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Section 4.2 Second Definition of the Cross Product
Writing the vectors in components gives another definition for the cross product.
Definition 4.8 . The Cross Product (Second Definition).
If \(\mathbf u = \langle u_1, u_2, u_3\rangle\) and \(\mathbf v = \langle v_1, v_2, v_3\rangle\text{,}\) then
\begin{equation}
\mathbf u \times \mathbf v =
\langle u_2v_3 - u_3v_2,\; u_3v_1 - u_1v_3,\; u_1v_2 - u_2v_1\rangle.\tag{4.3}
\end{equation}
One way to remember the second definition of the cross product is to write it as a determinant:
\begin{equation}
\mathbf u \times \mathbf v =
\begin{vmatrix}
\mathbf i \amp \mathbf j \amp \mathbf k \\
u_1 \amp u_2 \amp u_3 \\
v_1 \amp v_2 \amp v_3
\end{vmatrix}
=
\begin{vmatrix} u_2 \amp u_3 \\ v_2 \amp v_3 \end{vmatrix}\mathbf i
- \begin{vmatrix} u_1 \amp u_3 \\ v_1 \amp v_3 \end{vmatrix}\mathbf j
+ \begin{vmatrix} u_1 \amp u_2 \\ v_1 \amp v_2 \end{vmatrix}\mathbf k,\tag{4.4}
\end{equation}
where
\begin{equation}
\begin{vmatrix} a \amp b \\ c \amp d \end{vmatrix} = ad - bc.\tag{4.5}
\end{equation}
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 4.9 .
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Figure 4.9. The parallelogram built on the vectors \(\mathbf u\) and \(\mathbf v\text{,}\) with base \(|\mathbf v|\) and height \(|\mathbf u||\sin(\theta)|\text{.}\)
Fact 4.10 . Area of the Parallelogram.
\begin{equation}
\text{Area of the parallelogram}
= |\mathbf u||\mathbf v||\sin\theta|
= |\mathbf u \times \mathbf v|.\tag{4.6}
\end{equation}
Example 4.11 . Computing a cross product.
Calculate the cross product
\(\mathbf u \times \mathbf v\) if
\(\mathbf u = \langle 1,-1,2\rangle\) and
\(\mathbf v = \langle 3,-2,1\rangle\text{.}\)
Solution.
\begin{align*}
\mathbf u \times \mathbf v \amp=
\begin{vmatrix}
\mathbf i \amp \mathbf j \amp \mathbf k \\
1 \amp -1 \amp 2 \\
3 \amp -2 \amp 1
\end{vmatrix}
=
\begin{vmatrix} -1 \amp 2 \\ -2 \amp 1 \end{vmatrix}\mathbf i
- \begin{vmatrix} 1 \amp 2 \\ 3 \amp 1 \end{vmatrix}\mathbf j
+ \begin{vmatrix} 1 \amp -1 \\ 3 \amp -2 \end{vmatrix}\mathbf k\\
\amp= (-1+4)\mathbf i - (1-6)\mathbf j + (-2+3)\mathbf k
= 3\mathbf i + 5\mathbf j + \mathbf k
= \langle 3,5,1\rangle.
\end{align*}
Checkpoint 4.12 .
Show that
\((\mathbf u - \mathbf v) \times (\mathbf u + \mathbf v) = 2(\mathbf u \times \mathbf v)\text{.}\)
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