# Thread: proof for rectangular parallelepiped

1. ## proof for rectangular parallelepiped

Prove that if two diagonals of a rectangular parallelepiped are perpendicular, then its dimensions are equivalent to the sides of a right triangle and vice versa

2. Hello, mathwizard325!

I've proved it one way so far . . . a vector proof.

Prove that if two diagonals of a rectangular parallelepiped are perpendicular,
then its dimensions are equivalent to the sides of a right triangle and vice versa.

The "box" has dimensions: . $a$ in the $x$ direction,
. . $b$ in the $y$ direction, $c$ in the $z$ direction.

Orient it in the first octant with vertex $P$ at the origin.

The graph looks like this:

Code:
        T   |                W
(0,0,c)* - - - - - * (0,b,c)
/|          /|
/ |         / |
U     /  |        /  |
(a,0,c) * - - - - - * V |
|   |       |   |    S
| P * - - - | - * (0,b,0)
|  /        |  /
| /         | /
Q    |/          |/
(a,0,0) * - - - - - * R
/

One diagonal is: . $\overrightarrow{QW} \:=\:\langle\, -a,\,b,\,c\,\rangle$

. . . . Another is: . $\overrightarrow{SU} \:=\:\langle\, a,\,-b,\,c\,\rangle$

Since $\overrightarrow{QW} \perp \overrightarrow{SU}\!:\;\;\overrightarrow{QW}\cdot \overrightarrow{SU} \:=\:0 \quad\Rightarrow\quad \langle -a,\,b,\,c\rangle\cdot \langle a,\,-b,\,c\rangle\:=\:0$

Therefore: . $-a^2 - b^2 + c^2 \:=\:0 \quad\Rightarrow\quad a^2 + b^2 \:=\:c^2$