Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Vector math is something I find very interesting. However, we have never been told the link between vectors in physics (usually represented as arrows, e.g. a force vector) and in algebra (e.g. represented like a column matrix). It was really never explained well in classes.

Here are the things I can't wrap my head around:

  • How can a vector (starting from the algebraic definition) be represented as an arrow? Is it correct to assume that a vector (in a 2-dimensional space) $v = [1,1]$ could be represented as an arrow from the origin $[0,0]$ to the point $[1,1]$?
  • If the above assumption is correct, what does it mean in the physics representation to normalize a vector?
  • If I have a vector $[1,1]$, would the vector $[-1,1]$ be orthogonal to that first vector? (Because if you draw the arrows they are perpendicular).
  • How can one translate an object along a vector? Is that simply scalar addition?

These questions probably sound really odd, but they come from a lack of decent explanation in both physics and algebra.

share|improve this question
1  
Related questions: here, here, and here. –  Zev Chonoles May 25 '13 at 19:41
add comment

3 Answers

up vote 1 down vote accepted

Physicists tend to emphasize a geometric interpretations of vectors, which mathematicians need not do. This is because one of the main uses of vectors in physics is to talk about the geometry of some system, while vectors in general can be used in more abstract senses (and perhaps with no geometric interpretation at all).

Concerning your first bullet point, your interpretation is correct: this is a model of a vector space as an algebra of directions. Each arrow has a starting point and an ending point, which determines both length and orientation. These are the properties of a "direction".

Normalizing vectors is a way to lose the length information while maintaining directionality; it makes vectors unit length.

Yes, you're correct in saying those vectors would be orthogonal. This is part of the geometric interpretation.

I'm not sure what you refer to by "translating an object about a vector". It could be taking the location of the object and simply moving every point in the direction of the vector. This is better described by vector addition.

share|improve this answer
    
You can also show they're orthogonal in the "algebraic" interpretation by computing their inner product, $-1+1=0$. –  oldrinb May 25 '13 at 19:52
1  
Indeed, if there is a bilinear form to compute an inner product with. And indeed, sometimes matrix operations really have a Euclidean metric in mind. The transpose, for instance, is a special case of a more general adjoint of a linear operator--for Euclidean metrics, the adjoint is the transpose, but anyone who's worked with complex numbers will tell you the adjoint is the conjugate transpose instead, and so on for other signatures and metrics. Matrices don't need metrics to exist, but knowing the metric or signature can tell us what matrix operations are geometrically meaningful. –  Muphrid May 25 '13 at 20:06
add comment

You can represent a vector as an arrow in cartesian coordinates by drawing an arrow from (0,0) to the vector (row vector) for example <3,2> taken as a point on the plane (3,4). In other words, your first bullet point is correct.

Vector normalization is the process by which one takes an arbitrary vector (a, b) and converts it to a new vector (a', b') where the length of (a', b') is 1. For example, normalize(3, 4) = (3/5, 4/5). And we can verify that |(3/5, 4/5)| = sqrt(9/25 + 16/25) = sqrt(25/25) = 1.

That is correct. But keep in mind that for any vector (x, y), the vector (-x, y) will only be perpendicular to (x, y) if x = y.

share|improve this answer
    
Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. –  Zev Chonoles May 25 '13 at 19:50
    
Why would you normalize $v = [3,4]$ as $v' = [frac{3}{5},frac{4}{5}]$? Because those are simply numbers between 0 and 1? Or is 5 the magnitude of the vector? –  RaptorDotCpp May 25 '13 at 19:55
    
Yes, 5 is the magnitude of the vector. You can verify this by drawing a right triangle and using the pythagorean theorem. –  Muphrid May 25 '13 at 20:11
add comment

Actually, I think your intuition is very good here. An "arrow" has two bits of information: its length and direction. When you have a vector $v = [a, b]$, it's describing an arrow that moves $a$ in the horizontal direction and $b$ in the vertical direction. Since this defines a right triangle, this can be converted into length and direction information.

Normalizing a vector simply means adjusting the length of a vector to be $1$ without changing its direction.

To address your orthogonality question, yes, as you mentioned, $[-1, 1]$ and $[1,1]$ are indeed orthogonal, because of the geometric reasoning you correctly described. However, they are only orthogonal with respect to the standard inner product. This is imply the dot product, which has a lot of geometric connections. However, in mathematics, we can define many other types of inner products that don't have as nice geometric connections.

share|improve this answer
    
Thank you for the information. I have a (more practical) question though: say I have an object at a position (x,y) and another one at a position (i,j). If I want a vector to 'point at' (i,j) from (x,y) with a magnitute of 1, how would that vector be defined? (It's origin would lay in the point (x,y) and it would be orientated towards (i,j). –  RaptorDotCpp May 25 '13 at 20:00
    
Vectors only measure the total change, not the actual final or beginning positions. So it does not make sense to have a vector start at $(i,j)$. Vectors always begin at the origin. –  Christopher A. Wong May 26 '13 at 4:00
    
There is no problem with starting a vector anywhere in space. To get a vector pointing from point A to point B, you would subtract the vector pointing from origin to point A from the vector pointing from origin to point B. Basically its the vector difference between them, the result is a vector pointing from A to B, divide by the length to form a unit vector –  Dan May 29 '13 at 9:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.