# Can we get area vector of any flat shape using the cross product?

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

## 3 Answers

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: $$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin(\vec{a},\vec{b}).$$

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

Any small segment of an area of any shape (in 3D) can be represented by a vector perpendicular to the surface. It may or may not be convenient to do this with a cross product. (The two vectors you start with must lie in the surface.) A dot product does not yield a vector result. If you are working with a closed surface, it may be customary to define the positive vectors as pointing outward.