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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.
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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