The Cross Product

Section 4 The Cross Product

Let's start with the definition of the cross product.

Definition 4.1.

Let \(\mathbf{u} = \langle a,b,c \rangle \) and let \(\mathbf{v} = \langle x,y,z \rangle \text{.}\) The cross product, denoted \(\mathbf{u} \times \mathbf{v} \text{,}\) is a binary operation of vectors defined to be,

\begin{equation*} \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & b & c \\ x & y & z \\ \end{vmatrix} = (b(z) - c(y)) \mathbf{i}- (a(z) - c(x))\mathbf{j} + (a(y) - b(x))\mathbf{k} \end{equation*}

A few points of note before we start getting into how to use this product and what it is for. The cross product:

... is a vector.
... must be applied to vectors with three components. (Vectors from \(\mathbb{R}^3 \))
... is perpendicular to the other two vectors. (We will prove this below)
... follows the right hand rule. (more on this below as well)

Subsection 4.1 Cross Product Examples

Let's start by looking at a few examples of the cross product process. Then we will look more into the vectors themselves, and finally we will start to look at some interesting applications (leading to torque).

Example 4.2.

Consider the vectors, \(\mathbf{u} = \langle 2, 3, 4 \rangle \) and \(\mathbf{v} = \langle 0, 1, -2 \rangle \text{.}\) Find \(\mathbf{u} \times \mathbf{v} \) and \(\mathbf{v} \times \mathbf{u} \text{.}\) Are they the same?

\begin{equation*} \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 0 & 1 & -2 \\ \end{vmatrix} \\ = (3(-2) - 4(1)) \mathbf{i}- (2(-2) - 4(0))\mathbf{j} + (2(1) - 3(0))\mathbf{k} = \langle -10, 4, 2 \rangle \end{equation*}

\begin{equation*} \mathbf{v} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & -2 \\ 2 & 3 & 4 \\ \end{vmatrix} \\ = (1(4) - (-2)(3)) \mathbf{i}- (0(4) - (-2)(2))\mathbf{j} + (0(3) - 1(2))\mathbf{k} = \langle 10, -4, -2 \rangle \end{equation*}

From the last example we can see that the cross product is not the same when we change the order that the cross product is computed in. That is, \(\mathbf{u} \times \mathbf{v} \) may not equal \(\mathbf{v} \times \mathbf{u} \text{.}\) But all is not lost here. If we take another look with this thought in mind, we can see that they are similar. It turns out that the cross product is anticommutative. That is to say that, \(\mathbf{u} \times \mathbf{v} = - \mathbf{v} \times \mathbf{u} \text{.}\)

Proof.

Let \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \(\langle v_1, v_2, v_3 \rangle \text{.}\) Start by considering

\begin{equation*} \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ \end{vmatrix} \\ = (u_2(v_3) - u_3(v_2)) \mathbf{i}- (u_1(v_3) - u_3(v_1))\mathbf{j} + (u_1(v_2) - u_2(v_1))\mathbf{k} \end{equation*}

Now, we are going to factor out a negative from each of the above components. Note that we have also switched the order so that the negative term was on the right.

\begin{equation*} -(u_3(v_2) - u_2(v_3)) \mathbf{i} + (u_3(v_1) - u_1(v_3))\mathbf{j} - (u_2(v_1) - u_1(v_2))\mathbf{k} \end{equation*}

Next we will to consider what we would have gotten if we had crossed the vectors in the other order.

\begin{equation*} \mathbf{v} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ v_1 & v_2 & v_3 \\ u_1 & u_2 & u_3 \\ \end{vmatrix} \\ = (v_2(u_3) - v_3(u_2)) \mathbf{i}- (v_1(u_3) - v_3(u_1))\mathbf{j} + (v_1(u_2) - v_2(u_1))\mathbf{k} \end{equation*}

Finally, let's consider both of the results we found.

\begin{equation*} \mathbf{u} \times \mathbf{v} = -(u_3(v_2) - u_2(v_3)) \mathbf{i} + (u_3(v_1) - u_1(v_3))\mathbf{j} - (u_2(v_1) - u_1(v_2))\mathbf{k} \end{equation*}

\begin{equation*} \mathbf{v} \times \mathbf{u} = (v_2(u_3) - v_3(u_2)) \mathbf{i}- (v_1(u_3) - v_3(u_1))\mathbf{j} + (v_1(u_2) - v_2(u_1))\mathbf{k} \end{equation*}

We note from these collected equations that the two product orders are off by a sign. Thus, \(\mathbf{u} \times \mathbf{v} = - \mathbf{v} \times \mathbf{u} \)

Take a few minutes to try a few of these problems out. At this point the goal is to be able to correctly employ the cross product. We will start looking into the why in the next subsections.

Checkpoint 4.3. The Cross Product.

Checkpoint 4.4. The Cross Product.

Checkpoint 4.5. The Cross Product.

Subsection 4.2 The Geometry of the Cross Product

In this next section we are going to consider what all three of the vectors in the cross product are going to look like if and when we consider them visually. When we think of this product, the three vectors that we need to be considering are, \(\mathbf{u} \text{,}\) \(\mathbf{v} \) and the vector "made" from the product \(\mathbf{u} \times \mathbf{v} \text{.}\) We can choose to call this vector \(\mathbf{u} \times \mathbf{v} \) or we can give it a new name like \(\mathbf{w} \text{.}\)

Let's start with the vectors \(\mathbf{u} \) and \(\mathbf{v} \text{.}\)

Figure 4.6. Vectors \(\mathbf{u} \) and \(\mathbf{v} \) with common base point. LaTeX

Now we use what is traditionally called the right hand rule. To utilize this tool, we point our open hand in the direction of the first vector in the product (So, \(\mathbf{u} \) in \(\mathbf{u} \times \mathbf{v} \)). Then we close our hand in the direction of the second vector in the product, (So, \(\mathbf{v} \) in \(\mathbf{u} \times \mathbf{v} \)) to make a fist. Now our thumb is oriented in direction of the third vector, \(\mathbf{u} \times \mathbf{v} \text{.}\)

So for the image above,

Figure 4.7. Vectors \(\mathbf{u} \text{,}\) \(\mathbf{v} \) and \(\mathbf{u} \times \mathbf{v} \) with common base point. LaTeX

Checkpoint 4.8.

With the right hand rule described above, what would \(\mathbf{v} \times \mathbf{u} \text{?}\)

Solution

Figure 4.9. Vectors \(\mathbf{u} \text{,}\) \(\mathbf{v} \) and \(\mathbf{v} \times \mathbf{u} \) with common base point. LaTeX

With these images in mind, this is a good time to reconsider a note that we had above: The cross product is orthogonal to the other two vectors. Below the images is a proof of this fact.

Figure 4.10. Vectors \(\mathbf{u} \text{,}\) \(\mathbf{v} \) and \(\mathbf{u} \times \mathbf{v} \) with common base point with included right angles.

Proof.

Let \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \(\langle v_1, v_2, v_3 \rangle \text{.}\) Start by considering

Now we are going to use the dot product to check orthogonality. We will showcase one of the products and leave the other as an exercise.

\begin{equation*} \mathbf{u} \cdot (\mathbf{u} \times \mathbf{v}) \\ = \langle u_1, u_2, u_3 \rangle \cdot \langle (u_2(v_3) - u_3(v_2)), (u_3(v_1) - u_1(v_3)), (u_1(v_2) - u_2(v_1)) \rangle \\ = u_1(u_2(v_3) - u_3(v_2)) + u_2(u_3(v_1) - u_1(v_3)), u_3(u_1(v_2) - u_2(v_1)) \\ = (u_1u_2v_3)_a - (u_1u_3v_2)_b + (u_2u_3v_1)_c - (u_2u_1v_3)_a + (u_3u_1v_2)_b - (u_3u_2v_1)_c \\ = 0 \end{equation*}

(Subscripts used to showcase what terms cancel)

So moving forward, we want to think of the cross product of two vectors is a third vector that is protruding from the space that \(\mathbf{u} \) and \(\mathbf{v} \) are in and this vector \(\mathbf{u} \times \mathbf{v} \) is perpendicular to, \(\mathbf{u} \) and \(\mathbf{v} \text{.}\) This gives us a strong visual guide to lean on as we start to build applications around the cross product.

Subsection 4.3 Cool Geometric Application

As a last topic on the cross product before we look at torque, we are going to look at an interesting application of the cross product. Let's start with the result,

The Area of a Parallelogram and the Cross Product.

Let \(\mathbf{u} \) and \(\mathbf{v} \) be vectors that make up the adjacent sides of a parallelogram, then \(|| \mathbf{u} \times \mathbf{v}|| \) is equal to the area of the parallelogram.

This is a really interesting result that will give us a few important interconnections between the cross product and some concepts of trigonometry. To start exploring this result, let's look at an image of the parallelogram that we are considering.

Figure 4.11. Parallelogram with \(\mathbf{u} \) and \(\mathbf{v} \) as sides.