Reasoning behind the cross products used to find area

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.

Solution 1:

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).

Solution 2:

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.