# 2.4 The cross product  (Page 2/16)

 Page 2 / 16

Notice what this means for the direction of $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}.$ If we apply the right-hand rule to $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u},$ we start with our fingers pointed in the direction of $\text{v},$ then curl our fingers toward the vector $\text{u}.$ In this case, the thumb points in the opposite direction of $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}.$ (Try it!)

## Anticommutativity of the cross product

Let $\text{u}=⟨0,2,1⟩$ and $\text{v}=⟨3,-1,0⟩.$ Calculate $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ and $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}$ and graph them.

We have

$\begin{array}{ccc}\hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & ⟨\left(0+1\right),\text{−}\left(0-3\right),\left(0-6\right)⟩=⟨1,3,-6⟩\hfill \\ \hfill \text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}& =\hfill & ⟨\left(-1-0\right),\text{−}\left(3-0\right),\left(6-0\right)⟩=⟨-1,-3,6⟩.\hfill \end{array}$

We see that, in this case, $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right)$ ( [link] ). We prove this in general later in this section.

Suppose vectors $\text{u}$ and $\text{v}$ lie in the xy -plane (the z -component of each vector is zero). Now suppose the x - and y -components of $\text{u}$ and the y -component of $\text{v}$ are all positive, whereas the x -component of $\text{v}$ is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ point?

Up (the positive z -direction)

The cross products of the standard unit vectors $\text{i},\text{j},$ and $\text{k}$ can be useful for simplifying some calculations, so let’s consider these cross products. A straightforward application of the definition shows that

$\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}=\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}=\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}=0.$

(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar $0.\right)$ It’s up to you to verify the calculations on your own.

Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of $\text{i}$ and $\text{j}$ is parallel to $\text{k}.$ Similarly, the vector product of $\text{i}$ and $\text{k}$ is parallel to $\text{j},$ and the vector product of $\text{j}$ and $\text{k}$ is parallel to $\text{i}.$ We can use the right-hand rule to determine the direction of each product. Then we have

$\begin{array}{cccccccc}\hfill \text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}& =\hfill & \text{k}\hfill & & & \hfill \text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}& =\hfill & \text{−}\text{k}\hfill \\ \hfill \text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}& =\hfill & \text{i}\hfill & & & \hfill \text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}& =\hfill & \text{−}\text{i}\hfill \\ \hfill \text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}& =\hfill & \text{j}\hfill & & & \hfill \text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}& =\hfill & \text{−}\text{j}.\hfill \end{array}$

These formulas come in handy later.

## Cross product of standard unit vectors

Find $\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right).$

We know that $\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}=\text{i}.$ Therefore, $\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right)=\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}=0.$

Find $\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}\right).$

$\text{−}\text{i}$

As we have seen, the dot product is often called the scalar product because it results in a scalar. The cross product results in a vector, so it is sometimes called the vector product    . These operations are both versions of vector multiplication, but they have very different properties and applications. Let’s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.

## Properties of the cross product

Let $\text{u},\text{v},$ and $\text{w}$ be vectors in space, and let $c$ be a scalar.

$\begin{array}{cccccccc}\text{i.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & \text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right)\hfill & & \text{Anticommutative property}\hfill \\ \text{ii.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{v}+\text{w}\right)& =\hfill & \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}+\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\hfill & & \text{Distributive property}\hfill \\ \text{iii.}\hfill & & & \hfill c\left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)& =\hfill & \left(c\text{u}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(c\text{v}\right)\hfill & & \text{Multiplication by a constant}\hfill \\ \text{iv.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0& =\hfill & 0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}=0\hfill & & \text{Cross product of the zero vector}\hfill \\ \text{v.}\hfill & & & \hfill \text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & 0\hfill & & \text{Cross product of a vector with itself}\hfill \\ \text{vi.}\hfill & & & \hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =\hfill & \left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)·\text{w}\hfill & & \text{Scalar triple product}\hfill \end{array}$

## Proof

For property $\text{i}.,$ we want to show $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right).$ We have

$\begin{array}{cc}\hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =⟨{u}_{1},{u}_{2},{u}_{3}⟩\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}⟨{v}_{1},{v}_{2},{v}_{3}⟩\hfill \\ & =⟨{u}_{2}{v}_{3}-{u}_{3}{v}_{2},\text{−}{u}_{1}{v}_{3}+{u}_{3}{v}_{1},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}⟩\hfill \\ & =\text{−}⟨{u}_{3}{v}_{2}-{u}_{2}{v}_{3},\text{−}{u}_{3}{v}_{1}+{u}_{1}{v}_{3},{u}_{2}{v}_{1}-{u}_{1}{v}_{2}⟩\hfill \\ & =\text{−}⟨{v}_{1},{v}_{2},{v}_{3}⟩\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}⟨{u}_{1},{u}_{2},{u}_{3}⟩\hfill \\ & =\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right).\hfill \end{array}$

Unlike most operations we’ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule.

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