Question

Consider the two plane $P_1:2x-y+2z+3=0,P_2:3x-2y+6z+8=0,$ then which of the following statement(s) is/are correct for angle bisector between plane $P_1$ and $P_2$

• A. $5x-y-4z-3=0$ is acute angle bisector
• B. $5x-y-4z-3=0$ is obtuse angle bisector
• C. $23x-13y+32z+45=0$ is acute angle bisector
• D. $23x-13y+32z+45=0$ is obtuse angle bisector

Answer 1

Answer: (B), (C)

$23x-13y+32z+45=0$ is acute angle bisector

Given :
$P_1:2x-y+2z+3=0,P_2:3x-2y+6z+8=0,$
$\Rightarrow D.r^{\prime }s$ of normal to $P_1=(a_1,b_1,c_1)=(2,-1,2)$
$D.r^{\prime }s$ of normal to $P_2=(a_2,b_2,c_2)=(3,-2,6)$ and $d_1,d_2>0$
now, $a_1{a}_2+b_1{b}_2+c_1{c}_2>0$
So, The angle bisector is :
$\frac{P_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=\pm \frac{P_2}{\sqrt{a_2^2+b_2^2+c_2^2}}$
taking negative sign for acute angle bisector,
$\Rightarrow \frac{2x-y+2z+3}{\sqrt{4+1+4}}=-\frac{3x-2y+6z+8}{\sqrt{9+4+36}}$
on solving we get,
$23x-13y+32z+45=0$
and taking positive sign for obtuse angle bisector:
$\Rightarrow \frac{2x-y+2z+3}{\sqrt{4+1+4}}=\frac{3x-2y+6z+8}{\sqrt{9+4+36}}$
on solving we get the obtuse angle bisector is,
$5x-y-4z-3=0$