Notice what this means for the direction of
$\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}.$ If we apply the right-hand rule to
$\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u},$ we start with our fingers pointed in the direction of
$\text{v},$ then curl our fingers toward the vector
$\text{u}.$ In this case, the thumb points in the opposite direction of
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}.$ (Try it!)
Anticommutativity of the cross product
Let
$\text{u}=\u27e80,2,1\u27e9$ and
$\text{v}=\u27e83,\mathrm{-1},0\u27e9.$ Calculate
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}$ and
$\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}$ and graph them.
We see that, in this case,
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}=\text{\u2212}\left(\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}\right)$ (
[link] ). We prove this in general later in this section.
Suppose vectors
$\text{u}$ and
$\text{v}$ lie in the
xy -plane (the
z -component of each vector is zero). Now suppose the
x - and
y -components of
$\text{u}$ and the
y -component of
$\text{v}$ are all positive, whereas the
x -component of
$\text{v}$ is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}$ point?
The cross products of the standard unit vectors
$\text{i},\text{j},$ and
$\text{k}$ can be useful for simplifying some calculations, so let’s consider these cross products. A straightforward application of the definition shows that
(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar
$0.)$ It’s up to you to verify the calculations on your own.
Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of
$\text{i}$ and
$\text{j}$ is parallel to
$\text{k}.$ Similarly, the vector product of
$\text{i}$ and
$\text{k}$ is parallel to
$\text{j},$ and the vector product of
$\text{j}$ and
$\text{k}$ is parallel to
$\text{i}.$ We can use the right-hand rule to determine the direction of each product. Then we have
We know that
$\text{j}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{k}=\text{i}.$ Therefore,
$\text{i}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\left(\text{j}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{k}\right)=\text{i}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{i}=0.$
As we have seen, the dot product is often called the
scalar product because it results in a scalar. The cross product results in a vector, so it is sometimes called the
vector product . These operations are both versions of vector multiplication, but they have very different properties and applications. Let’s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.
Properties of the cross product
Let
$\text{u},\text{v},$ and
$\text{w}$ be vectors in space, and let
$c$ be a scalar.
For property
$\text{i}.,$ we want to show
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}=\text{\u2212}\left(\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}\right).$ We have
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
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?
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it