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 $P ?$

(- 2, 0, - \frac{1}{2})

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(0, 2, - 4)

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(4, 0, - 2)

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(3, 1, - \frac{1}{2})

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Solution

The correct option is A $(- 2, 0, - \frac{1}{2})$
$P_{1} \equiv x - 2 y - 2 z + 1 = 0, P_{2} \equiv 2 x - 3 y - 6 z + 1 = 0$
Pair of bisectors be
$\frac{x - 2 y - 2 z + 1}{3} = \pm \frac{2 x - 3 y - 6 z + 1}{7}$
As $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 1 (2) + (- 2) (- 3) + (- 2) (- 6) > 0$

$-$ ve sign gives acute angle bisector
i.e., $7 (x - 2 y - 2 z + 1) = - 3 (2 x - 3 y - 6 z + 1)$
$\Rightarrow 13 x - 23 y - 32 z + 10 = 0$
Clearly $(- 2, 0, - \frac{1}{2})$ satisfy above plane.

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