Question

The equation of the plane bisecting the acute angle between the planes $x-y+z-1=0$ and $x+y+z=2$ is

• A. $x+z=\frac{3}{2}$
• B. $2y=1$
• C. $x-y-z=3$
• D. None of these

Answer 1

The correct option is A $x+z=\frac{3}{2}$

Given planes are $x-y+z-1=\mathrm{0.....}\left(1\right)$ and $x+y+z-2=\mathrm{0.........}\left(2\right)$
Therefore equation of plane bisecting these planes are
$\frac{x-y+z-1}{\sqrt{3}}=\pm \frac{x+y+z-2}{\sqrt{3}}$
$\Rightarrow x+z=\frac{3}{2}.......\left(3\right)$ and $y=\frac{1}{2}.......\left(4\right)$
If $\theta$ is the angle between (2) and (4) then,
$\cos \theta =\frac{1/2}{\left(1/2\right).\left(\sqrt{3}\right)}=\frac{1}{\sqrt{3}}$
$\Rightarrow \theta >{45}^{\circ }$
Hence plane (4) bisects the obtuse angle between the given planes.
Therefore equation of plane bisecting acute angle between given plane is
$x+z=\frac{3}{2}$