Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I recall learning the phrase "a vector is something with magnitude and direction" many years ago. It dawned on me that "magnitudes" are well defined as numbers (or numbers with units), but I have never heard of a definition of a "direction." I know unit vectors can be treated like directions, but they are still technically vectors. I figure a direction would have the following properties:

  • A direction times a number is a vector
  • The inner product of a direction with a direction is a number
  • Directions cannot be added or subtracted

The motivation is to mathematically define directions like "north" or "east." It makes sense to say "4 meters north" or "6 meters east," and makes sense to say "4 meters north and 6 meters east" but it makes no sense to say "north and east" because you never specified how much north or east. However, it is perfectly reasonable to say $north\cdot east=0$ because north and east are perpendicular.

One could go further and define outer or exterior products between directions and inner products between the resulting entities.

So it seems to me there is decent motivation to formalize the notion of a direction, but I am unaware of whether anyone has done it.

share|cite|improve this question
$\frac{v}{\|v\|}$ – Jp McCarthy Jan 17 '13 at 3:01
Yes I know. Those however are still technically vectors and can be added and subtracted as such. I was wondering whether there was any point to defining a mathematical object analogous to a unit vector without a well defined notion of addition and subtraction. – Alex Eftimiades Jan 17 '13 at 3:04
I could say that the reason that a unit vector is still a vector and not a direction is because a vector divided by a magnitude is still a vector. The reals are closed under division, so any time you divide a vector by a magnitude you get a new vector with a different magnitude, not a direction. – Alex Eftimiades Jan 17 '13 at 3:09
I think the concept of an "angle" is pretty close to a formulation of the concept of "direction." – Gyu Eun Lee Jan 17 '13 at 3:13
up vote 5 down vote accepted

There are several ways you could define direction. Here's one that I think is intuitively clear:

Two non-zero vectors $\vec{u}, \vec{v}$ in $\mathbb{R}^n$ are equivalent if there exists $\lambda > 0$ such that $\vec{v} = \lambda \vec{u}$. It is an easy exercise to check that this is indeed an equivalence relation.

A direction is an equivalence class of non-zero vectors. This captures the notion that direction is "which way" the arrow is pointing, but it doesn't matter how long the arrow is.

The "multiplication" of a positive real number $\lambda$ and a direction should be the (unique) vector in the equivalence class of the direction that has length $\lambda$. The "product" of two directions (I'd prefer to call it the cosine of the angle between them) should be $\frac{\vec{u}}{||\vec{u}||} \cdot \frac{\vec{v}}{||\vec{v}||}$, where $\vec{u}$ and $\vec{v}$ are any representatives of the equivalence classes corresponding to the two directions. (Again, it's easy to check that you get the same number no matter which representatives you select.) Finally, note that adding or subtracting directions is not well-defined; in this case, the result of adding or subtracting representatives of the equivalence class give different values depending on which representatives you choose.

share|cite|improve this answer
I had to read the Wikipedia article on equivalence relations (I have a relatively narrow math background), but once I understood it, I realize your definition fits my notion of a direction quite well. +1 – Alex Eftimiades Jan 17 '13 at 3:52
I thought I would mention that on a Minkowski metric, I could say that the product of the "time" direction with itself is -1, so one does not have to equate the "direction product" with the cosine of an angle. I presume in its broadest sense, it is just a mapping from pairs of directions to the interval [-1,1]. – Alex Eftimiades Jan 17 '13 at 3:58

Your Answer


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.