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?
Solution
${P_1}:x - 2y - 2z + 1 = 0$<br><br>${P_2}:2x - 3y - 6z + 1 = 0$<br><br>$$\left| {{{x - 2y - 2z + 1} \over {\sqrt {1 + 4 + 4} }}} \right| = \left| {{{2x - 3y - 6z + 1} \over {\sqrt {{2^2} + {3^2} + {6^2}} }}} \right|$$<br><br>${{x - 2y - 2z + 1} \over 3} = \pm {{2x - 3y - 6z + 1} \over 7}$<br><br>Since ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 20 > 0$<br><br>$\therefore$ Negative sign will give acute bisector<br><br>$7x - 14y - 14z + 7 = - [6x - 9y - 18z + 3]$<br><br>$\Rightarrow 13x - 23y - 32z + 10 = 0$<br><br>$\left( { - 2,0, - {1 \over 2}} \right)$ satisfy it $\therefore$ Ans. (b)
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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