Proof 23: Cross-product

$$\vec c=\vec a \times \vec b$$

A cross-product is defined with two features:

  • It’s magnitude $c$ is the perpendicular part of one vector multiplied with the other vector’s magnitude.
  • It’s direction is perpendicular to both vectors.

We are looking for the formula for each of the coordinates $\vec c=(c_1,c_2,c_3)$ in the cross-product.

The cross-product magnitude $c$

Those coordinates are with Pythagoras’ Theorem tied to its magnitude $c$:

$$c=\sqrt{c_1^2+c_2^2+c_3^2}\tag 1$$

This is our 1st equation with 4 unknowns.

The perpendicular part of one vector, say $\vec a$, is $a_\perp$. It can be found with the sine rule, when split into components parallel and perpendicular to the other vector, giving us the imagined right-angled triangle:

$$\sin(\theta)= \frac{\text{opposite side}}{ \text{hypotenuse} } =\frac{a_\perp}{a}\quad\Leftrightarrow\quad a_\perp=\sin(\theta)a$$

The cross-product magnitude $c$ is the perpendicular part of one vector multiplied with the other vector’s magnitude $b$:

$$c=b\;a_\perp= b\;\underbrace{a \sin(\theta)}_{a_\perp}\tag 2 $$

We would get the same result by taking the perpendicular part of $\vec b$. This result is general. This is our 2nd equation with 3 new unknowns, so in total 7 unknowns: $a$, $b$, $c$, $\theta$ as well as $c_1$, $c_2$ and $c_3$.

The angle $\theta$ between $\vec a$ and $\vec b$ and their magnitudes $a$ and $b$

To get rid of the angle $\theta$ between $\vec a$ and $\vec b$ we must use the formula for the angle between two vectors, which is our 3rd equation:

$$\cos(\theta)=\frac{\vec a \cdot \vec b}{ab} \tag 3$$

The vectors $\vec a$ and $\vec b$ have magnitudes $a$ and $b$ according to Pythagoras’ Theorem, giving us our 4th and 5th equation:

$$ \begin{align} a=\sqrt{a_1^2+a_2^2+a_3^2} \tag 4\\
b=\sqrt{b_1^2+b_2^2+b_3^2} \tag 5 \end{align} $$

No new unknowns are introduced with these three.

The cross-product direction

The cross-product $\vec c$ is perpendicular to both $\vec a$ and $\vec b$. Perpendicular means that the dot-products are zero, giving us our 6th and 7th equations:

$$ \begin{align} \vec a\cdot \vec c=a_1c_1 + a_2c_2 + a_3c_3 = 0\tag 6 \\
\vec b\cdot \vec c=b_1c_1 + b_2c_2 + b_3c_3 = 0\tag 7 \end{align} $$

Wrapping it up

These 7 equations are enough to find the in total 7 unknowns:

$$\begin{align}c&=\sqrt{c_1^2+c_2^2+c_3^2} \tag 1 \\
c&=ba \sin(\theta) \tag 2\\
\cos(\theta)&=\frac{\vec a \cdot \vec b}{ab} \tag 3\\
a&=\sqrt{a_1^2+a_2^2+a_3^2} \tag 4\\
b&=\sqrt{b_1^2+b_2^2+b_3^2} \tag 5\\
0&= a_1c_1 + a_2c_2 + a_3c_3 \tag 6\\
0&= b_1c_1 + b_2c_2 + b_3c_3 \tag 7\end{align}$$

Solving them (isolating unknowns and inserting into the others) gives the following result for $c_1$, $c_2$ and $c_3$:

$$c_1=a_2b_3-a_3b_2\\ c_2=a_3b_1-a_1b_3\\ c_3=a_1b_2-a_2b_1$$

which can be put back into the cross-product vector:

$$\vec c=\vec a\times \vec b=\begin{pmatrix} a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 \\ a_1b_2-a_2b_1 \end{pmatrix} \qquad\blacksquare$$

See all the steps layed out in [1]


  1. Derivation of the cross product formula’ (web page), Heaven's in the backyard, 2011, web.archive.org/web/20180624123849/http://heaveninthebackyard.blogspot.com/2011/12/derivation-of-cross-product-formula.html (archived version, accessed Feb. 6th, 2020)