The equation of the parallel plane lying midway between the parallel planes $2 x - 3 y + 6 z - 7 = 0$ and $2 x - 3 y + 6 z + 7 = 0$

2 x - 3 y + 6 z + 1 = 0

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2 x - 3 y + 6 z - 1 = 0

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2 x - 3 y + 6 z = 0

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2 x - 3 y + 6 z = 11

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Solution

The correct option is C $2 x - 3 y + 6 z = 0$
Since, the plane is parallel to the $2$ given planes, it will have the same normal.
Hence, the plane is $2 x - 3 y + 6 z = d$
Since, it lies midway between the two planes $2 x - 3 y + 6 z = 7$ and $2 x - 3 y + 6 z = - 7$ , it has to be equidistant from them.
Hence, $d = 0$

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is ___

A line having direction ratios $3, 4, 5$ cuts $2$ planes $2 x - 3 y + 6 z - 12 = 0$ and $2 x - 3 y + 6 z + 2 = 0$ at point P & Q, then Find length of PQ

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The equation of the parallel plane lying midway between the parallel planes 2x−3y+6z−7=0 and 2x−3y+6z+7=0

The correct option is C 2x−3y+6z=0Since, the plane is parallel to the 2 given planes, it will have the same normal.Hence, the plane is 2x−3y+6z=dSince, it lies midway between the two planes 2x−3y+6z=7 and 2x−3y+6z=−7, it has to be equidistant from them. Hence, d=0

The equation of the parallel plane lying midway between the parallel planes $2 x - 3 y + 6 z - 7 = 0$ and $2 x - 3 y + 6 z + 7 = 0$