Understanding the definition of the cross product

Question

I know how to use the cross product, I know what it means and how it relates to the dot product.

$$|a \times b| = ||a||b| \cdot \sin(\theta) \vec{n}|\\ a \cdot b = |a||b| \cdot \cos(\theta)$$

I also understand why and how you can calculate the area of two 3d vectors with the cross product.

What I don't understand where this is coming from

$$\displaystyle{(a_x, a_y, a_z) \times (b_x, b_y, b_z) = (a_y b_z - a_z b_y, a_z b_x - a_x b_z, a_x b_y - a_y b_x) = \|a\| \|b\| \sin(\theta) \vec{n}}$$

I mean I could imagine that I could invent it myself with the following properties and a lot of trial and error.

$a \times b = c$

$a \cdot c = 0$

$b \cdot c = 0$

But I don't really have an intuition of why it works the way it does.

Answer 1

First notice that if we define $(u \wedge v ) \cdot w = \det (u,v,w)$ and let $i,j,k,l = 1,2,3$ then $$(e_i \wedge e_j) \cdot (e_k \wedge e_l) = \begin{vmatrix}e_i \cdot e_k & e_j \cdot e_k \\e_i \cdot e_l & e_j \cdot e_l\end{vmatrix}$$

Consequently $$(u \wedge v)\cdot (u \wedge v)=|u \wedge v |^2 = \begin{vmatrix}u \cdot u & v \cdot u \\ u \cdot v & v \cdot v\end{vmatrix} = |u|^2 |v|^2 (1 - \cos ^2 \theta) = A^2$$

where $\theta$ is the angle between $u $ and $v$ and $\{u,v, u \wedge v\}$ is a positive basis. As for the intuition $A$ is the area of the paralelogram generated by $u$ and $v$.

Note: Sometimes $u \wedge v$ can be written as $u \times v$.

Answer 2

Let $i_x, i_y, i_z$ be the basis vectors of a right-handed coordinate system. Now we can write \begin{equation} i_x \times i_y = i_z \quad i_y \times i_z = i_x \quad i_z \times i_x = i_y \end{equation} By the definition of vector product if the vectors are orthogonal the $sin (\theta) = 1$ while if the vectors are parallel $sin (\theta) = 0$ then \begin{equation} i_x \times i_x = 0 \quad i_y \times i_y = 0 \quad i_z \times i_z = 0 \end{equation}

Now consider the vectors $A$ and $ B$. By using the above expressions, we can easily express the vector product of two vectors $A$ and $B$ in terms of their components: \begin{equation} C = A \times B = (A_x i_x + A_y i_y + A_z i_z) \times (B_x i_x + B_y i_y + B_z i_z) = (A_y B_z-A_z B_y)i_x+(A_z B_x-A_x B_z)i_y+(A_x B_y-A_y B_x)i_z \end{equation} Which can be written by using the determinant of a matrix: \begin{equation} C = A \times B = det (\begin{bmatrix} i_x & i_{y} & i_{z} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{bmatrix}) \end{equation}

Answer 3

For three unknowns we want to solve three equations simultaneously: $$a \cdot n = 0 $$ $$ b \cdot n = 0 $$ $$||n|| = 1 $$ The first two constraints are defined because the cross product vector $n$, will be perpendicular to both $a$ and $b$ when their dot product is 0. The third constraint is used as a trick to fill out the system of equations so that it can be solvable for three unknowns, so we just define the length of the cross product will be of unit length.

$$a_1n_1 + a_2n_2 + a_3n_3 = 0$$ $$n_1 = \dfrac{-a_2n_2}{a_1} - \dfrac{a_3n_3}{a_1}$$ $$b_1n_1 + b_2n_2 + b_3n_3 = 0$$ $$b_1(\dfrac{-a_2n_2}{a_1} - \dfrac{a_3n_3}{a_1}) + b_2n_2 + b_3n_3 = 0$$ $$\dfrac{-a_2b_1n_2}{a_1} - \dfrac{a_3b_1n_3}{a_1} + b_2n_2 + b_3n_3 = 0$$ $$\dfrac{-a_2b_1n_2}{a_1} + b_2n_2 = \dfrac{a_3b_1n_3}{a_1} - b_3n_3 $$

$$ n_2(b_2 - \dfrac{a_2b_1}{a_1}) = \dfrac{a_3b_1n_3}{a_1} - b_3n_3$$

$$ n_2(\dfrac{a_1b_2 - a_2b_1}{a_1}) = \dfrac{a_3b_1n_3 - a_1b_3n_3}{a_1}$$ $$ n_2(a_1b_2 - a_2b_1) = a_3b_1n_3 - a_1b_3n_3$$ $$ n_2 = \dfrac{a_3b_1n_3 - a_1b_3n_3}{a_1b_2 - a_2b_1}$$ $$ n_2 = n_3\dfrac{a_3b_1 - a_1b_3}{a_1b_2 - a_2b_1}$$ sub $n_2$ back into $n_1$: $$n_1 = \dfrac{-a_2(n_3\dfrac{a_3b_1 - a_1b_3}{a_1b_2 - a_2b_1})}{a_1} - \dfrac{a_3n_3}{a_1}$$ $$n_1 = \dfrac{n_3}{a_1}(\dfrac{-a_3a_2b_1 + a_1a_2b_3}{a_1b_2 - a_2b_1} - a_3) $$ $$ n_1 = \dfrac{n_3}{a_1} \dfrac{-a_3a_2b_1 + a_1a_2b_3 -a_1a_3b_2 + a_2a_3b_1}{a_1b_2 - a_2b_1}$$ $$ n_1 = \dfrac{n_3}{a_1} \dfrac{a_1a_2b_3 -a_1a_3b_2}{a_1b_2 - a_2b_1}$$ $$ n_1 = n_3\dfrac{a_2b_3 -a_3b_2}{a_1b_2 - a_2b_1}$$ sub into third constraint: $$ n_1^2 + n_2^2 + n_3^2 = 1$$ $$ (n_3\dfrac{a_2b_3 -a_3b_2}{a_1b_2 - a_2b_1})^2 + (n_3\dfrac{a_3b_1 - a_1b_3}{a_1b_2 - a_2b_1})^2 + n_3^2 = 1$$ $$ n_3^2((\dfrac{a_2b_3 -a_3b_2}{a_1b_2 - a_2b_1})^2 + (\dfrac{a_3b_1 - a_1b_3}{a_1b_2 - a_2b_1})^2 + (\dfrac{a_1b_2 - a_2b_1}{a_1b_2 - a_2b_1})^2)) = 1$$ $$ n_3^2\dfrac{(a_2b_3 -a_3b_2)^2 + (a_3b_1 - a_1b_3)^2 + (a_1b_2 - a_2b_1)^2}{(a_1b_2 - a_2b_1)^2} = 1$$ let $$Z = (a_2b_3 -a_3b_2)^2 + (a_3b_1 - a_1b_3)^2 + (a_1b_2 - a_2b_1)^2$$ $$1 = n_3^2 \dfrac{Z}{(a_1b_2 - a_2b_1)^2}$$ $$n_3 = \dfrac{a_1b_2 - a_2b_1}{\sqrt{Z}}$$ sub into $n_1$ and $n_2$: $$n_1 = \dfrac{a_1b_2 - a_2b_1}{\sqrt{Z}} \cdot \dfrac{a_2b_3 - a_3b_2}{a_1b_2 - a_2b_1}$$ $$n_1 = \dfrac{a_2b_3 - a_3b_2}{\sqrt{Z}} $$ $$n_2 = \dfrac{a_1b_2 - a_2b_1}{\sqrt{Z}} \cdot \dfrac{a_3b_1 - a_1b_3}{a_1b_2 - a_2b_1} $$ $$n_2 = \dfrac{a_3b_1 - a_1b_3}{\sqrt{Z}} $$

so $$n = \dfrac{1}{\sqrt{Z}} \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{pmatrix}$$ where $\sqrt{Z}$ is equal to $\sqrt{(a_2b_3 -a_3b_2)^2 + (a_3b_1 - a_1b_3)^2 + (a_1b_2 - a_2b_1)^2}$ but you can see that $Z$ is of the form $\sqrt{n_1^2 + n_2^2 + n_3^2}$ so then $\dfrac{1}{\sqrt{Z}}$ has the effect of normalizing the cross product vector $n$, but for simplicity we can ignore the output is normalized and we get the expected equation:

$$n = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{pmatrix}$$

References:

• https://richbeveridge.wordpress.com/2015/03/04/derivation-of-the-cross-product/
• https://web.archive.org/web/20180624123849/http://heaveninthebackyard.blogspot.com/2011/12/derivation-of-cross-product-formula.html

Answer 4

Here's the best way to think about the cross product, in my opinion. Let $b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}, c = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} \in \mathbb R^3$, and let $L:\mathbb R^3 \to \mathbb R$ be the function defined by $$ L(a) = \det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} . $$ The function $L$ is linear, so there exists a vector $w \in \mathbb R^3$ such that $L(a) = a \cdot w$ for all $a \in \mathbb R^3$. This vector $w$ is called the cross product of $b$ and $c$, and is denoted $b \times c$. So the defining equation for the cross product is $$ \tag{$\heartsuit$}\det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} = a \cdot (b \times c) $$ for all $a \in \mathbb R^3$.

The geometric interpretation of the determinant (it is the signed volume of the parallelepiped spanned by $a,b$, and $c$) leads directly to the geometric interpretation of the cross product. For example, the parallelepiped spanned by $b, b$, and $c$ is degenerate, so its volume is $0$. It follows that $$ b \cdot (b \times c) = 0. $$ Similarly, $c \cdot(b \times c) = 0$. So $b \times c$ is orthogonal to both $b$ and $c$.

What about the norm of $b \times c$? Let $A$ be the area of the parallelogram spanned by $b$ and $c$, and let $\theta$ be the angle between $b \times c$ and $a$. (Assume for simplicity that $0 \leq \theta \leq \pi/2$.) The volume $V$ of the parallelepiped spanned by $a, b$, and $c$ is \begin{align} V &= \text{"area of the base times the height"} \\ &= A \|a \| \cos(\theta). \end{align} Comparing this formula with $$ V = a \cdot (b \times c) = \| a \| \| b \times c \| \cos(\theta) $$ reveals that $\| b \times c \| = A$. So the norm of $b \times c$ is the area of the parallelogram spanned by $b$ and $c$.