Question

Equation of the plane bisecting the acute angle between the planes $x+2y-2z-9=0,3x-4y+12z-26=0$ is

• A. $2(4x+17y-31z)+36=0$
• B. $8x-16y+4z+27=0$
• C. $16x-32y+8z-27=0$
• D. None of these
  

Answer 1

The correct option is D None of these

The equation of the given planes are

$3x-4y+12z-26=0$ ...$\left(1\right)$

$x+2y-2z-9=0$ ...$\left(2\right)$

$\therefore$the equations of the planes bisecting the angles between them are $\frac{3x-4y+12z-26}{\sqrt{9+16+144}}=\pm \frac{x+2y-2z-9}{\sqrt{1+4+4}}$

$\Rightarrow 3(3x-4y+12z-26)=\pm 13(x+2y-2z-9)$

$\Rightarrow 4x+38y-62z-39=0$ ...$\left(3\right)$

and $22x+14y+102-195=0$ ...$\left(4\right)$

If $\theta$ isthe angle between the planes $\left(4\right)$ and $\left(2\right)$, we have

$\cos \theta =\frac{1\left(22\right)+2\left(14\right)-2\left(10\right)}{\sqrt{1+4+4}.\sqrt{484+196+100}}=\sqrt{\frac{5}{39}}$

$\Rightarrow \sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{5}{39}}=\sqrt{\frac{34}{39}}$

$\Rightarrow \tan \theta =\sqrt{\frac{34}{5}}>1\Rightarrow \theta >{45}^O$

Hence, the plane $\left(4\right)$ bisects the obtuse angle between the given plane. Thus the other plane $\left(3\right)$ bisects the acute angle.

$\therefore 4x+38y-62z-36=0$