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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 ?
Option: 1 (0,2,-4)
Option: 2 \left(-2,0,-\frac{1}{2}\right)
Option: 3 (4,0,-2)
Option: 4 \left(3,1,-\frac{1}{2}\right)

Answers (1)

best_answer

The angle bisectors are

\frac{x-2y-2z+1}{\sqrt{1^{1}+2^{2}+2^{2}}}= \pm \frac{2x-3y-6z+1}{\sqrt{2^{2}+3^{2}+6^{2}}}

\Rightarrow 7\left ( x-2y-2z+1 \right )= \pm 3\left ( 2x-3y-6z+1 \right )\Rightarrow7x-14y-14z+7= \pm \left ( 6x-9y-18z+3 \right )

\Rightarrow 13x-23y-32y+10= 0----\left ( 1 \right )
    OR     x-5y+4z+4= 0---\left ( 2 \right )

Take a point (-1,0,0) on plane 1 and find its distance from both angle bisectors
d_{1}= \left | \frac{-13+10}{\sqrt{13^{2}+23^{2}+32^{2}}} \right |= \frac{3}{\sqrt{1722}}
d_{2}= \left |\frac{-1}{\sqrt{1^{2}+5^{2}+4^{2}}}\right |= \frac{1}{\sqrt{42}}= \frac{3}{\sqrt{368}}

Clearly d_{1}< d_{2}

So plane in equation (1) is acute angle Bisector
13x-23y-32z+10= 0
Out of the given option, \left ( -2,0,\frac{-1}{2} \right ) satisfies this plane

Posted by

Kuldeep Maurya

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