# 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.

where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
can you provide the details of the parametric equations for the lines that defince doubly-ruled surfeces (huperbolids of one sheet and hyperbolic paraboloid). Can you explain each of the variables in the equations?