Introduction.
There is a special situation in three dimensions that any two non-parallel vectors \(\vec{u}\) and \(\vec{v}\) determine a plane through the origin, in that all the combinations \(a\vec{u} + b\vec{v}\) describe points in a plane, and it is then interesting to find the unique direction perpendicular to that plane. This is done by a special product \(\vec{w} = \vec{u} \times \vec{v}\) to be defined soon.
The magnitude of this product is also significant: it will give the area of the triangle determined by the factors \(\vec{u}\) and \(\vec{v}\) as \(\|\vec{u} \times \vec{v}\|/2\text{.}\) That is, for the points \(U\) and \(V\) in \(\mathbb{R}^3\) such that \(\vec{u} = \vector{OU}\) and \(\vec{v} = \vector{OV}\text{,}\) the triangle \(\triangle OUV\) has area \(\|\vec{u} \times \vec{v}\|/2\text{.}\) Note that this means that the parallelogram with corners given by \(\vec{0}\text{,}\) \(\vec{u}\text{,}\) \(\vec{v}\) and \(\vec{u}+\vec{v}\) has area \(\|\vec{u} \times \vec{v}\|\text{.}\) This will be useful in calcuating areas and volumes, and describing magnetic and torque forces.
The final detail missing is the direction of this product, since for any vector \(\vec{w}\) meeting the above conditions, its negation does also. That choice will be resolved by appealing to the "right-handedness" discussed with 3D coordinate systems, which will be sufficient to determine the value of this new product uniquely: it will be required that the triple \(\vec{u}\)–\(\vec{v}\)–\(\vec{w}\) be right-handed, in the way that, for example, the triple \(\veci\)–\(\vecj\)–\(\veck\) is.
As with the dot product, there is an algebraic formula that gives the product vector; this time, it will be possible to derive this formula from the above list of desired properties with the help of expecting it to be distributive over addition.
Much as with the dot product or scalar product, this is often called the cross product based solely on the notation, but also the vector product, due to its value being a vector.
A warning: this deviates from what you might expect in several ways: this "product" is neither commutative nor associative.
Subsection 2.4.1 Definition and a Derivation
The cross product can be defined as
Definition 2.4.1.
\begin{equation}
\vec{u} \times \vec{v} = \vector{u_1 , u_2 , u_3}\times \vector{v_1 , v_2 , v_3}
= \vector{ u_2 v_3-u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2-u_2 v_1}\tag{2.4.1}
\end{equation}
This is also known as the
vector product, for contrast to the scalar product as in
Definition 2.3.1.
One way to see why this has to be the desired formula is to start with the above geometric requirements, see what they say about products of the three standard basis vectors \(\veci\text{,}\) \(\vecj\) and \(\veck\text{,}\) and then combine by assuming distributivity (which can then be verified in hind-sight.)
The first simple step—which does most of the work—is to see that
\begin{equation*}
\veci \times \vecj = \veck:
\end{equation*}
To be perpendicular to \(\veci\) and \(\vecj\text{,}\) the product must be a multiple of \(\veck\text{.}\)
The parallelogram formed by \(\veci\) and \(\vecj\) is a square of area 1, so the product has length one, and so is \(\pm \veck\text{.}\)
The right hand rule then forces the result to be \(\veck\text{,}\) as claimed.
Similarly we get
\begin{equation*}
\vecj \times \veck = \veci,
\quad \veck \times \veci = \vecj
\end{equation*}
The second key step is reversing the order of the factors; right-handedness forces negations:
\begin{equation*}
\vecj \times \veci = -\veck,\quad
\veck \times \vecj = -\veci,\quad
\veci \times \veck = -\vecj
\end{equation*}
So commutativity fails: in fact the pattern of negation seen here is general: the cross-product is anti-commutative.
Finally, what about parallel vectors? They determine a "parallelogram" of area zero, so
\begin{equation*}
\veci \times \veci = \vecj \times \vecj = \veck \times \veck = \vec{0}.
\end{equation*}
Exercise 2.4.2.
Assuming that this product is distributive over addition, expand
\begin{equation*}
\vec{u} \times \vec{v} = (u_1 \veci + u_2 \vecj + u_3 \veck) \times (v_1 \veci + v_2 \vecj + v_3 \veck)
\end{equation*}
into a sum of nine pieces and use the above nine basic products to verify Equation
(2.4.1).
Subsection 2.4.2 Determinants
To complete the justification of this definition, and for other purposes, it is convenient to introduce the concept of a determinant from linear algebra.
A determinant of order 2 is defined by
\begin{equation}
\left| \begin{array}{cc} u_1 \amp u_2 \\ v_1 \amp v_2 \end{array} \right| = u_1v_2-u_2v_1.\tag{2.4.2}
\end{equation}
A determinant of order 3 is defined using this as
\begin{equation}
\begin{split}
\left| \begin{array}{ccc} u_1 \amp u_2 \amp u_3 \\ v_1 \amp v_2 \amp v_3 \\ w_1 \amp w_2 \amp w_3 \end{array} \right|
\amp= u_1\left| \begin{array}{cc} v_2 \amp v_3 \\ w_2 \amp w_3 \end{array} \right|
- u_2\left| \begin{array}{cc} v_1 \amp v_3 \\ w_1 \amp w_3 \end{array} \right|
+ u_3\left| \begin{array}{cc} v_1 \amp v_2 \\ w_1 \amp w_2 \end{array} \right|,
\\
\amp= u_1 v_2 w_3-u_1 v_3 w_2 - u_2 v_1 w_3 + u_2 v_3 w_1 + u_3 v_1 w_2 - u_3 v_2 w_1.
\end{split}\tag{2.4.3}
\end{equation}
The factors \(u_i\) multiplying each order 2 determinant come from the top row; the coefficient of each is the determinant of what you get when you delete from the 3x3 array of numbers the row and column that contain the factor \(u_i\text{;}\) the signs alternate. (This pattern also describes the order 2 determinant, but then all that is left after deleting one row and one column is a single number.)
The cross product as a determinant.
The coefficients in the cross product are given by order two determinants:
\begin{equation}
\vec{u} \times \vec{v}
= \veci \left| \begin{array}{cc} u_2 \amp u_3 \\ v_2 \amp v_3 \end{array} \right|
- \vecj \left| \begin{array}{cc} u_1 \amp u_3 \\ v_1 \amp v_3 \end{array} \right|
+ \veck \left| \begin{array}{cc} u_1 \amp u_2 \\ v_1 \amp v_2 \end{array} \right|,
% Omit this nice but unnecesary and non-standard cyclic alternative:
%\\
%\amp=\amp \veci \left| \begin{array}{cc} u_2 \amp u_3 \\ v_2 \amp v_3 \end{array} \right|
%+ \vecj \left| \begin{array}{cc} u_3 \amp u_1 \\ v_3 \amp v_1 \end{array} \right|
%+ \veck \left| \begin{array}{cc} u_1 \amp u_2 \\ v_1 \amp v_2 \end{array} \right|.
%\nonumber\tag{2.4.4}
\end{equation}
This is like the above order 3 determinant except with the components of \(\vec{u}\) changed to standard basis vectors, each \(v_i\) replaced by \(u_i\) and each \(w_i\) replaced by \(v_i\text{.}\) Thus
\begin{equation}
\vec{u} \times \vec{v} = \left| \begin{array}{ccc} \veci \amp \vecj \amp \veck \\ u_1 \amp u_2 \amp u_3 \\ v_1 \amp v_2 \amp v_3 \end{array} \right|\tag{2.4.5}
\end{equation}
where we allow vectors to appear where previously we had numbers.
Theorem 2.4.3.
The cross product is distributive over addition:
\begin{equation*}
(\vec{u} + \vec{v}) \times \vec{w} = (\vec{u} \times \vec{w}) + (\vec{v} \times \vec{w})
\end{equation*}
Proof.
This is shown by inserting the definition in all three places and simplifying.
Exercise 2.4.4.
Show that for any vector \(\vec{u}\text{,}\) \(\vec{u} \times \vec{u} = \vec{0}.\)
(Think about why this makes inverses impossible with the vector product.)
Exercise 2.4.5.
Use the above fact to show that the cross product is anti-commutative: \(\vec{u} \times \vec{v} = - \vec{v} \times \vec{u}\)
Hint.
Expand \((\vec{u} + \vec{v}) \times (\vec{u} + \vec{v})\)
Theorem 2.4.6.
The vector \(\vec{u} \times \vec{v}\) is orthogonal to both \(\vec{u}\) and \(\vec{v}\text{.}\)
Proof.
This is shown by computing \((\vec{u} \times \vec{v}) \cdot \vec{u}\) and simplifying until nothing is left; likewise for \((\vec{u} \times \vec{v}) \cdot \vec{v}\text{.}\)The Geometrical Meaning of the Cross Product.
Theorem 2.4.7. The Magnitude of the Cross Product.
With \(\theta\) the angle between vectors \(\vec{u}\) and \(\vec{v}\text{,}\) their cross product has magnitude
\begin{equation*}
\|\vec{u} \times \vec{v}\| = \|\vec{u}\| \|\vec{v}\| \sin \theta
\end{equation*}
Proof.
Thus, just as \(\vec{u} \cdot \vec{v} = 0\) test for two vectors being orthogonal, we have:
Corollary 2.4.8. A Test for Parallel Vectors.
Two vectors \(\vec{u}\) and \(\vec{v}\) are parallel if and only if their cross product vanishes: \(\vec{u} \times \vec{v} = \vec{0}.\) (This includes the case of them pointing in opposite directions, sometimes called "anti-parallel".)
