Let
$L$ be a line in the plane and let
$M$ be any point not on the line. Then, we define distance
$d$ from
$M$ to
$L$ as the length of line segment
$\stackrel{\u2014}{MP},$ where
$P$ is a point on
$L$ such that
$\stackrel{\u2014}{MP}$ is perpendicular to
$L$ (
[link] ).
When we’re looking for the distance between a line and a point in space,
[link] still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let
$P$ be an arbitrary point on line
$L$ and let
$\text{v}$ be a direction vector for
$L$ (
[link] ).
By
[link] , vectors
$\overrightarrow{PM}$ and
$\text{v}$ form two sides of a parallelogram with area
$\Vert \overrightarrow{PM}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}\Vert .$ Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:
We can use this formula to find a general formula for the distance between a line in space and any point not on the line.
Distance from a point to a line
Let
$L$ be a line in space passing through point
$P$ with direction vector
$\text{v}.$ If
$M$ is any point not on
$L,$ then the distance from
$M$ to
$L$ is
Find the distance between t point
$M=\left(1,1,3\right)$ and line
$\frac{x-3}{4}=\frac{y+1}{2}=z-3.$
From the symmetric equations of the line, we know that vector
$\text{v}=\u27e84,2,1\u27e9$ is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point
$P\left(3,\mathrm{-1},3\right)$ lies on the line. Then,
Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be
skew lines (
[link] ).
To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point (
[link] ).
Classifying lines in space
For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.
Line
${L}_{1}$ has direction vector
${\text{v}}_{1}=\u27e82,1,1\u27e9;$ line
${L}_{2}$ has direction vector
${\text{v}}_{2}=\u27e81,3,\mathrm{-2}\u27e9.$ Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point,
$\left(x,y,z\right),$ that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:
There is no single point that satisfies the parametric equations for
${L}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{L}_{2}$ simultaneously. These lines do not intersect, so they are skew (see the following figure).
Line
L
_{1} has direction vector
${\text{v}}_{1}=\u27e81,\mathrm{-1},1\u27e9$ and passes through the origin,
$\left(0,0,0\right).$ Line
${L}_{2}$ has a different direction vector,
${\text{v}}_{2}=\u27e82,1,1\u27e9,$ so these lines are not parallel or equal. Let
$r$ represent the parameter for line
${L}_{1}$ and let
$s$ represent the parameter for
${L}_{2}\text{:}$
Solve the system of equations to find
$r=1$ and
$s=-1.$ If we need to find the point of intersection, we can substitute these parameters into the original equations to get
$\left(1,\mathrm{-1},1\right)$ (see the following figure).
Lines
${L}_{1}$ and
${L}_{2}$ have equivalent direction vectors:
$\text{v}=\u27e86,\mathrm{-2},3\u27e9.$ These two lines are parallel (see the following figure).
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
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.