Section 1.5 Extra Topic: Cross Product and its Properties
In [cross-reference to target(s) "Section-Dot-Product-and-its-Properties" missing or not unique] we introduced the dot product, one of two important products for vectors. We will now introduce the second type of product, called the cross product. There are several important distinctions to keep in mind. First, the dot product is defined for two vectors of \(\R^n\text{,}\) for any natural number \(n\text{;}\) the cross-product will only be defined for vectors of \(\R^3\text{.}\) Second, the dot product is a scalar; the cross product of two vectors will be a vector. Finally, we will find that unlike the dot product, the cross product is not commutative.
The cross product has many applications in physics and engineering. It also has important geometric properties which will be addressed in this section and in [cross-reference to target(s) "Section-Determinants-as-Areas-and-Volumes" missing or not unique].
Subsection 1.5.1 Preliminaries
In order to define the cross product in a convenient way we need to define \(2\times 2\) and \(3\times 3\) determinants. If you know how to find such determinants you may skip this section and proceed directly to the definition.
Definition 1.5.1. \(2\times 2\) Determinant.
A \(2\times 2\) determinant is a number associated with a \(2\times 2\) matrix
\begin{equation*}
\det{\begin{bmatrix}
a \amp b\\
c \amp d
\end{bmatrix}}=\begin{vmatrix}
a \amp b\\
c \amp d
\end{vmatrix} = ad-bc
\end{equation*}
Example 1.5.2.
\begin{equation*}
\begin{vmatrix}
2 \amp 4\\
-5 \amp 3
\end{vmatrix} =(2)(3)-(4)(-5)=6+20=26
\end{equation*}
Definition 1.5.3. \(3\times 3\) Determinant.
A \(3\times 3\) determinant is a number associated with a \(3\times 3\) matrix
\begin{equation*}
\det{\begin{bmatrix}
a_1 \amp a_2 \amp a_3\\
b_1 \amp b_2 \amp b_3\\
c_1 \amp c_2 \amp c_3
\end{bmatrix}}=
\begin{vmatrix}
a_1 \amp a_2 \amp a_3\\
b_1 \amp b_2 \amp b_3\\
c_1 \amp c_2 \amp c_3
\end{vmatrix} =a_1
\begin{vmatrix}
b_2 \amp b_3\\
c_2 \amp c_3
\end{vmatrix} -a_2
\begin{vmatrix}
b_1 \amp b_3\\
c_1 \amp c_3
\end{vmatrix} +a_3
\begin{vmatrix}
b_1 \amp b_2\\
c_1 \amp c_2
\end{vmatrix}
\end{equation*}
Example 1.5.4.
\begin{align*}
\begin{vmatrix}
4 \amp 2 \amp 1\\
5 \amp -3 \amp 0\\
-2 \amp 3 \amp 2
\end{vmatrix}\amp =4
\begin{vmatrix}
-3 \amp 0\\
3 \amp 2
\end{vmatrix} -2
\begin{vmatrix}
5 \amp 0\\
-2 \amp 2
\end{vmatrix} +1
\begin{vmatrix}
5 \amp -3\\
-2 \amp 3
\end{vmatrix}\\
\amp =4\Big ((-3)(2)-(0)(3)\Big)-2\Big((5)(2)-(0)(-2)\Big)+\Big((5)(3)-(-3)(-2)\Big) \\
\amp =(4)(-6)-(2)(10)+(15-6) \\
\amp =-35
\end{align*}
For more help on
\(3 \times 3\) determinants, see
Example 7.1.4 and the video located there.
Subsection 1.5.2 Definition of the Cross Product
Definition 1.5.5.
Let \(\mathbf{u=\begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix}}\) and \(\mathbf{v}=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}\) be vectors in \(\R^3\text{.}\) The cross product of \(\mathbf{u}\) and \(\mathbf{v}\text{,}\) denoted by \(\mathbf{u}\times\mathbf{v}\text{,}\) is given by
\begin{align*}
\mathbf{u}\times\mathbf{v}\amp =(u_2v_3-u_3v_2)\mathbf{i}-(u_1v_3-u_3v_1)\mathbf{j}+(u_1v_2-u_2v_1)\mathbf{k} \\
\amp =(u_2v_3-u_3v_2)\begin{bmatrix}1\\0\\0\end{bmatrix}-(u_1v_3-u_3v_1)\begin{bmatrix}0\\1\\0\end{bmatrix}+(u_1v_2-u_2v_1)\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}u_2v_3-u_3v_2\\-u_1v_3+u_3v_1\\u_1v_2-u_2v_1\end{bmatrix}
\end{align*}
This formula is much easier to remember when stated symbolically in terms of determinants.
\begin{equation*}
\mathbf{u}\times \mathbf{v}=\begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix}\times\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\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}=\mathbf{i}\begin{vmatrix}u_2\amp u_3\\v_2\amp v_3\end{vmatrix}-\mathbf{j}\begin{vmatrix}u_1\amp u_3\\v_1\amp v_3\end{vmatrix}+\mathbf{k}\begin{vmatrix}u_1\amp u_2\\v_1\amp v_2\end{vmatrix}
\end{equation*}
Example 1.5.6.
Find the cross product of \(\mathbf{u}=\begin{bmatrix}3\\ -10\\ 2\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}-2\\ 4\\ 7\end{bmatrix}\text{.}\)
\begin{align*}
\mathbf{u}\times \mathbf{v}\amp =
\begin{vmatrix}
\mathbf{i} \amp \mathbf{j} \amp \mathbf{k}\\
3 \amp -10 \amp 2\\
-2 \amp 4 \amp 7
\end{vmatrix} =\mathbf{i}
\begin{vmatrix}
-10 \amp 2\\
4 \amp 7
\end{vmatrix} -\mathbf{j}
\begin{vmatrix}
3 \amp 2\\
-2 \amp 7
\end{vmatrix} +\mathbf{k}
\begin{vmatrix}
3 \amp -10\\
-2 \amp 4
\end{vmatrix}\\
\amp =\mathbf{i}\Big((-10)(7)-(2)(4)\Big)-\mathbf{j}\Big((3)(7)-(2)(-2)\Big)+\mathbf{k}\Big((3)(4)-(-10)(-2)\Big) \\
\amp =-78\mathbf{i}-25\mathbf{j}-8\mathbf{k}=\begin{bmatrix}-78\\ -25\\ -8\end{bmatrix}
\end{align*}
Subsection 1.5.3 Properties of the Cross Product
Exploration 1.5.1.
What would happen if we took the cross product of the vectors in
Example 1.5.6 but reversed the order?
Let \(\mathbf{v}=\begin{bmatrix}-2\\ 4\\ 7\end{bmatrix}\) and \(\mathbf{u}=\begin{bmatrix}3\\ -10\\ 2\end{bmatrix}\text{.}\) Recall that \(\mathbf{u}\times\mathbf{v}=\begin{bmatrix}-78\\-25\\-8\end{bmatrix}\text{.}\) We need to compute \(\mathbf{v}\times\mathbf{u}\text{.}\) If you have already studied the effect that switching two rows of a matrix has on its determinant, you should be able to guess the outcome of the upcoming computation.
\begin{align*}
\mathbf{v}\times \mathbf{u}\amp =
\begin{vmatrix}
\mathbf{i} \amp \mathbf{j} \amp \mathbf{k}\\
-2 \amp 4 \amp 7\\
3 \amp -10 \amp 2
\end{vmatrix} =\mathbf{i}
\begin{vmatrix}
4 \amp 7\\
-10 \amp 2
\end{vmatrix} -\mathbf{j}
\begin{vmatrix}
-2 \amp 7\\
3 \amp 2
\end{vmatrix} +\mathbf{k}
\begin{vmatrix}
-2 \amp 4\\
3 \amp -10
\end{vmatrix}\\
\amp =\mathbf{i}\Big((4)(2)-(7)(-10)\Big)-\mathbf{j}\Big((-2)(2)-(7)(3)\Big)+\mathbf{k}\Big((-2)(-10)-(4)(3)\Big) \\
\amp =78\mathbf{i}+25\mathbf{j}+8\mathbf{k}=\begin{bmatrix}78\\ 25\\ 8\end{bmatrix}=-(\mathbf{u}\times\mathbf{v})
\end{align*}
This computation shows that the cross product is an operation that is not commutative. It also suggests that switching the order of the vectors changes the sign of the result.
The next theorem lists two additional properties of the cross product. Proofs of these properties are routine and are left to the reader. (See
Exercise 1.5.6.6 and
Exercise 1.5.6.7)
Theorem 1.5.9.
Let \(\mathbf{u}\text{,}\) \(\mathbf{v}\) and \(\mathbf{w}\) be vectors in \(\mathbb{R}^3\text{,}\) and \(k\) be a scalar, then\\
Scalar Associativity
\begin{equation*}
(k\mathbf{u})\times \mathbf{v}=\mathbf{u}\times (k\mathbf{v})=k(\mathbf{u}\times \mathbf{v})
\end{equation*}
Distributivity
\begin{equation*}
(\mathbf{u}+\mathbf{v})\times \mathbf{w}=\mathbf{u}\times \mathbf{w}+\mathbf{v}\times \mathbf{w}
\end{equation*}
\begin{equation*}
\mathbf{u}\times (\mathbf{v}+\mathbf{w})=\mathbf{u}\times \mathbf{v}+\mathbf{u}\times \mathbf{w}
\end{equation*}
The cross product has several important geometric properties. The following problems give us a glimpse of these properties.
Exploration 1.5.2.
Compute the following products:
\begin{equation*}
\mathbf{i}\times\mathbf{j}=\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix},\quad\mathbf{j}\times\mathbf{k}=\begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}
\end{equation*}
For the two vectors in each product, sketch the vectors together with the product vector. What do you observe about the relationship between the cross product and the plane determined by the two vectors in the product?
Hint: \(\mathbf{i}\times\mathbf{j}=\mathbf{k}\text{.}\) Vector \(\mathbf{k}\) is orthogonal to both \(\mathbf{i}\) and \(\mathbf{j}\)
Subsection 1.5.4 Orthogonality Property
Theorem 1.5.10.
Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors of \(\R^3\text{,}\) then \(\mathbf{u}\times\mathbf{v}\) is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\text{.}\)
Proof.
This proof can be done by direct computation and is left to the reader. (See
Exercise 1.5.6.8)
Subsection 1.5.5 Cross Product and the Angle between Vectors
Recall that the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) is related to the angle \(\theta\) between \(\mathbf{u}\) and \(\mathbf{v}\) by the following formula
\begin{equation}
\mathbf{u}\cdot\mathbf{v}=\norm{\mathbf{u}}\norm{\mathbf{v}}\cos\theta\tag{1.5.1}
\end{equation}
We will derive an analogous result for the cross product. To do so, we will need the following Lemma.
Lemma 1.5.11.
Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in \(\R^3\text{.}\) Then
\begin{equation*}
\norm{\mathbf{u}\times\mathbf{v}}^2=\norm{\mathbf{u}}^2\norm{\mathbf{v}}^2-(\mathbf{u}\cdot\mathbf{v})^2
\end{equation*}
Proof.
The following theorem establishes a relationship between the magnitude of the cross product, the magnitudes of the two vectors involved in the cross product and the angle between the two vectors. It is important to note that the identity in this theorem involves the magnitude of the cross product, not the cross product itself.
Theorem 1.5.12.
Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in \(\R^3\text{.}\) Let \(\theta\) be the angle between \(\mathbf{u}\) and \(\mathbf{v}\) such that \(0\leq\theta\leq \pi\text{.}\) Then
\begin{equation*}
\norm{\mathbf{u}\times\mathbf{v}}=\norm{\mathbf{u}}\norm{\mathbf{v}}\sin\theta.
\end{equation*}
Proof.
\begin{align*}
\norm{\mathbf{u}\times\mathbf{v}}^2\amp =\norm{\mathbf{u}}^2\norm{\mathbf{v}}^2-(\mathbf{u}\cdot\mathbf{v})^2 \\
\amp =\norm{\mathbf{u}}^2\norm{\mathbf{v}}^2-(\norm{\mathbf{u}}\norm{\mathbf{v}}\cos \theta)^2 \\
\amp =\norm{\mathbf{u}}^2\norm{\mathbf{v}}^2-\norm{\mathbf{u}}^2\norm{\mathbf{v}}^2\cos^2\theta \\
\amp =\norm{\mathbf{u}}^2\norm{\mathbf{v}}^2(1-\cos^2\theta) \\
\amp =\norm{\mathbf{u}}^2\norm{\mathbf{v}}^2\sin^2\theta
\end{align*}
Observe that all magnitudes are non-negative. Also, \(\sin\theta\geq 0\) because \(0\leq\theta\leq \pi\text{.}\) Taking the square root of both sides give us
\begin{equation*}
\norm{\mathbf{u}\times\mathbf{v}}=\norm{\mathbf{u}}\norm{\mathbf{v}}\sin\theta
\end{equation*}
Exercises 1.5.6 Exercises
1.
2.
Exercise Group.
Find the cross product \(\mathbf{u}\times\mathbf{v}\text{,}\) and verify that \(\mathbf{u}\times\mathbf{v}\) is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\text{.}\)
3.
\(\mathbf{u}=\begin{bmatrix}2\\-1\\4\end{bmatrix}\text{,}\) \(\mathbf{v}=\begin{bmatrix}0\\3\\1\end{bmatrix}\text{.}\)
Answer.
\begin{equation*}
\mathbf{u}\times\mathbf{v}=\begin{bmatrix} -13\\ -2\\ 6\end{bmatrix}
\end{equation*}
4.
\(\mathbf{u}=\begin{bmatrix}-1\\5\\-3\end{bmatrix}\text{,}\) \(\mathbf{v}=\begin{bmatrix}2\\-1\\-4\end{bmatrix}\text{.}\)
Answer.
\begin{equation*}
\mathbf{u}\times\mathbf{v}=\begin{bmatrix} -23\\ -10\\ -9\end{bmatrix}
\end{equation*}
6.
7.
8.
9.
Prove that the cross product of any vector with itself is the zero vector.
10.
Suppose that \(\mathbf{u}\) is a non-zero vector. Let \(\mathbf{v}=k\mathbf{u}\) for \(k\neq 0\text{.}\) Argue that \(\mathbf{u}\times \mathbf{v}=\mathbf{0}\) in two different ways:
