# Vectors of zero magnitude

#### mrspeedybob

I read a post in an unrelated thread to the effect that some vectors do not have a direction. I assume the poster was referring to vectors with zero magnitude. At first this makes sense but the more I thought about it the more it seems to me that zero magnitude vectors do still have direction, at least in some cases.

My logic is as follows...
The purpose of mathematics is to describe reality. A velocity vector for example should tell me what I need to know about the direction and speed of an object. If the object is a ball sitting still in the middle of a field a velocity magnitude of zero tells me everything I need to know. If the object is a bullet with a velocity magnitude of zero I am still very much interested in which direction the bullet is pointed.

Based on this reasoning it seems to me that there are 2 classes of vectors, those which do and those which do not still have a direction if their magnitude is zero.

Is this reasoning in line with how vectors are usually used? If so, what kind of terminology and notation is used to distinguish the 2? If not, how should the difference be described mathematically

#### SteveL27

The purpose of mathematics is to describe reality.
That's actually a misconception. As a striking example, mathematics has Euclidean geometry and non-Euclidean geometry. Both are consistent logical systems; but they can't both be true of the physical universe.

What math does is provide abstract models and tools so that applied scientists can pick and choose whatever they need.

In math, a vector is simply an element of a vector space. That's an abstract definition that has no physical meaning at all. The number zero is a vector in the 1-dimensional vector space of the real numbers. You can add real numbers and multiply them by scalars (which are other real numbers). Zero's a perfectly reasonable vector.

In any event, even in the physical world, if two vectors of equal magnitude and opposite direction are composed with each other via vector addition, the result is the zero vector, right?

#### Studiot

I expect it was my comment.

Consider the vector {0,0} in the x-y plane.

The 'direction' of this vector is $\arctan \left( {\frac{0}{0}} \right)$

Which is what exactly?

#### Number Nine

Ignoring the fact that vectors are abstract concepts that don't really have to describe reality at all; if your bullet has velocity = 0, then in what direction is the velocity supposed to be pointing, exactly? Sounds to me like you want a vector denoting the orientation of the bullet, which is something completely different.

#### SteveL27

Ignoring the fact that vectors are abstract concepts that don't really have to describe reality at all; if your bullet has velocity = 0, then in what direction is the velocity supposed to be pointing, exactly? Sounds to me like you want a vector denoting the orientation of the bullet, which is something completely different.
Excellent point. The bullet in a loaded gun pointed at you has exactly the same zero velocity vector as the bullet in a loaded gun pointed away from you. But clearly these are not the same situation!

Perhaps zero has an orientation after all :-)

#### pwsnafu

If the object is a ball sitting still in the middle of a field a velocity magnitude of zero tells me everything I need to know. If the object is a bullet with a velocity magnitude of zero I am still very much interested in which direction the bullet is pointed.
Orientation is not the same as position. Take a bullet shell could flying at velocity (0,0,-1), which probably would be interpreted as falling, but it could rotating as it does so. Just because a bullet a heading towards does not mean the bullet oriented at you.

#### mrspeedybob

Ignoring the fact that vectors are abstract concepts that don't really have to describe reality at all; if your bullet has velocity = 0, then in what direction is the velocity supposed to be pointing, exactly? Sounds to me like you want a vector denoting the orientation of the bullet, which is something completely different.
In the example I gave, yes, the orientation of the bullet. I can think of other examples though.

To put it in more abstract terms let's suppose I have a vector described by angle a and magnitude m. m is a function of time so m=f(t). at t=now f(t)=0 and so this vector would have no dirrection but at t≠now f(t)≠0 and so the dirrection would = a. This scenerio is easy to describe in polar co-ordinates but doesn't translate easily into component vectors. if I describe the same vector in terms of components I have to say that at t=now fx(t)=0 and fy(t)=0 and so unless I define exactly what the function is I loose the a information.

#### Vorde

In the example I gave, yes, the orientation of the bullet. I can think of other examples though.

To put it in more abstract terms let's suppose I have a vector described by angle a and magnitude m. m is a function of time so m=f(t). at t=now f(t)=0 and so this vector would have no dirrection but at t≠now f(t)≠0 and so the dirrection would = a. This scenerio is easy to describe in polar co-ordinates but doesn't translate easily into component vectors. if I describe the same vector in terms of components I have to say that at t=now fx(t)=0 and fy(t)=0 and so unless I define exactly what the function is I loose the a information.
I don't see your point. Would you mind clarifying/elaborating? I don't see why what you said makes some zero vectors require direction: of course not knowing what the function is makes you lose information.

#### pwsnafu

In the example I gave, yes, the orientation of the bullet. I can think of other examples though.

To put it in more abstract terms let's suppose I have a vector described by angle a and magnitude m. m is a function of time so m=f(t). at t=now f(t)=0 and so this vector would have no dirrection but at t≠now f(t)≠0 and so the dirrection would = a. This scenerio is easy to describe in polar co-ordinates but doesn't translate easily into component vectors. if I describe the same vector in terms of components I have to say that at t=now fx(t)=0 and fy(t)=0 and so unless I define exactly what the function is I loose the a information.
You bring up polar coordinates. In this space, we define the origin as one point. That mean (0,0) is the same as (0,∏) which is the same as (0,a).

I will make this blunt: we do not use ℝn to describe orientation. We use other mathematical objects, in this case, probably SO(3). If you wanted to describe the direction and orientation of a bullet you need $\mathbb{R}^3 \oplus SO(3)$.

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