# Are parallel vectors always scalar multiple of each others?

I read this in a tutorial of a university course :

We note that the vectors V, cV are parallel, and conversely, if two vectors are parallel (that is, they have the same direction), then one is a scalar multiple of the other.

Q1. There is an implication in the statement that two vectors are parallel if they are in same direction. Isn't it half right ? I mean in 3D space, two lines could not be in same direction and still be parallel right ?

Q2. If the above statement doesn't hold, then saying

one is a scalar multiple of the other.

is also wrong ?

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"Parallel" is defined as "have the same direction" (up to negatives). – Daniel Fischer Aug 6 '13 at 13:45
@DanielFischer You mean to say 'Parallel lines' and 'Parallel vectors' are two different things ? – Amit Tomar Aug 6 '13 at 13:46
The statement is correct, and is usually the definition of 'parallel'. Edit: @AmitTomar YES. This is correct. – Daniel Littlewood Aug 6 '13 at 13:46
Agree with @Daniel Fischer, except where it's simply defined as being linearly dependent. – Jonathan Y. Aug 6 '13 at 13:47
You're confusing concept of parallel lines and parallel vectors. Vectors are defined as so called class or relation of equivalence, therefore WLOG all vectors start from one point. From this standpoint, parallel vectors are always have either same or opposite directions. – Kaster Aug 6 '13 at 13:48

Parallel vectors on a $K$-Vector space $V$, by definiton, means: $$u \parallel v :\Leftrightarrow \exists \lambda \in K: \lambda \cdot u = v$$ Also, Parallel lines are defined by parallelicity of their respective direction vectors, wich, when fixing $0_V$ as an element of the line, implies equality of the two lines.

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Okay, third time's the charm.

First, two vectors are parallel when one is a scalar multiple of each other: given $\mathbf{u}$, $\mathbf{v}$ vectors, they are parallel if there exists $\lambda \neq 0$ such that $\mathbf{u} = \lambda \mathbf{v}$. Geometrically, this can be interpreted as follows: if $\lambda$ is positive, then the two vectors (which, remember, must both be drawn starting at the origin) point in the same direction and thus overlap. If $\lambda$ is negative, then they point in opposite directions.

Now, vectors are not the same as lines. The obvious distinction is that vectors are arrows that start at the origin (the point $(0,0,0)$ in $\mathbb{R}^3$, for example) and end somewhere, while lines are, well, lines, and they can be anywhere in space and have any direction. Any line can be described by a vector $\mathbf{v}$, called its direction vector, and a point $P$ through which the line passes. We define two lines to be parallel when their direction vectors are parallel.

You mention that two lines could not be in the same direction and still be parallel. According to our definition, that's wrong. What you're thinking of is called skew lines; they are lines that don't intersect but are also not parallel.

To answer your question, then, the statement you quote is correct. Two vectors are parallel if and only if one is a nonzero multiple of the other.

I hope this helps. If you have any doubts still, then maybe posting a drawing of what you're thinking of could help.

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How are lines parallel if their direction vectors aren't linearly dependent? – Jonathan Y. Aug 6 '13 at 13:48
@user142526: I din't word that very well. Let me fix it. – Javier Aug 6 '13 at 13:48
@JavierBadia those are called skew lines, not parallel. – Kaster Aug 6 '13 at 13:54
@Kaster: I think I may be misunderstanding the OP's question. Let me ask. – Javier Aug 6 '13 at 14:05

There is an implicit assumption in this discussion that the vector space is over the real numbers and that λ is real. If u = λv and λ is complex, according to the definition you are giving u and v would be "parallel". But geometrically that is no longer true. They could be at any angle to one another.

So at the least it might be wise to define "parallel" as meaning u = λv where λ is a real number?

However, the underlying field does not need to be either the real or complex numbers -- it can be any field at all; and the vectors do not have to be arrows; so what would "parallel" mean then?

I think this aproach would work better: Let u be a vector in the space X and W the collection of all vectors in X which are orthogonal to u. I would say u and v are "parallel" if (v,w) [inner product] = 0 for every w in W; i.e. that v is also orthogonal to every vector in W.

There are some details to attend to here. For example, we need to show that W is itself a subspace of X, with dimension one less than that of X (not hard to do). And u itself is the basis for a one-dimensional subspace of X. So now what we're saying that v is "parallel" to u if v is in the subspace generated by u.

It seems to me this approach removes the geometric considerations from the problem, and clarifies the notion of "parallel" (if you want this notion) for any vector space.

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