Taking}\text{-ive}$

$\Rightarrow 7(2x-y+2z+3)=-3(3x-2y+6z+8)$

$\Rightarrow 23x-13y+32z+45=0$

Hence, this is the answer.