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Question
Planes are drawn parallel to the coordinate planes through the points (3, 0, –1) and (–2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.
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Solution
Let P\[\equiv\](3, 0, −1), Q\[\equiv\](−2, 5, 4)
PE = Distance between the parallel planes ABCP and FQDE
=\[\left| 4 + 1 \right| = 5\]
(These planes are perpendicular to the z-axis)
PA = Distance between the parallel planes ABQF and PCDE
= \[\left| - 2 - 3 \right| = 5\]
(These planes are perpendicular to the x-axis)
Similarly, PC =\[\left| 5 - 0 \right| = 5\]
Thus, the length of the edges of the parallelepiped are 5, 5 and 5
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