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
= $|4 + 1| = 5$
(These planes are perpendicular to the z-axis)
PA = Distance between the parallel planes ABQF and PCDE
= $|- 2 - 3| = 5$
(These planes are perpendicular to the x-axis)
Similarly, PC = $|5 - 0| = 5$
Thus, the length of the edges of the parallelepiped are 5, 5 and 5

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