# Circular area as vector

It might sound lame, but can all the area be defined as vector quantity? I understand how the area of a parallelogram or a triangle is a vector. But when it comes to a circle, I don't understand. Say it is $\pi r^2$.

Is that something $r\cdot r$ = $r^2\sin0$ = r^2?

And also in parallelogram we define area as vector cross product. So can the cross product only define area as a vector? And if yes, why can't dot products do the same?

Its merely the formalism of vector algebra that allows us to do this. This is only true in 3d, in higher dimensions this can't be maintained: there is no cross product for example in 4d or higher. So we can't take the product of vectors and get a vector in general.

This means areas, volumes and the like are not really represented by vectors.

Instead what are used are wedge products and these give the signed area of a parallelogram as a 'bivector' such as $u \wedge v$, (where $u$ and $v$ are vectors indicating the two sides) or the signed volume of a parallelopid as a 'trivector' $u \wedge v \wedge w$ (where $u$, $v$ and $w$ are vectors indicating the three sides). In a sense, this is more natural as it follows our intuition more closely rather than representing an area by a 1d vector.

You may be confusing two different aspects of a vector cross product.

The absolute value of a vector cross product is: $\vec{a}x\vec{b}=|\vec{a}||\vec{b}|sin(\vec{a},\vec{b}).$

This"happens" to be the area of a parallelogram with edges $\vec{a}$ and $\vec{b}$. Proof is immediate, because $|\vec{b}|sin(\vec{a},\vec{b})$ is height to the base $\vec{a}$. You can push this idea further and calculate area of a triangle as half the vector cross product of two edges. But that's about it. You cannot use vector cross product for calculating general areas such as a circle.

The second aspect relates to the fact that a vector cross product (not its absolute value) is a vector by itself with a direction of its own. The direction is perpendicular to both vectors ($\vec{a}$ and $\vec{b}$) or perpendicular to the plane containing these two vectors. This aspect is useful in many fields but not necessarily related to areas.

A dot product is defined as $|\vec{a}||\vec{b}|cos(\vec{a},\vec{b})$ and this cannot be used for calculating an area.

• It is a very good explanation.. but i am thinking then can't we define "any" area as a vector? What if the direction of the area is important. – sdebarun Jan 16 '18 at 15:48
• @D.Saha Yes you can define any plane area as a vector $A\hat{n}$ where $A$ is the magnitude of area and $\hat{n}$ is the 'outwards' unit normal. – sammy gerbil Jan 17 '18 at 14:43