# 2.4 The cross product

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• Calculate the cross product of two given vectors.
• Use determinants to calculate a cross product.
• Find a vector orthogonal to two given vectors.
• Determine areas and volumes by using the cross product.
• Calculate the torque of a given force and position vector.

Imagine a mechanic turning a wrench to tighten a bolt. The mechanic applies a force at the end of the wrench. This creates rotation, or torque, which tightens the bolt. We can use vectors to represent the force applied by the mechanic, and the distance (radius) from the bolt to the end of the wrench. Then, we can represent torque by a vector oriented along the axis of rotation. Note that the torque vector is orthogonal to both the force vector and the radius vector.

In this section, we develop an operation called the cross product, which allows us to find a vector orthogonal to two given vectors. Calculating torque is an important application of cross products, and we examine torque in more detail later in the section.

## The cross product and its properties

The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product of two vectors that generates a third vector orthogonal to the first two. Consider how we might find such a vector. Let $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩$ and $\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩$ be nonzero vectors. We want to find a vector $\text{w}=⟨{w}_{1},{w}_{2},{w}_{3}⟩$ orthogonal to both $\text{u}$ and $\text{v}$ —that is, we want to find $\text{w}$ such that $\text{u}·\text{w}=0$ and $\text{v}·\text{w}=0.$ Therefore, ${w}_{1},$ ${w}_{2},$ and ${w}_{3}$ must satisfy

$\begin{array}{ccc}\hfill {u}_{1}{w}_{1}+{u}_{2}{w}_{2}+{u}_{3}{w}_{3}& =\hfill & 0\hfill \\ \hfill {v}_{1}{w}_{1}+{v}_{2}{w}_{2}+{v}_{3}{w}_{3}& =\hfill & 0.\hfill \end{array}$

If we multiply the top equation by ${v}_{3}$ and the bottom equation by ${u}_{3}$ and subtract, we can eliminate the variable ${w}_{3},$ which gives

$\left({u}_{1}{v}_{3}-{v}_{1}{u}_{3}\right){w}_{1}+\left({u}_{2}{v}_{3}-{v}_{2}{u}_{3}\right){w}_{2}=0.$

If we select

$\begin{array}{ccc}\hfill {w}_{1}& =\hfill & {u}_{2}{v}_{3}-{u}_{3}{v}_{2}\hfill \\ \hfill {w}_{2}& =\hfill & \text{−}\left({u}_{1}{v}_{3}-{u}_{3}{v}_{1}\right),\hfill \end{array}$

we get a possible solution vector. Substituting these values back into the original equations gives

${w}_{3}={u}_{1}{v}_{2}-{u}_{2}{v}_{1}.$

That is, vector

$\text{w}=⟨{u}_{2}{v}_{3}-{u}_{3}{v}_{2},\text{−}\left({u}_{1}{v}_{3}-{u}_{3}{v}_{1}\right),{u}_{1}{v}_{2}-{u}_{2}{v}_{1}⟩$

is orthogonal to both $\text{u}$ and $\text{v},$ which leads us to define the following operation, called the cross product.

## Definition

Let $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩.$ Then, the cross product     $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ is vector

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

From the way we have developed $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v},$ it should be clear that the cross product is orthogonal to both $\text{u}$ and $\text{v}.$ However, it never hurts to check. To show that $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ is orthogonal to $\text{u},$ we calculate the dot product of $\text{u}$ and $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}.$

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

In a similar manner, we can show that the cross product is also orthogonal to $\text{v}.$

## Finding a cross product

Let $\text{p}=⟨-1,2,5⟩\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{q}=⟨4,0,-3⟩$ ( [link] ). Find $\text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}.$

Substitute the components of the vectors into [link] :

$\begin{array}{cc}\hfill \text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}& =⟨-1,2,5⟩\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}⟨4,0,-3⟩\hfill \\ & =⟨{p}_{2}{q}_{3}-{p}_{3}{q}_{2},{p}_{1}{q}_{3}-{p}_{3}{q}_{1},{p}_{1}{q}_{2}-{p}_{2}{q}_{1}⟩\hfill \\ & =⟨2\left(-3\right)-5\left(0\right),\text{−}\left(-1\right)\left(-3\right)+5\left(4\right),\left(-1\right)\left(0\right)-2\left(4\right)⟩\hfill \\ & =⟨-6,17,-8⟩.\hfill \end{array}$

Find $\text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}$ for $\text{p}=⟨5,1,2⟩$ and $\text{q}=⟨-2,0,1⟩.$ Express the answer using standard unit vectors.

$\text{i}-9\text{j}+2\text{k}$

Although it may not be obvious from [link] , the direction of $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ is given by the right-hand rule. If we hold the right hand out with the fingers pointing in the direction of $\text{u},$ then curl the fingers toward vector $\text{v},$ the thumb points in the direction of the cross product, as shown.

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