Alright, so I do not have any issues with calculating the area between two vectors. That part is easy. Everywhere that I looked seemed to explain how to calculate the area, but not why the cross product is used instead of the dot product.

I was hoping math.se could explain this, it has been irking me too much. I understand the role of the cross product in torque and magnetism, but this one escapes me.


5 Answers 5


Here is the "right way" to think about the cross product:

For any two vector $A,B \in \mathbb{R}^3$, there is a linear map $L_{A,B} : \mathbb{R}^3 \to \mathbb{R}$ defined by $L_{A,B}(C) = Det(A,B,C)$.

To be explicit, if $A = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} $, $B = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} $, $C = \begin{bmatrix} x \\ y \\ z \end{bmatrix} $,

then $L_{A,B}(C) = \begin{vmatrix} a_1 & b_1 & x \\ a_2 & b_2 & y \\ a_3 & b_3 & z \end{vmatrix} = (a_2b_3 - a_3b_2)x + (a_3b_1 - a_1b_3)y+(a_1b_2 - a_2b_1)z$

Any linear map from $\mathbb{R}^n \to \mathbb{R}$ can be represented by a dot product against some fixed vector, and this one is no exception. In fact, from the formula above we see that

$$ L_{A,B}(C) = \begin{bmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} $$

This is a pretty special vector, and it depends only on $A$ and $B$, so let's define $$A \times B = \begin{bmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{bmatrix}$$

Now we can rewrite this as $L_{A,B}(C) = (A\times B)\cdot C$. Recalling the definition of $L_{A,B}$, we see that

$$\color{blue}{Det(A,B,C) = (A\times B)\cdot C}$$

This is the defining feature of the cross product, and the most important thing to realize about it. In words:

$$\color{blue}{\text{The signed volume of the parallelepiped formed by $A,B$ and $C$}}$$

$$\color{blue}{\text{is the dot product of $A\times B$ with $C$ }}$$

Now we can immediately see that $A\times B$ is perpendicular to both $A$ and $B$:

$$(A\times B)\cdot B = Det(A,B,B) = 0$$ $$\text{and}$$ $$(A \times B)\cdot A = Det(A,B,A) = 0$$

Now we can try to compute the length of $A\times B$. $$ \left|\left|(A\times B)\right|\right|^2 = (A\times B) \cdot (A\times B) = Det(A,B,A\times B)$$

But geometrically, $Det(A,B,A\times B)$ is the signed volume of the parallelepiped formed by $A$, $B$, and $A\times B$. By the above equation, this sign is positive (since it is equal to a square). Since $A\times B$ is perpendicular to both $A$ and $B$, then by geometry, the volume of the parallelepiped is the length of $A \times B$ times the area of the parallelogram formed by $A$ and $B$. So

$$\left|\left|(A\times B)\right|\right|^2 = \left|\left|(A\times B)\right|\right| (\text{Area of parallelogram spanned by $A$ and $B$})$$

So finally,

$$\left|\left|(A\times B)\right|\right| = \text{Area of parallelogram spanned by $A$ and $B$}$$

This is a little longer than the other answers, but hopefully it show you where the cross product really comes from and why it is natural. In particular, the characterizing property in blue above is THE thing you should remember about cross products. All the other properties flow from that one (perpendicularity, pointing in positive direction, and length).


For geometric reasons the area of the parallelogram $P$ spanned by two vectors ${\bf a}$ and ${\bf b}$ scales with $|{\bf a}|$ and $|{\bf b}|$, whereby the proportionality factor depends in some way on the enclosed angle $\phi$, like so: $${\rm area}(P)=f(\phi)\cdot|{\bf a}|\cdot|{\bf b}|\ .$$ Now we all learn in high school that in fact $f(\phi)=\sin\phi$ $\>(0\leq\phi\leq\pi)$, so that the scalar product $\langle{\bf a},{\bf b}\rangle$ cannot come into the game. On the other hand the elementary geometric explanation of the cross product has $|{\bf a}\times{\bf b}|=|{\bf a}|\cdot|{\bf b}|\cdot\sin\phi$, as required.

A more sophisticated explanation would be the following: The cross product ${\bf a}\times{\bf b}$ is by definition tied to the volume form ${\rm vol}$ by $${\rm vol}({\bf a},{\bf b},{\bf x})=\langle{\bf a}\times{\bf b},{\bf x}\rangle\qquad({\bf x}\in{\mathbb R}^3)\ .$$ Now the volume of the prism spanned by ${\bf a}$, ${\bf b}$ and ${\bf x}$ as a function of ${\bf x}$ is proportional to the area of the base parallelogram $P$. Therefore it is plausible that $|{\bf a}\times{\bf b}|$ is proportional to this area.

  • $\begingroup$ See my answer for why the defining equation not only shows it is plausible that it is proportional to the area, but that the constant of proportionality must be 1. $\endgroup$ Dec 19, 2013 at 14:18

I think the signed part of area is the most difficult to assign some intuitive meaning.

Consider two vectors in $\mathbb{R^2}$, and let $A : \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ be the signed area. Then $A$ should be linear in each variable separately, since we should have $A(\lambda x,y) = \lambda A(x,y)$, and $A(x_1+x_2,y) = A(x_1,y)+A(x_2,y)$ (see the following diagram to convince yourself why this should be the case):

enter image description here

The area between a vector and itself should be zero (that is $A(z,z) = 0$ for all $z$), so the linearity requirement gives $A(x+y,x+y) = A(x,x)+A(x,y)+A(y,x) + A(y,y) = A(x,y)+A(y,x) = 0$, and so $A(x,y) = -A(y,x)$.

Then $A(x,y) = A(\sum_i x_i e_i, \sum_j y_j e_j) = \sum_i \sum_j x_i y_i A(e_i,e_j) = (x_1 y_1 -x_2y_2 )A(e_1,e_2)$.

It seems reasonable to assign an area of one to the area spanned by $e_1,e_2$, hence $A(e_1,e_2) = 1$, which gives $A(x,y) = x_1 y_1 -x_2y_2 $ (which, of course, equals $\det \begin{bmatrix} x & y \end{bmatrix}$).


I think one way to approach this is to generalize to higher dimensions. Given $k$ vectors in $n$ dimensional space, you can form an $n \times k$ matrix who columns are the $k$ vectors. Then you can pick $\left( n\atop k \right)$ submatrices of size $k\times k$. Calculate the determinant of each of these $k \times k$ matrices. Then take the sum of the squares of these numbers. This will be the square of the $k$-dimensional volume of the parallelepiped generated by the $k$-vectors in $n$-dimensional space.

Why is this true? First you want to learn about alternating forms, also called exterior algebras: http://en.wikipedia.org/wiki/Exterior_algebra. Then you will see that if you perform an orthogonal transformation on $\mathbb R^n$ to place the $k$ vectors so that only their first $k$-coordinates are non-zero, that this will also induce an orthogonal transformation on the exterior algebra $\Lambda^k(\mathbb R^n)$. Then only one of the coordinates of the $k$-form will be non-zero, and its value will the the determinant of the $k\times k$ matrix of the transformed $k$-vectors,

This is a very brief, and somewhat sophisticated description. But I have to say it took me years to understand this. I don't know if I can easily reduce it to a short answer here. Not at least without several hours of thinking of how to write it.


What I'm missing in the other answers is that a surface area, in physics, separates the inside of a body from its outside, most of the time. Therefore it is natural to have a vector at the surface area which points to (the inside or) the outside. Which is precisely the outer product (called the normal on the surface if the area is taken apart as a scalar). It would be hard to describe e.g. conservation of mass in Fluid Flow without such an "area with a direction". A sample - and simple - application is found here in MSE.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.