Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane P. Then which of the following points lies on P?
(A) $(0,2,-4)$
(B) $(4,0,-2)$
(C) $(-2,0,-\frac {1}{2})$
(D) $(3,1,-\frac {1}{2})$
Solution
Bisectors are given by,
$\frac{{x – 2y – 2z + 1}}{3} = \pm \frac{{2x – 3y – 6z + 1}}{7}$
$ \Rightarrow 7x – 14y – 14z + 7 = \pm (6x – 9y – 18z + 3)$
Hence, $x-5y+4z+4=0$ & $13x-23y-32z+10=0$
Let $\theta $ be the angle between $x-2y-2z+1=0$ & $x-5y+4z+4=0$
$\cos \theta = \frac{{|1 + 10 – 8|}}{{3 \times \sqrt {42} }} = \frac{1}{{\sqrt {42} }} < \frac{1}{{\sqrt 2 }}$
$\theta > 45^\circ $
So, $x-5y+4z+4=0$ is the obtuse angle bisector.
$\therefore $ Acute angle bisector $P \equiv 13x – 23y – 32z + 10 = 0$
The point $(-2,0,-\frac {1}{2})$ satisfies P.
Answer: (C)