InterviewSolution
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InterviewSolution
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Show that the distance between planes `2x-2y+z+3=0 and 4x-4y+2z+5=0 is 1/6` |
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Answer» Equation of first plane `3x+2y-z + 1 = 0` Let `(x_(1),y_(1),z_(1))` be any point on this plane. `:. 3x_(1)+2y_(1)-z_(1) + 1 = 0 "……"(1)` Perpendcicular distance from point `(x_(1),y_(1)z_(1))` to the second plane `3x+2y-z+5 = 0` `=|(3x_(1)+2y_(1)-z_(1)+5)/(sqrt(3^(2)+2^(2)+(-1)^(2)))|` , [From eq.(1)] `= |((-1)+5)/(sqrt(9+4+1))|=4/(sqrt(14))` units Therefore the distance between parallel planes `= (4)/(sqrt(14))` units. |
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