Then, by property i.,
$0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}=0$ 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:
Use the cross product properties to calculate
$\left(2\text{i}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}3\text{j}\right)\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{j}.$
Use the properties of the cross product to calculate
$\left(\text{i}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{k}\right)\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\left(\text{k}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{j}\right).$
So far in this section, we have been concerned with the direction of the vector
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v},$ but we have not discussed its magnitude. It turns out there is a simple expression for the magnitude of
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}$ 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,
$\Vert \text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}\Vert =\Vert \text{u}\Vert \xb7\Vert \text{v}\Vert \xb7\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$
Proof
Let
$\text{u}=\u27e8{u}_{1},{u}_{2},{u}_{3}\u27e9$ and
$\text{v}=\u27e8{v}_{1},{v}_{2},{v}_{3}\u27e9$ be vectors, and let
$\theta $ denote the angle between them. Then
Taking square roots and noting that
$\sqrt{{\text{sin}}^{2}\theta}=\text{sin}\phantom{\rule{0.2em}{0ex}}\theta $ for
$0\le \theta \le 180\text{\xb0},$ we have the desired result:
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
$\text{u}=\u27e80,4,0\u27e9$ and
$\text{v}=\u27e80,0,\mathrm{-3}\u27e9.$
Use
[link] to find the magnitude of
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v},$ where
$\text{u}=\u27e8\mathrm{-8},0,0\u27e9$ and
$\text{v}=\u27e80,2,0\u27e9.$
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
$2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}2$ determinant is defined by
A
$3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}3$ determinant is defined in terms of
$2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}2$ determinants as follows:
[link] is referred to as the
expansion of the determinant along the first row . Notice that the multipliers of each of the
$2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}2$ determinants on the right side of this expression are the entries in the first row of the
$3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}3$ determinant. Furthermore, each of the
$2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}2$ determinants contains the entries from the
$3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}3$ 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
$2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}2$ determinant contains the entries that remain if you cross out the first row and first column of the
$3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}3$ determinant. Similarly, for the second term, the multiplier is
${a}_{2},$ and the
$2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}2$ determinant contains the entries that remain if you cross out the first row and second column of the
$3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}3$ determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.
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?