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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?

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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – sdebarun Jan 16 '18 at 15:48
  • $\begingroup$ @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. $\endgroup$ – sammy gerbil Jan 17 '18 at 14:43

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