# Parts of an 'information' in phyiscs.

1. Jul 8, 2009

### dE_logics

Suppose a physical quantity needs to be represented in pictorial form...for example distance.
Such value represented should have 3 parts -
1)Orientation -
Or the angle that it makes to the axes of the Cartesian coordinate system under which things are being observed. In the 2 images below, there's a represented quantity by a line segment, the only different between the 2 is the orientation WRT the axes -

http://img221.imageshack.us/img221/3521/orentation2.png [Broken]

http://img107.imageshack.us/img107/6857/orentation1.png [Broken]

2)Position -
The exact position of the plotted value; if another value only has to differ WRT this vector by position, this value should be moved translational to give the other value. These 2 images below shows difference between the 2 plotted values by virtue of position -

http://img395.imageshack.us/img395/1133/position2.png [Broken]

http://img221.imageshack.us/img221/3977/position1.png [Broken]

3)Direction -
Given the above 2 factors, this one will be left as binary. It actually shows the direction at which the magnitude of a plotted value has progressed in. In the 2 images below, difference lies between the direction solely -

http://img240.imageshack.us/img240/6527/direction2.png [Broken]

http://img35.imageshack.us/img35/8817/direction1.png [Broken]

Are these 3 factors correct? If so, how much does a scalar and a vector define?

Last edited by a moderator: May 4, 2017
2. Jul 9, 2009

No one?

3. Jul 9, 2009

### dE_logics

Are you all yet not able to understand the problem?

4. Jul 9, 2009

### Gear300

What do you mean by "how much?"

5. Jul 10, 2009

### dE_logics

By how much I mean how much out of the all 3 factors does a scalar and vector define?

For instance a vector might define all 3 factors while a scalar might give info about only 1........just for example.

6. Jul 12, 2009

### dE_logics

I can't believe no one can answer!

7. Jul 12, 2009

### AUMathTutor

To define a 2-dimensional vector, you by definition only need 2 pieces of information. There are as many ways to do this as you can come up with, but the two most popular ones are: polar coordinates (angle and distance) and Cartesian coordinates (ordinate and abscissa - please correct my spelling as appropriate).

It seems like what you are specifying is a directed line segment. A directed line segment can be represented as a four-dimensional vector, and is therefore requires four distinct numbers to represent in its entirety. Below are some schemes for encoding this:

(x1, y1, x2, y2) - a Cartesian-Cartesian system, with the segment starting at (x1, y1) and finishing at (x2, y2).

(x1, y1, r, theta) - a Cartesian-Polar system, with the segment starting at (x1, y1) and then forming an angle theta with the +x axis and having length r.

(r1, theta1, r2, theta2) - Polar Polar

(r, theta, x2, y2) - Polar Cartesian

8. Jul 14, 2009

### dE_logics

aaa...that sorta didn't answer the question.

So finally what does a vector define?...if a line segment represented as a vector has to be plotted by representation of exact 'points'...it should define all 3.

If this is so, on the same graph we cannot represent 2 equal vectors...the other vector should be seperated by translation.

9. Jul 14, 2009

### ibcnunabit

A vector, by itself, only has magnitude ("length") and direction. It doesn't specify a position, but it can be applied to a point which does have a position. (A vector is a different thing entirely from a line segment, although it can be represented diagrammatically as one.) The initial (starting) point is the thing you are missing. Applying the vector to that point will give the other point that's at the end of your line segment (the terminal point).

A vector is a variable that has magnitude and direction. A scalar only has magnitude.
An example of a scalar would be traveling at 50 mph (speed). A vector would be going 50 mph West (velocity). But even "50 mph West" doesn't tell you where you're going 50 mph FROM, which is why you need an initial point. From that information, you can determine exactly where'd you'd be in an hour. Hope that helps.

--Mike from Shreveport

10. Jul 14, 2009

### AUMathTutor

A vector is an ordered n-tuple of scalar values over some scalar field. We're using the real numbers, I imagine. It is a list of numbers.

A line segment is defined by two points. If I give you two points, you can always draw a line segment, and if I give you a line segment, you can always pick out the two endpoints.

What is the "third point" for a line segment? If I give you the two endpoints, you can calculate the locus of all points on the line segment. If I give you any two well-defined points* on the line segment (distinct points), you can compute the locus. (well defined in this case means you have the numbers - the information - and where the numbers fall on the segment - the coding scheme)

I think the error here is your use of the word "vector". A vector is just a list of numbers; the significance you attach to the list is more or less arbitrary. For instance, you seem to be describing two-dimensional vectors that can be translated in the 2-dimensional Cartesian coordinate plane. This is easy to accomodate: the result is a 4-dimensional vector no different from the line segment above, which we can consider in several encoding schemes:

(x1, y1, x2, y2) - start point is #1, end point is #2
(xb, yb, dx, dy) - start point is b, then move dx and dy
etc.

In any of these schemes, to see whether two translated vectors v1 and v2 are equal just requires checking their components:

(1, 2, 3, 4) = (1, 2, 3, 4) != (1, 3, 4, 2)

Now, if what you're asking for is a comparison operator $which returns true if the translated vectors point in the same direction, or have the same magnitude, or both, well then all you have to do is the following: same magnitude & direction: v1$ v2 <=> (v1.x2 - v1.x1) = v2.dx AND (v1.y2 - v1.y1) = v2.dy

same direction:
v1 $v2 <=> (v1.y2-v1.y1)/(v1.x2-v1.x1) = (dy / dx) same magnitude: v1$ v2 <=> (v1.y2-v1.y1)^2 + (v1.x2-v1.x1)^2 = dx^2 + dy^2