Volume of parallelepiped defined by vectors \begin{align} \begin{bmatrix}a\\b\\c\end{bmatrix}&,\quad \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix},\quad\begin{bmatrix}w_1\\w_2\\w_3\end{bmatrix} = \det\begin{pmatrix}\begin{bmatrix}a&v_1&w_1\\b&v_2&w_2\\c&v_3&w_3\end{bmatrix}\end{pmatrix}\\ &= a(v_2w_3 - v_3w_2) + b(v_3w_1 - v_1w_3) + c(v_1w_2 - v_2w_1)\\ &= a(\text{Area of parallelogram obtained by projecting parallelepiped on the $yz$ plane})\\ &+b(\text{Area of parallelogram obtained by projecting parallelepiped on the $zx$ plane}) \\ &+c(\text{Area of parallelogram obtained by projecting parallelepiped on the $xy$ plane}) \end{align}

I can visualize the case when any one component of vectors $v$ and $w$ are $0$. Suppose the $z$ component is $0$. So there would not be any parallelogram formed on the $yz$ and $zx$ plane. So the volume would simply be $$=c(\text{Area of parallelogram obtained by projecting parallelepiped on the $xy$ plane})$$ which is simple enough to visualize. But, I am having trouble visualizing when all the three components of $v$ and $w$ are non-zero. Can anyone help?

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  • You may find [this answer](https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant/669#669) useful. – dxiv Aug 20 '18 at 21:18

2 Answers2


I'm not sure there's an intuitive way to visualize a volume calculation as a linear combination of areas, except in the case you have already. It's just the $$ \text{Volume} = \text{Area of base}\cdot \text{height} $$ formula from high school geometry.

But I would encourage you go the other direction, and induct from this case to the more general one. There is a rotation of space which brings $v$ and $w$ into the $xy$-plane. This rotation preserves determinants. So the volume formula has to be the same.

Intuitively, volume is a little more intrinsic to space than the axes or coordinates we install to describe space. So we can choose our coordinates/axes to make the visualization convenient.

Matthew Leingang
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First consider the case for a cuboid. The three edges are orthogonal to each other. Lets represent these edges as columns of a matrix $A$.

Now $A^T$ will have the rows as these edges and multiplying both we get, $$ A^TA = \begin{bmatrix}e_1\\e_2\\e_3\end{bmatrix}\begin{bmatrix}e_1&e_2&e_3\end{bmatrix} $$ where $e_i$ are edge vectors with lengths $l_i$ i.e. $\Vert e_i^Te_i \Vert = l_i^2$

This gives $$ A^TA = \begin{bmatrix}l_1^2&&0 \\& \ddots &\\0&&l_3^2\end{bmatrix} $$

Now $det(A^TA)$ is product of diagonal elements for a diagonal matrix and we get $det(A) = l_1^2l_2^2l_3^2$

But we know the volume of cuboid is $l_1l_2l_3$ and $det(A^T) = det(A)$. So we get $$ det(A^TA) = det(A)^2 = (l_1l_2l_3)^2 $$ And so, volume of a cuboid = $det(A)$.

Now if the edges are not orthogonal to each other, we get a parallelopiped.

$$ det(A) = \begin{vmatrix}e_1\\e_2\\e_3\end{vmatrix} = \begin{vmatrix}e_1\\e_2-q_2\\e_3-q_3\end{vmatrix} $$

where $q_i$ are the projection vectors onto the edges, so that $e_i-q_i$ is made perpendicular to the edge $e_{i-1}$. Now these new vectors are made orthogonal to each other while preserving the determinant value, which gives the determinant of edge matrix gives the volume of parellelopiped and can be extended to any number of dimensions.

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