Question

Let two planes $p_1:2x-y+z=2$, and $p_2:x+2y-z=3$ are given. The equation of the acute angle bisector of planes $P_1$ and $P_2$ is

• A. $x-3y+2z+1=0$
• B. $3x+y-5=0$
• C. $x+3y-2z+1=0$
• D. $3x+z+7=0$

Answer 1

Answer: (B)

$3x+y-5=0$

Given, $2x-y+z=2$ ...$\left(i\right)$

and $x+2y-z=3$ ...$\left(ii\right)$

$\therefore$ Equation of the planes bisecting the angles between them are. $\frac{2x-y+z-2}{\sqrt{4+1+1}}=\pm \frac{x+2y-z-3}{\sqrt{1+4+1}}$

$\Rightarrow 2x-y+z-2=\pm x+2y-z-3$ ...$\left(iii\right)$

and $3x+y-5=0$ ...$\left(iv\right)$

If $\theta$ be the angle between the plane $\left(iv\right)$ and $\left(ii\right)$, we have $\cos \theta =\frac{1\left(3\right)+2\left(1\right)-2\left(0\right)}{\sqrt{1+4+1}\sqrt{9+1+25}}=\frac{5}{\sqrt{210}}$

$\Rightarrow \tan \theta =\frac{5}{\sqrt{185}}<1$

$\therefore \theta <{45}^o$

Hence, equation of the acute angle of bisects is $3x+y-5=0$.