# 2.4 The cross product  (Page 7/16)

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Think about using a wrench to tighten a bolt. The torque $\tau$ applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound. Although torque is dimensionally equivalent to work (it has the same units), the two concepts are distinct. Torque is used specifically in the context of rotation, whereas work typically involves motion along a line.

## Evaluating torque

A bolt is tightened by applying a force of $6$ N to a 0.15-m wrench ( [link] ). The angle between the wrench and the force vector is $40\text{°}.$ Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.

Substitute the given information into the equation defining torque:

$‖\tau ‖=‖\text{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{F}‖=‖\text{r}‖‖\text{F}‖\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =\left(0.15\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(6\phantom{\rule{0.2em}{0ex}}\text{N}\right)\text{sin}\phantom{\rule{0.2em}{0ex}}40\text{°}\approx 0.58\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}.$

Calculate the force required to produce $15\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}$ torque at an angle of $30º$ from a 150-cm rod.

$20$ N

## Key concepts

• The cross product $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ of two vectors $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩$ and $\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩$ is a vector orthogonal to both $\text{u}$ and $\text{v}.$ Its length is given by $‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖=‖\text{u}‖·‖\text{v}‖·\text{sin}\phantom{\rule{0.2em}{0ex}}\theta ,$ where $\theta$ is the angle between $\text{u}$ and $\text{v}.$ Its direction is given by the right-hand rule.
• The algebraic formula for calculating the cross product of two vectors,
$\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩,$ is
$\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\left({u}_{2}{v}_{3}-{u}_{3}{v}_{2}\right)\text{i}-\left({u}_{1}{v}_{3}-{u}_{3}{v}_{1}\right)\text{j}+\left({u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right)\text{k}.$
• The cross product satisfies the following properties for vectors $\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w},$ and scalar $c\text{:}$
• $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right)$
• $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{v}+\text{w}\right)=\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}+\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}$
• $c\left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)=\left(c\text{u}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(c\text{v}\right)$
• $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0=0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}=0$
• $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=0$
• $\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)=\left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)·\text{w}$
• The cross product of vectors $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩$ and $\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩$ is the determinant $|\begin{array}{ccc}\text{i}\hfill & \text{j}\hfill & \text{k}\hfill \\ {u}_{1}\hfill & {u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill & {v}_{3}\hfill \end{array}|.$
• If vectors $\text{u}$ and $\text{v}$ form adjacent sides of a parallelogram, then the area of the parallelogram is given by $‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖.$
• The triple scalar product of vectors $\text{u},$ $\text{v},$ and $\text{w}$ is $\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right).$
• The volume of a parallelepiped with adjacent edges given by vectors $\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}$ is $V=|\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)|.$
• If the triple scalar product of vectors $\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}$ is zero, then the vectors are coplanar. The converse is also true: If the vectors are coplanar, then their triple scalar product is zero.
• The cross product can be used to identify a vector orthogonal to two given vectors or to a plane.
• Torque $\tau$ measures the tendency of a force to produce rotation about an axis of rotation. If force $\text{F}$ is acting at a distance $\text{r}$ from the axis, then torque is equal to the cross product of $\text{r}$ and $\text{F}\text{:}$ $\tau =\text{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{F}.$

## Key equations

• The cross product of two vectors in terms of the unit vectors
$\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\left({u}_{2}{v}_{3}-{u}_{3}{v}_{2}\right)\text{i}-\left({u}_{1}{v}_{3}-{u}_{3}{v}_{1}\right)\text{j}+\left({u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right)\text{k}$

For the following exercises, the vectors $\text{u}$ and $\text{v}$ are given.

1. Find the cross product $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ of the vectors $\text{u}$ and $\text{v}.$ Express the answer in component form.
2. Sketch the vectors $\text{u},\text{v},$ and $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}.$

$\text{u}=⟨2,0,0⟩,$ $\text{v}=⟨2,2,0⟩$

a. $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=⟨0,0,4⟩;$
b. $\text{u}=⟨3,2,-1⟩,$ $\text{v}=⟨1,1,0⟩$

$\text{u}=2\text{i}+3\text{j},$ $\text{v}=\text{j}+2\text{k}$

a. $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=⟨6,-4,2⟩;$
b. $\text{u}=2\text{j}+3\text{k},$ $\text{v}=3\text{i}+\text{k}$

Simplify $\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}-2\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}-4\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}+3\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}.$

$-2\text{j}-4\text{k}$

Simplify $\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}+2\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}-3\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}+5\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right).$

In the following exercises, vectors $\text{u}$ and $\text{v}$ are given. Find unit vector $\text{w}$ in the direction of the cross product vector $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}.$ Express your answer using standard unit vectors.

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