Let the acute angle bisector


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?

  1. $\left(3,1,-\frac{1}{2}\right)$

  2. $\left(-2,0,-\frac{1}{2}\right)$

  3. $(0,2,-4)$

  4. $(4,0,-2)$

Correct Option: , 2


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

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

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

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

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)

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