# 2.4 The cross product  (Page 7/16)

 Page 7 / 16

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.

what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how to synthesize TiO2 nanoparticles by chemical methods
Zubear
what's the program
Jordan
?
Jordan
what chemical
Jordan
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!