# volume of a parallelogram [duplicate]

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Determinants and volume of parallelotopes

Can you give me a direction about how to prove that $|det(UVW)|$ is the 3D volume of the parallelogram that defined by $U, V$ and $W$, for $U, V, W \in \mathbb{R}^{3}$ ?

Thanks.

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## marked as duplicate by Phira, t.b., David Speyer, Asaf Karagila, Ｊ. Ｍ.Nov 25 '11 at 0:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The determinant works for rectangular axis-parallel boxes, and is unchanged under the operation "add a multiple of one column to another" (which ought to leave the volume unchanged by geometric considerations). This determines its value for all nontrivial parallelipeds. –  Henning Makholm Nov 20 '11 at 7:16
A nit-pick: The 3-dimensional analog of a parallelogram is called a "parallelopiped". –  Austin Mohr Nov 20 '11 at 7:39
No, parallelepiped. There's the Greek "epi" in there. –  Phira Nov 20 '11 at 8:04
Not a duplicate as far as I can see. The earlier question asks whether there's a generalization to dimension $>3$; this one seeks a proof for the 3-dimensional case. An answer to one is not the same as an answer to the other. –  Henning Makholm Nov 21 '11 at 2:03

## 1 Answer

First note that $a=\frac{V\times W}{\|V\times W)\|}$ is an unit vector orthogonal to the plane spanned by $V$ and $W$. If $\theta$ is the angle between $U$ and $a$, then $U\cdot a=\|U\|\|a\|\cos\theta=\|U\|\cos\theta$ since $a$ is an unit vector. Notice that $a\cdot U$ is the height $h$ of the parallelepiped formed by $U$, $V$ and $W$ with the parallelogram spanned by $V$ and $W$ as the base (It would be easier to understand if you draw a picture), i.e. $$h=a\cdot U.$$ On the other hand, the area $A$ of the base is equal to the area the parallelogram spanned by $V$ and $W$, that is $$A=\|V\times W\|.$$ Therefore, the volume of the parallelepiped is given by $$hA=(U\cdot a)\|V\times W\|=U\cdot (V\times W)$$ since $a=\frac{V\times W}{\|V\times W)\|}$. Now the result follows from $$\det(U,V,W)=U\cdot(V\times W).$$

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That's answer my question, thanks. –  ofer Nov 20 '11 at 8:21
Is there a elegant way to expand this prove to $\mathbb{R}^n$? –  ofer Nov 20 '11 at 10:55