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Question
Planes are drawn through the points (5, 0, 2) and (3, –2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed.
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Solution
Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.
\[\therefore\]PA = Distance between planes PBEC and QDAF
= 5\[-\]3
= 2
PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and\[-\]2,respectively, from the zx-plane.
\[\therefore\]PB = 0
\[-\](\[-\]2)
= 2
PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.
\[\therefore\]PC = 2\[-\]5
=\[-\]3
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