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 parallelopiped so formed.

Answer 1

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)

PE = Distance between the parallel planes ABQF and PCDE = -2 -3 = 5

(These planes are perpendicular to the z-axis)

PA = Distance between are 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 paralleloepiped are 5.5 and 5.