Let the acute angle bisector of the two planes x-2y-2z+1=0 and 2x-3y-6z+1=0 be the plane P…

Question

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 $P$?

Accepted Answer

${P_1}:x - 2y - 2z + 1 = 0$

${P_2}:2x - 3y - 6z + 1 = 0$

$$\left| {{{x - 2y - 2z + 1} \over {\sqrt {1 + 4 + 4} }}} \right| = \left| {{{2x - 3y - 6z + 1} \over {\sqrt {{2^2} + {3^2} + {6^2}} }}} \right|$$

$${{x - 2y - 2z + 1} \over 3} =  \pm {{2x - 3y - 6z + 1} \over 7}$$

Since $${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 20 > 0$$

$$\therefore$$ Negative sign will give acute bisector

$$7x - 14y - 14z + 7 =  - [6x - 9y - 18z + 3]$$

$$ \Rightarrow 13x - 23y - 32z + 10 = 0$$

$$\left( { - 2,0, - {1 \over 2}} \right)$$ satisfy it $$\therefore$$ Ans. (b)

Solutions · 1