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The distance between the parallel planes `2x-3y+6z=5` and `6x-9y+18z+20=0`, is
A. `5/3` units
B. `5sqrt(3)` units
C. `8/5` units
D. `8sqrt(5)` units

1 Answer

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Correct Answer - a
The given parallel planes are
`2x-3y+6z=5`…………..(i)
`6x-9y+18z+20=0`……………(ii)
Let `P(x_(1),y_(1),z_(1))` be a point on (i). Then `2x_(1)-3y_(1)+6z_(1)=5`…………(iii)
Distance between the given planes
= Length of perpendicular from `P(x_(1),y_(1),z_(1))` on (ii)
`=|6x_(1)-9y_(1)+18z_(1)+20|/sqrt(6^(2)+(-9)^(2)+18^(2)) = |3(2x_(1)-3y_(1)+6z_(1))+20|/sqrt(441)`
`=|(3 xx 5) +20|/21=35/21` units `=5/3` units [using (iii)]

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