Question

Let the acute angle bisector of the two planes $x-2 y-2 z+1=0$ and $2 x-3 y-6 z+1=0$ be the plane P. Then which of the following points lies on $P$ ?

• A. $\left(3,1,-\frac{1}{2}\right)$
• B. $\left(-2,0,-\frac{1}{2}\right)$
• C. $(0,2,-4)$
• D. $(4,0,-2)$

Answer 1

Answer: (B)

$P _{1}: x -2 y -2 z +1=0$
$P _{2}: 2 x -3 y -6 z +1=0$
$\left|\frac{x-2 y-2 z+1}{\sqrt{1+4+4}}\right|=\left|\frac{2 x-3 y-6 z+1}{\sqrt{2^{2}+3^{2}+6^{2}}}\right|$
$\frac{x-2 y-2 z+1}{3}=\pm \frac{2 x-3 y-6 z+1}{7}$
Since $a_{1}\, a_{2}+b_{1} \,b_{2}+c_{1} \,c_{2}=20>0$
$\therefore $ Negative sign will give acute bisector
$7 x-14 y-14 z+7$
$=-[6 x-9 y-18 z+3]$
$\Rightarrow 13 x-23 y-32 z+10=0$
$\left(-2,0,-\frac{1}{2}\right)$ satisfy it