# How is “area” a vector?

"We consider Area as a vector."

How is an area a vector? Why is that the vector is always normal to the area element?

• Short answer: because the space of bivectors has the same dimension as the space of vectors in 3D space. You can't cheat this way in any other number of dimensions. – Dietrich Epp Jan 3 '13 at 23:45
• – Qmechanic Jun 17 '13 at 19:16

It's just a definition that is mostly useful in physics. If you want to integrate the flow of a liquid or an electromagnetic field out of a given surface (say the surface of a ball), you would like to multiply each differential surface area by the corresponding flow. But what if the flow isn't perpendicular to the surface? in this case we want to multiply the flow (which is a vector) with the surface normal.

So instead of writing: $${\bf{v}}\cdot{\bf{n}}\ dA$$ We can now write: $${\bf{v}}\cdot{\bf{dA}}$$

• Excellent explanation. – Inquisitive Jan 3 '13 at 19:57

It's a cheat, a mathematical sleight of hand. In 3d, a planar subspace will always have a unique normal direction, and so you can get away with using those normal vectors instead of introducing a lot of extra mathematical framework. Using normal vectors makes sense because these vectors are unique up to a sign and magnitude.

In spaces with more than three dimensions, planes no longer have normal vectors, and dealing with other frameworks to handle such objects is unavoidable.

Edit: using clifford algebra, you can directly handle areas (planes) as bivectors. The product of vectors that produces such objects is not the cross product but the wedge product. The wedge product is anticommutative, so $a \wedge b = - b \wedge a$, similar to the cross product, and in addition $a \wedge a = 0$.

Here's an example. Let $u = 3\hat x + 2 \hat y$ and $v = 5 \hat y + 2 \hat z$. The area of a parallelogram spanned by these two vectors is

$$A = u \wedge v = (3 \hat x + 2 \hat y) \wedge (5 \hat y + 2 \hat z) = 15 \, \hat x \wedge \hat y + 6 \, \hat x \wedge \hat z + 4 \, \hat y \wedge \hat z$$

Compare this with the corresponding calculation using the cross product. The components are all the same, but the interpretation of the object is different. Again, the main reason for doing this instead of using pseudovectors is that this approach is valid in any number of dimensions. In addition, some aspects of linear transformations are not obvious using pseudovectors (why should a pseudovector transform differently than the vectors used to build it?) but can be seen more intuitively using bivectors. One such case is inversion through the origin.

• Muphrid: "and so you can get away with using those normal vectors instead of introducing a lot of extra mathematical framework." Do you have some links to show how same calculations can be done without considering Area as a vector. – Inquisitive Jan 3 '13 at 19:56
• @Inquisitive I've added a section on how to use bivectors. There are many approaches to going beyond vector algebra. Consider searching for information on tensors in general, differential forms, and clifford algebra. These are somewhat advanced, though. – Muphrid Jan 3 '13 at 20:27

Actually, you can have a n-dimensional cross-product which generalizes the 3-dimensional cross-product.

If $A_i, i=1, n-1$ are $n-1$ $n$-dimensional vectors with the elements of $A_i$ being $a_{i,1}, a_{1,2}, ..., a_{i, n}$, and the $n$ unit vectors are $e_1, ..., e_n$, then the following determinant is a n-dimensional vector orthogonal to all the $A_i$:

$\left| \begin{array}{cccc} e_1 & e_2 & ... & e_n \\ A_{1,1} & A_{1,2} & ... & A_{1,n} \\ A_{2,1} & A_{2,2} & ... & A_{2,n} \\ ... & ... & ... & ... \\ A_{k,1} & A_{k,2} & ... & A_{k,n} \\ ... & ... & ... & ... \\ A_{n-1,1} & A_{n-1,2} & ... & A_{n-1,n} \\ \end{array} \right|$

• Indeed. What's going on here is that one builds up the $n-1$ dimensional object spanned by the vectors, which has only one direction orthogonal to it. The cross product is an operation built on duality, which is a well-founded notion in any number of dimensions. – Muphrid Jan 3 '13 at 22:49