The volume of a parallelepiped $p_2$ spanned by the face diagonals of another parallelepiped $p_1$ is twice the volume of $p_1$.

Question

I would like to purely geometrically (without referring to the dot and cross vector product) prove the following:

The volume of a parallelepiped $p_2$ spanned by the face diagonals of another parallelepiped $p_1$ is twice the volume of the $p_1$, i. e. $V_{p_2}=2V_{p_1}$.

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The statement follows easily from the definition:

Let $\vec a,\vec b,\vec c$ be vectors of the sides with the same origin in a vertex of the parallelepiped $p_1$

$$\begin{aligned}V_{p_2}&=\left(\vec b+\vec c\right)\cdot\left(\left(\vec a+\vec c\right)\times\left(\vec a+\vec b\right)\right)\\&=\left(\vec b+\vec c\right)\cdot\left(\underbrace{\vec a\times\vec a}_{=0}+\vec a\times\vec b+\vec c\times\vec a+\vec c\times\vec b\right)\\&=\underbrace{\vec b\cdot\left(\vec a\times\vec b\right)}_{=0}+\vec b\left(\vec c\times\vec a\right)+\underbrace{\vec b\cdot\left(\vec c\times\vec b\right)}_{=0}+\vec c\cdot\left(\vec a\times\vec b\right)+\underbrace{\vec c\cdot\left(\vec c\times\vec a\right)}_{=0}+\underbrace{\vec c\cdot\left(\vec c\times\vec b\right)}_{=0} \end{aligned}$$

Obviously $\vec c\cdot\left(\vec a\times\vec b\right)=\vec c\cdot\left(\left(\vec a+\vec c\right)\times\left(\vec a+\vec b\right)\right)$, so the determinant won't be changed by adding these rows, i. e., $$\begin{vmatrix}c_1&c_2&c_3\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}=\begin{vmatrix}c_1&c_2&c_3\\a_1+c_1&a_2+c_2&a_3+c_3\\a_1+b_1&a_2+b_2&a_3+b_3\end{vmatrix}.$$ This means the volume will remain the same as long as the new parallelepiped is spanned by at least one former vector side.

We can interpret it like this: Let $ABCDEFGH$ be an arbitrary parallelepiped and let

$\begin{aligned}\vec a&=\overrightarrow{AB}\\\vec b&=\overrightarrow{BC}=\overrightarrow{AD}\\\vec c&=\overrightarrow{AE}\\\vec a+\vec b&=\overrightarrow{AC}\\\vec b+\vec c&=\overrightarrow{ AH}\end{aligned}$.

Let $I\in CD$ s. t. $\overrightarrow{CI}=\vec a$ then the area of the parallelogram $ABIC$ spanned by the vectors $\vec a$ and $\vec a+\vec b$ equals the area of the parallelogram $ABCD$ spanned by the vectors $\vec a,\vec b$.

Next, let $J, K$ be points s .t. $\overrightarrow{AB}=\overrightarrow{IJ}=\overrightarrow{CK}$.

Then the parallelepipeds $ABCDEFGH$ and $ABICHGJK$ have equal heights and bases, and hence, equal volumes.

Let points $L,M,N$ be s. t. $\overrightarrow{NL}=\overrightarrow{AH}=\overrightarrow{BG}$.

Then $ACKH\cong BIJG\cong FNLM$.

Picture:

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However, I don't know how to continue proving the volume of $AFMHCNLK$ is twice the volume of $ABICHGJK$.

May I ask for advice on solving this task?

Thank you in advance!

Answer 1

Here's a start on a more purely geometric proof:

Put the origin at one vertex and change coordinates so that the parallelepiped becomes the unit cube. You can do that by choosing the three edges at the origin as the basis vectors.

That change of coordinates scales all volumes the same way, so preserves the proportion you are interested in.

For the unit cube, the face diagonals are $(0,1,1)$ , $(1,0,1)$ and $(1,1,0)$. The parallelepiped they determine has volume $2$ because the Jacobian of the change of coordinate transformation to that coordinate system is the determinant $$ \begin{array}{|ccc|} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array} = 2. $$

Answer 2

You have used the formula $\vec{u}\cdot(\vec{v}\times\vec{w})$ for the volume of the parallelopiped spanned by $\vec{u}$, $\vec{v}$, and $\vec{w}$. Indeed this is how you calculated the volume of $p_2$. So in the final expression for $V_{p_2}$ you can recognize the terms $\vec{b}\cdot(\vec{c}\times\vec{a})$ and $\vec{c}\cdot(\vec{a}\times\vec{b})$ both as the volume of $V_{p_1}$, which should conclude the proof.

Note that the triple product is invariant under cyclic permutation, so these two expressions are both equal to $\vec{a}\cdot(\vec{b}\times\vec{c})$. This should make sense, because the order in which we name the three edges of a parallelopiped intersecting at a vertex shouldn't matter, as long as the orientation is the same. Alternatively, you can also make sense of this from the fact that determinants are invariant under cyclic permutation of rows.