Planes are drawn parallel to the coordinate planes through the point `P(x_1, y_1,\ z_1)` and `Q(x_2, y_2, z_2)` . Find the length of the edges of the parallelopiped so formed.

Options

A. `x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)`
B. `x_(2)+x_(1),y_(2)+y_(1),z_(2)+z_(1)`
C. `x_(1)x_(2),y_(1)y_(2),z_(1)z_(2)`
D. none of these

Answer

A

Explanation

Clearly, `PA,PB` and `PC` are the lengths of the edges of the parallelopiped shown in fig.
Clearly `PBEC QDAC` are planes parallel to yz-plane such that their distances from yz-plane are `x_(1)` and `x_(2)` respectively. So,
`PA=` Distance between the planes `PBEC` and `QDAF`
`=x_(2)-x_(1)`

PB is the distance between the planes `PAFC` and `BDQE` which are parallel to zx-plane and are at distances `y_(1)` and `y_(2)`, respectively, from zx-plane.
`:.PB=y_(2)-y_(1)`
Similarly, PC is the distance between the parallel planes PBDA and CEQF which are at distances `z_(1)` and `z_(2)` respectively, from xy-plane.
`:. PC=z_(2)-z_(1)`