Question

Equation of the plane bisecting the acute angle between the planes $x-2y+2z+3=0$ and $3x-6y-2z+2=0$

• A. $2(8x-16y+4z)+27=0$
• B. $8x-16y+4z+27=0$
• C. $16x-32y+8z-27=0$
• D. $16x+32y+8z+27=0$

Answer 1

The correct option is A $2(8x-16y+4z)+27=0$

Equations of planes bisecting the angles between two given planes are

$\frac{x-2y+2z+3}{\sqrt{1^2+2^2+2^2}}=\pm \frac{3x-6y-2z+2}{\sqrt{3^2+6^2+2^2}}$

$\frac{x-2y+2z+3}{3}=\pm \frac{3x-6y-2z+2}{7}$

$2x-4y-20z-15=0$ and $16x-32y+8z+27=0$

Lets find out angle between one of the bisector and one given plane:

Angle between $16x-32y+8z+27=0$ and $x-2y+2z+3=0$ is

$\cos \theta =\frac{Aa+Bb+Cc}{\sqrt{\sum A^2}.\sqrt{\sum a^2}}$

$\cos \theta =\frac{16+64+16}{24\sqrt{21}}=\frac{4}{\sqrt{21}}>\frac{1}{\sqrt{2}}$

$\theta <{45}^0$

Hence, $16x-32y+8z+27=0$ is the equation of required acute angle bisector.