Question

Planes are drawn through the points d(5, 0, 2) and (3, -2, 5)

parallel to the coordinate planes. Find the lengths of the edges of

the rectangular parallelopiped so formed.

Answer 1

Clearly, PBEC and QDAF are the plane parallel to 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 distance 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 distance 2 and 5, respectively, from the xy-plane.

$\therefore$ PC = 2 - 5 = -3