Question

The equation of the plane bisecting the obtuse angle between the planes $2x-3y+z+2=0$ and $x+2y-3z+1=0$ is

• A. $3x-y-2z+3=0$
• B. $3x+y+2z-3=0$
• C. $x-5y+4z+1=0$
• D. $Noneofthese$

Answer 1

Answer: (A), (C)

The correct options are
A $3x-y-2z+3=0$
C $x-5y+4z+1=0$

We have

$2x-3y+z+2=0......\left(1\right)$

$x+2y-3z+1=0......\left(2\right)$

The bisecting planes are

$\frac{2x-3y+z+2}{\sqrt{2^2+{\left(3\right)}^2+{\left(1\right)}^2}}=\pm \frac{x-2y-3z+1}{\sqrt{1^2+2^2+{(-3)}^2}}$

$\Rightarrow \frac{2x-3y+z+2}{\sqrt{14}}=\pm \frac{x+2y-3z+1}{14}$

$\Rightarrow 2x-3y+z+2=\pm x+2+2y-3z+1$

Taking $+ive$ Sign.

$2x-3y+z+2=x+2+2y-3z+1$

$\Rightarrow 2x-x-3y-2y+z+32+2-1=0$

$\Rightarrow x-5y+4z+1=0$

Taking $–ive$ Sign and we get.

$2x-3y+z+2=-(x+2y-3z+1)$

$2z-3y+z+2+x+2y-3z+1=0$

$3x-y-2z+3=0$

Hence, this is the answer.