InterviewSolution
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InterviewSolution
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Find the equation of the plane mid-parallel to the planes `2x-2y+z+3=0` and `2x-2y+z+9=0`. |
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Answer» Let the required equation of the plane be `2x-2y+z+k=0`. This plane equidistant from each of the given planes. Lety `P(x_(1),y_(1),z_(1))` be any point on the plane `2x-2y+z+k=0`. Then, `2x_(1)-2y_(1)+z_(1)+k=0 rArr 2x_(1)-2y_(1)+z_(1)=-k`…………….(i) `therefore P(x_(1),y_(1),z_(1))`is equidistant from the planes `2x-2y+z+3=0` and `2x-2y+z+9=0`. `therefore |2x_(1)-2y_(1)+z_(1)+3|/sqrt(2^(2)+(-2)^(2)+1^(2))=|2x_(1)-2y_(1)+z_(1)+9|/sqrt(2^(2)+(-2)^(2)+1^(2))` `rArr |(2x_(1)-2y_(1)+z_(1)+3)|=|(2x_(1)-2y_(1)+z_(1)+9|` `rArr |-k+3|=(-k+9)` or `-(-k+3)=-k+9` `rArr k-3 =-k+9 rArr 2k=12 rArr k=6 [therefore (-k+3) ne (-k+9)]`. Hence, the required equation of the plane is `2x-2y+z+6=0`. |
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