What is the Direction of a Zero (Null) Vector?

Solution 1:

The zero vector has no particular direction; this is consistent with the fact that it is orthogonal to every other vector. (It doesn't really make sense to say it has "direction 0", since direction is not a magnitude; "direction 0" makes no more sense than "direction 1" or "direction 5.873".)

Alternatively, you could say that it points in every direction, but with zero magnitude, since if you take any vector and multiply it by zero, you get the zero vector. "Every direction" is the same as "no particular direction"; it's just a different way of phrasing things.

This is one flaw with the traditional description of a vector as being a pair consisting of a magnitude and a direction: For the zero vector, the magnitude is zero, but the direction is arbitrary.

Solution 2:

Since adding the zero vector to any non-zero vector does not change the direction of the latter, it cannot have a proper direction because of the parallelogram rule :-)

Solution 3:

Just a remark: there is some relevant basic algebraic topology here.

As Matt E points out in his answer, the "direction" of a vector $v \in \mathbb{R}^n$ is not a number. What then is it? A natural answer is that a direction is an element of the unit sphere $S^{n-1} = {(x_1,\ldots,x_n) \in \mathbb{R}^n \ | \ x_1^2 + \ldots + x_n^2 = 1 }$.

Then the map, say $D$, which assigns each $v \in \mathbb{R}^n \setminus {0}$ to its direction to its direction $D(v) = \frac{v}{||v||} \in S^{n-1}$ is a deformation retraction. It follows that $\mathbb{R}^n \setminus {0}$ has the same homotopy type as the sphere $S^{n-1}$. In particular, $\mathbb{R}^n \setminus {0}$ does not have the same homotopy as $\mathbb{R}^n$ -- i.e., it is not contractible. This can be taken as a precision of the idea that there is no natural way to extend $D$ so as to be defined at the zero vector. (For that matter, there is evidently no continuous extension of $D$ to $\mathbb{R}^n$, since $D$ is surjective on any deleted neighborhood of $0$.)

Note also that any algebraic geometer will inevitably be reminded of projective space, in which the zero vector must be excluded for similar reasons. In this context, (real) projective $n-1$-space is obtained by failing to distinguish between the directions of $v$ and $-v$. Topologically, this amounts to taking a quotient of $S^{n-1}$ by identifying antipodal points.

Solution 4:

First, we need a definition of direction.

How about this one: the direction of a vector $x$ is a vector $u$ of unit length (that is, $|u| = 1$) such that $cx = u$ for some positive real number $c$.

According to the definition, it is clear that if $x = 0$, then there exists no $c$ such that $cx$ is a unit vector.

Also, it is clear that the direction $d(x)$ of a vector $x$ is a function mapping $x$ to a unit vector. For any given $x$, the coefficient $c$ is just $\frac{1}{|x|}$, and so the function is this: $$d(x) = \frac{x}{|x|}$$.

It is clear that the function is not defined at $x = 0$, where it invokes division by zero.