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?
Correct Option: , 2
Solution:
$P_{1}: x-2 y-2 z+1=0$
$P_{2}: 2 x-3 y-6 z+1=0$
$\left|\frac{x-2 y-2 z+1}{\sqrt{1+4+4}}\right|=\left|\frac{2 x-3 y-6 z+1}{\sqrt{2^{2}+3^{2}+6^{2}}}\right|$
$\frac{x-2 y-2 z+1}{3}=\pm \frac{2 x-3 y-6 z+1}{7}$
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 $\therefore$ Ans (2)