Question

The equation of the plane bisecting the obtuse angle between the planes $x+y+z=1$ and $x+2y-4z=5$ is

• A. $(\sqrt{7}-1)x+(\sqrt{7}-2)y+(\sqrt{7}+4)z+5-\sqrt{7}=0$
• B. $(\sqrt{7}+1)x+(\sqrt{7}+2)y+(\sqrt{7}+4)z+5-\sqrt{7}=0$
• C. $(\sqrt{7}+1)x+(\sqrt{7}+2)y+(\sqrt{7}-4)z=\sqrt{7}$
• D. None of these

Answer 1

The correct option is D None of these

Given planes are $x+y+z-1=\mathrm{0.....}\left(1\right)$ and $x+2y-4z-5=\mathrm{0.........}\left(2\right)$
Therefore equation of planes bisecting these planes are
$\frac{x+y+z-1}{\sqrt{3}}=\pm \frac{x+2y-4z-5}{\sqrt{21}}$

$\Rightarrow x+y+z-1=\pm \frac{x+2y-4z-5}{\sqrt{7}}$

$\Rightarrow (\sqrt{7}-1)x+(\sqrt{7}-2)y+(\sqrt{7}+4)z=\sqrt{7}+5...\left(3\right)$ and $(\sqrt{7}+1)x+(\sqrt{7}+2)y+(\sqrt{7}-4)z=\sqrt{7}-5....\left(4\right)$
If $\theta$ is the angle between $\left(1\right)$ and $\left(3\right)$, then

$\cos \theta =\frac{(\sqrt{7}-1).1+(\sqrt{7}-2).1+(\sqrt{7}+4).1}{\left(\sqrt{(\sqrt{7}-1{)}^2+(\sqrt{7}-2{)}^2+(\sqrt{7}+4{)}^2}\right).\left(\sqrt{3}\right)}=\frac{3\sqrt{7}+2}{\left(\sqrt{40+2\sqrt{7}}\right).\left(\sqrt{3}\right)}>\frac{1}{2}$

$\Rightarrow \theta >{45}^{\circ }$
Hence, plane $\left(1\right)$ bisects the obtuse angle between the given planes.
Therefore equation of plane bisecting acute angle between given plane is
$(\sqrt{7}-1)x+(\sqrt{7}-2)y+(\sqrt{7}+4)z=\sqrt{7}+5$
Hence, option 'D' is correct.