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Section 4.1 Definition and Properties of the Cross Product

Definition 4.1 The Cross Product (First Definition).

Given two vectors \(\mathbf u\) and \(\mathbf v\) in \(\mathbb R^3\text{,}\) the cross product \(\mathbf u \times \mathbf v\) is the vector defined by
\begin{equation} \mathbf u \times \mathbf v = \left(\|\mathbf u\|\|\mathbf v\|\sin\theta\right)\mathbf n,\tag{4.1} \end{equation}
where \(\theta\text{,}\) with \(0 \le \theta \le \pi\text{,}\) is the angle between \(\mathbf u\) and \(\mathbf v\text{,}\) and \(\mathbf n\) is the unit vector orthogonal to both \(\mathbf u\) and \(\mathbf v\) whose direction is given by the right-hand rule.
The magnitude of \(\mathbf u \times \mathbf v\) is equal to the area of the parallelogram built on the two vectors \(\mathbf u\) and \(\mathbf v\text{,}\) as shown in Figure 4.2.
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Figure 4.2. The parallelogram built on the vectors \(\mathbf u\) and \(\mathbf v\text{,}\) with base \(\|\mathbf v\|\) and height \(\|\mathbf u\||\sin(\theta)|\text{.}\)
Note that \(\|\mathbf u\|\|\mathbf v\||\sin\theta|\) is the length of the cross product \(\mathbf u \times \mathbf v\text{.}\) Since \(\sin(\theta)\) can be negative, we need to include its absolute value.
The vector \(\mathbf n\) is the unit vector that determines the direction of \(\mathbf u \times \mathbf v\) using the right-hand rule. Note that \(\mathbf u \times \mathbf v\) is orthogonal (perpendicular) to both vectors, as shown in Figure 4.4.
Figure 4.4. The cross product \(\mathbf u \times \mathbf v\) points in the direction of the unit vector \(\mathbf n\text{,}\) given by the right-hand rule, and is orthogonal to both \(\mathbf u\) and \(\mathbf v\text{;}\) its length is \(\|\mathbf u\|\|\mathbf v\||\sin\theta|\text{.}\)
Index finger along u, middle finger along v, thumb along u cross v.
(a) Index–Middle–Thumb method.
Fingers curl from u toward v, thumb points along u cross v.
(b) Rotating-fingers method.
Figure 4.5. Two equivalent right-hand rules for \(\mathbf{u}\times\mathbf{v}\text{.}\)

Proof.
Although the cross product was defined geometrically above, all five properties are most easily checked using the equivalent component formula (4.4) established in the next section: if \(\mathbf a = \langle a_1,a_2,a_3\rangle\) and \(\mathbf b = \langle b_1,b_2,b_3\rangle\text{,}\) then \(\mathbf a \times \mathbf b = \langle a_2b_3-a_3b_2,\; a_3b_1-a_1b_3,\; a_1b_2-a_2b_1\rangle\text{.}\) Write \(\mathbf u = \langle u_1,u_2,u_3\rangle\text{,}\) \(\mathbf v = \langle v_1,v_2,v_3\rangle\text{,}\) and \(\mathbf w = \langle w_1,w_2,w_3\rangle\text{.}\)
(1) Each component of \((r\mathbf u)\times(s\mathbf v)\) has the form \((ru_i)(sv_j) - (ru_j)(sv_i) = rs(u_iv_j-u_jv_i)\text{,}\) which is \(rs\) times the corresponding component of \(\mathbf u \times \mathbf v\text{.}\) Hence \((r\mathbf u)\times(s\mathbf v) = (rs)(\mathbf u \times \mathbf v)\text{.}\)
(2) The first component of \(\mathbf u \times (\mathbf v + \mathbf w)\) is
\begin{equation*} u_2(v_3+w_3) - u_3(v_2+w_2) = (u_2v_3-u_3v_2) + (u_2w_3-u_3w_2), \end{equation*}
which is the sum of the first components of \(\mathbf u \times \mathbf v\) and \(\mathbf u \times \mathbf w\text{.}\) The same holds for the other two components, so \(\mathbf u \times (\mathbf v+\mathbf w) = \mathbf u \times \mathbf v + \mathbf u \times \mathbf w\text{.}\)
(3) By the same computation with the roles of the two factors exchanged, the first component of \((\mathbf v+\mathbf w)\times \mathbf u\) is \((v_2+w_2)u_3 - (v_3+w_3)u_2 = (v_2u_3-v_3u_2) + (w_2u_3-w_3u_2)\text{,}\) the sum of the first components of \(\mathbf v \times \mathbf u\) and \(\mathbf w \times \mathbf u\text{;}\) likewise for the other two components, so \((\mathbf v+\mathbf w)\times \mathbf u = \mathbf v \times \mathbf u + \mathbf w \times \mathbf u\text{.}\)
(4) The first component of \(\mathbf v \times \mathbf u\) is \(v_2u_3 - v_3u_2 = -(u_2v_3-u_3v_2)\text{,}\) the negative of the first component of \(\mathbf u \times \mathbf v\text{;}\) likewise for the other two components, so \(\mathbf v \times \mathbf u = -(\mathbf u \times \mathbf v)\text{.}\)
(5) Since \(\mathbf 0 = \langle 0,0,0\rangle\text{,}\) every component of \(\mathbf 0 \times \mathbf u\) is a difference of products each containing a factor of \(0\text{,}\) so \(\mathbf 0 \times \mathbf u = \mathbf 0\text{.}\)

Checkpoint 4.8.

Show that \((\mathbf u - \mathbf v) \times (\mathbf u + \mathbf v) = 2(\mathbf u \times \mathbf v)\text{.}\)

Example 4.9 Computing a cross product using the first definition.

Use Definition 4.1 to compute
  1. \(\mathbf i \times \mathbf j\text{,}\) and also \(\mathbf j \times \mathbf i\text{;}\)
  2. \(\mathbf u \times \mathbf v\text{,}\) where \(\mathbf u = 4\mathbf i\) and \(\mathbf v\) is a vector of length \(3\) lying in the \(xy\)-plane, making an angle of \(30^\circ\) with \(\mathbf u\) measured counterclockwise.
Solution.
Part 1. The vectors \(\mathbf i\) and \(\mathbf j\) are unit vectors, so \(\|\mathbf i\| = \|\mathbf j\| = 1\text{,}\) and they are perpendicular, so \(\theta = 90^\circ\) and \(\sin\theta = 1\text{.}\) Both lie in the \(xy\)-plane, so the unit vector orthogonal to both is either \(\mathbf k\) or \(-\mathbf k\text{;}\) curling the fingers of the right hand from \(\mathbf i\) toward \(\mathbf j\) points the thumb along \(+\mathbf k\text{,}\) so \(\mathbf n = \mathbf k\) and
\begin{equation*} \mathbf i \times \mathbf j = \left(\|\mathbf i\|\|\mathbf j\|\sin 90^\circ\right)\mathbf k = (1)(1)(1)\,\mathbf k = \mathbf k. \end{equation*}
For \(\mathbf j \times \mathbf i\) the two lengths and the angle are unchanged, but the right-hand rule now runs from \(\mathbf j\) toward \(\mathbf i\text{,}\) which points the thumb along \(-\mathbf k\text{.}\) Hence \(\mathbf j \times \mathbf i = -\mathbf k = -(\mathbf i \times \mathbf j)\text{,}\) in agreement with property 4 of Theorem 4.7.
Part 2. Here \(\|\mathbf u\| = 4\text{,}\) \(\|\mathbf v\| = 3\text{,}\) and \(\theta = 30^\circ\text{,}\) so the length of the cross product is
\begin{equation*} \|\mathbf u \times \mathbf v\| = \|\mathbf u\|\|\mathbf v\|\sin 30^\circ = (4)(3)\left(\tfrac12\right) = 6. \end{equation*}
Both vectors lie in the \(xy\)-plane, and turning from \(\mathbf u\) to \(\mathbf v\) is counterclockwise, so the right-hand rule again gives \(\mathbf n = \mathbf k\) and
\begin{equation*} \mathbf u \times \mathbf v = 6\,\mathbf k = \langle 0,0,6\rangle. \end{equation*}
As a check on the magnitude, Fact 4.3 says \(\|\mathbf u \times \mathbf v\|\) is the area of the parallelogram built on \(\mathbf u\) and \(\mathbf v\text{:}\) its base is \(\|\mathbf u\| = 4\) and its height is \(\|\mathbf v\|\sin 30^\circ = 1.5\text{,}\) giving an area of \(6\text{.}\)
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