Question
Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8 z = 12 is
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2 units
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4 units
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8 units
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Solution
D.

2x + 3y + 4z - 4 = 0 ...(1)
and 4x + 6y +8 z – 12 = 0 ...(2)
Here

∴ planes (1) and (2) are parallel.
Let (x1,y1.,z1) be any point on plane (1).
∴ 2x1 + 3y1 + 4 z1 – 4 = 0
or 2x + 31+ 4z1 = 4 ...(3)
Length of perpendicular from (x1, y1, z1,) to the plane (2) is
