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KEAM 2012 Math Paper

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Question : 96 of 120

Marks: +1, -0

The equation of the plane which is equidistant from the two parallel planes

2x-2y + z + 3 = 0

4x-4y + 2z + 9 = 0

is

$8x-8y+4z+15=0$
$8x-8y+4z-15=0$
$8x-8y+4z+3=0$
$8x-8y+4z-3=0$
$8x-8y+4z+4=0$

Solution:

Let the equation of plane parallel between

2x-2y+z+3=0

and

2x-2y+z+\frac{9}{2}=0

is

2x-2y+z+c=0

Now, distance between two given parallel lines

=\frac{\frac{9}{2}-3}{\sqrt{2^{2}+2^{2}+1^{2}}}=\frac{\frac{3}{2}}{3}=\frac{1}{2}

\because

Required plane is equidistant from given planes

\therefore \frac{c-3}{\sqrt{2^{2}+2^{2}+1^{2}}}=\frac{\frac{1}{2}}{2}

\Rightarrow \frac{c-3}{3}=\frac{1}{4}

\Rightarrow c=\frac{3}{4}+3

\Rightarrow c=\frac{15}{4}

\therefore

Required equation of plane is

\therefore 2x-2y+z+\frac{15}{4}=0

\Rightarrow 8x-8y+4z+15=0

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