# 2.4 The cross product  (Page 3/16)

 Page 3 / 16

For property $\text{iv}.,$ this follows directly from the definition of the cross product. We have

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

Then, by property i., $0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}=0$ as well. Remember that the dot product of a vector and the zero vector is the scalar $0,$ whereas the cross product of a vector with the zero vector is the vector $0.$

Property $\text{vi}.$ looks like the associative property, but note the change in operations:

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

## Using the properties of the cross product

Use the cross product properties to calculate $\left(2\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}.$

$\begin{array}{cc}\hfill \left(2\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}& =2\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\hfill \\ & =2\left(3\right)\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\hfill \\ & =\left(6\text{k}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\hfill \\ & =6\left(\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right)\hfill \\ & =6\left(\text{−}\text{i}\right)=-6\text{i}.\hfill \end{array}$

Use the properties of the cross product to calculate $\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right).$

$\text{−}\text{k}$

So far in this section, we have been concerned with the direction of the vector $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v},$ but we have not discussed its magnitude. It turns out there is a simple expression for the magnitude of $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ involving the magnitudes of $\text{u}$ and $\text{v},$ and the sine of the angle between them.

## Magnitude of the cross product

Let $\text{u}$ and $\text{v}$ be vectors, and let $\theta$ be the angle between them. Then, $‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖=‖\text{u}‖·‖\text{v}‖·\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$

## Proof

Let $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩$ and $\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩$ be vectors, and let $\theta$ denote the angle between them. Then

$\begin{array}{cc}\hfill {‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖}^{2}& ={\left({u}_{2}{v}_{3}-{u}_{3}{v}_{2}\right)}^{2}+{\left({u}_{3}{v}_{1}-{u}_{1}{v}_{3}\right)}^{2}+{\left({u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right)}^{2}\hfill \\ & ={u}_{2}^{2}{v}_{3}^{2}-2{u}_{2}{u}_{3}{v}_{2}{v}_{3}+{u}_{3}^{2}{v}_{2}^{2}+{u}_{3}^{2}{v}_{1}^{2}-2{u}_{1}{u}_{3}{v}_{1}{v}_{3}+{u}_{1}^{2}{v}_{3}^{2}+{u}_{1}^{2}{v}_{2}^{2}-2{u}_{1}{u}_{2}{v}_{1}{v}_{2}+{u}_{2}^{2}{v}_{1}^{2}\hfill \\ & ={u}_{1}^{2}{v}_{1}^{2}+{u}_{1}^{2}{v}_{2}^{2}+{u}_{1}^{2}{v}_{3}^{2}+{u}_{2}^{2}{v}_{1}^{2}+{u}_{2}^{2}{v}_{2}^{2}+{u}_{2}^{2}{v}_{3}^{2}+{u}_{3}^{2}{v}_{1}^{2}+{u}_{3}^{2}{v}_{2}^{2}+{u}_{3}^{2}{v}_{3}^{2}\hfill \\ & \phantom{\rule{2em}{0ex}}-\left({u}_{1}^{2}{v}_{1}^{2}+{u}_{2}^{2}{v}_{2}^{2}+{u}_{3}^{2}{v}_{3}^{2}+2{u}_{1}{u}_{2}{v}_{1}{v}_{2}+2{u}_{1}{u}_{3}{v}_{1}{v}_{3}+2{u}_{2}{u}_{3}{v}_{2}{v}_{3}\right)\hfill \\ & =\left({u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}\right)\left({v}_{1}^{2}+{v}_{2}^{2}+{v}_{3}^{2}\right)-{\left({u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3}\right)}^{2}\hfill \\ & ={‖\text{u}‖}^{2}{‖\text{v}‖}^{2}-{\left(\text{u}·\text{v}\right)}^{2}\hfill \\ & ={‖\text{u}‖}^{2}{‖\text{v}‖}^{2}-{‖\text{u}‖}^{2}{‖\text{v}‖}^{2}{\text{cos}}^{2}\theta \hfill \\ & ={‖\text{u}‖}^{2}{‖\text{v}‖}^{2}\left(1-{\text{cos}}^{2}\theta \right)\hfill \\ & ={‖\text{u}‖}^{2}{‖\text{v}‖}^{2}\left({\text{sin}}^{2}\theta \right).\hfill \end{array}$

Taking square roots and noting that $\sqrt{{\text{sin}}^{2}\theta }=\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ for $0\le \theta \le 180\text{°},$ we have the desired result:

$‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖=‖\text{u}‖‖\text{v}‖\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$

This definition of the cross product allows us to visualize or interpret the product geometrically. It is clear, for example, that the cross product is defined only for vectors in three dimensions, not for vectors in two dimensions. In two dimensions, it is impossible to generate a vector simultaneously orthogonal to two nonparallel vectors.

## Calculating the cross product

Use [link] to find the magnitude of the cross product of $\text{u}=⟨0,4,0⟩$ and $\text{v}=⟨0,0,-3⟩.$

We have

$\begin{array}{cc}\hfill ‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖& =‖\text{u}‖·‖\text{v}‖·\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \hfill \\ & =\sqrt{{0}^{2}+{4}^{2}+{0}^{2}}·\sqrt{{0}^{2}+{0}^{2}+{\left(-3\right)}^{2}}·\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi }{2}\hfill \\ & =4\left(3\right)\left(1\right)=12.\hfill \end{array}$

Use [link] to find the magnitude of $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v},$ where $\text{u}=⟨-8,0,0⟩$ and $\text{v}=⟨0,2,0⟩.$

$16$

## Determinants and the cross product

Using [link] to find the cross product of two vectors is straightforward, and it presents the cross product in the useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have an alternative. We can calculate the cross product of two vectors using determinant    notation.

A $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinant is defined by

$|\begin{array}{cc}{a}_{1}\hfill & {a}_{2}\hfill \\ {b}_{1}\hfill & {b}_{2}\hfill \end{array}|={a}_{1}{b}_{2}-{b}_{1}{a}_{2}.$

For example,

$|\begin{array}{cc}3\hfill & \hfill -2\\ 5\hfill & \hfill 1\end{array}|=3\left(1\right)-5\left(-2\right)=3+10=13.$

A $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant is defined in terms of $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinants as follows:

$|\begin{array}{ccc}{a}_{1}\hfill & {a}_{2}\hfill & {a}_{3}\hfill \\ {b}_{1}\hfill & {b}_{2}\hfill & {b}_{3}\hfill \\ {c}_{1}\hfill & {c}_{2}\hfill & {c}_{3}\hfill \end{array}|={a}_{1}|\begin{array}{cc}{b}_{2}\hfill & {b}_{3}\hfill \\ {c}_{2}\hfill & {c}_{3}\hfill \end{array}|-{a}_{2}|\begin{array}{cc}{b}_{1}\hfill & {b}_{3}\hfill \\ {c}_{1}\hfill & {c}_{3}\hfill \end{array}|+{a}_{3}|\begin{array}{cc}{b}_{1}\hfill & {b}_{2}\hfill \\ {c}_{1}\hfill & {c}_{2}\hfill \end{array}|.$

[link] is referred to as the expansion of the determinant along the first row . Notice that the multipliers of each of the $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinants on the right side of this expression are the entries in the first row of the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant. Furthermore, each of the $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinants contains the entries from the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant that would remain if you crossed out the row and column containing the multiplier. Thus, for the first term on the right, ${a}_{1}$ is the multiplier, and the $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinant contains the entries that remain if you cross out the first row and first column of the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant. Similarly, for the second term, the multiplier is ${a}_{2},$ and the $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinant contains the entries that remain if you cross out the first row and second column of the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.

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 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!