# proof for rectangular parallelepiped

Printable View

• Apr 6th 2010, 04:18 PM
mathwizard325
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
• Apr 6th 2010, 09:21 PM
Soroban
Hello, mathwizard325!

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

Quote:

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: .$\displaystyle a$ in the $\displaystyle x$ direction,
. . $\displaystyle b$ in the $\displaystyle y$ direction, $\displaystyle c$ in the $\displaystyle z$ direction.

Orient it in the first octant with vertex $\displaystyle P$ 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: .$\displaystyle \overrightarrow{QW} \:=\:\langle\, -a,\,b,\,c\,\rangle$

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

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

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