Why Do We Draw Dimensionless Unit Vectors in Diagrams?

[etotheipi]

A unit vector, $\frac{\vec{v}}{|\vec{v}|}$, has dimensions of $\frac{L}{L} = 1$, i.e. it is dimensionless. It has magnitude of 1, no units.

For a physical coordinate system, the coordinate functions $x^i$ have some units of length, e.g. $\vec{x} = (3\text{cm})\hat{x}_1 + (6\text{cm})\hat{x}_2$. For instance, the axes might arbitrarily be labelled with values in "centimetres": however this choice itself is not too important since there is a one-to-one correspondence between the values of any given length measured in two different units.

We often draw the unit vectors inside the Euclidian space, like this:

However, if the unit vectors are dimensionless, why does it make sense to draw them on the diagram? That is to say, how do you judge how long a length of a "vector of magnitude 1 (dimensionless)" is?

 

[member 587159]

Vectors itself don't have dimensions (in the physical sense).

What would be the dimension of the vector $(1,2,3)= e_1 + 2e_2 + 3e_3 \in \mathbb{R}^3$?

 

[fresh_42]

It is not dimensionless. $|\vec{v}|$ is considered a dimensionless scalar. Otherwise $\left|\dfrac{\vec{v}}{|\vec{v}|}\right|\neq \dfrac{|\vec{v}|}{|\vec{v}|}=1$ since only scalars can pulled out of the absolute value function. Thus the unit vector keep the dimension of the vector.

 

[fresh_42]

Vectors don't have dimensions (in the physical sense).

What would be the dimension of the vector $(1,2,3) \in \mathbb{R}^3$?

They do in physics, e.g.velocity has a dimension $LT^{-1}$.

 

[member 587159]

They do in physics, e.g.velocity has a dimension $LT^{-1}$.

Yes, but can't you always reduce a vector to a scalar by taking the norm? I was never taught to assign a dimension to a non-scalar quantity. But I didn't have much physics to begin with, so I may be completely wrong here.

 

[etotheipi]

It is not dimensionless. $|\vec{v}|$ is considered a dimensionless scalar. Otherwise $\left|\dfrac{\vec{v}}{|\vec{v}|}\right|\neq \dfrac{|\vec{v}|}{|\vec{v}|}=1$ since only scalars can pulled out of the absolute value function. Thus the unit vector keep the dimension of the vector.

These notes suggested to me that a unit vector is dimensionless, and of magnitude 1, whilst the vector components and the magnitude of the vector have dimensions. Specifically,

A unit vector is a dimensionless vector one unit in length used only to specify a given direction

Doesn't it make sense to say $|\vec{v}| = 10\text{ms}^{-1}$?

 

[fresh_42]

Yes, but can't you always reduce a vector to a scalar by taking the norm? I was never taught to assign a dimension to a non-scalar quantity. But I didn't have much physics to begin with, so I may be completely wrong here.

IMO it should be mandatory to note all units throughout a physical calculation. Thus the vectors have units and therewith a dimension.

Maybe I was wrong to consider the length dimensionless, and the unit vector shouldn't have one and it is a pure direction. Then the solution of the paradoxon is that it only regains its dimension if drawn, i.e. we actually draw $\dfrac{\vec{v}}{|\vec{v}|}\cdot |\vec{1}|$ where $\vec{1}$ carries the units of our drawing.

The typical case is $|\vec{v}|=\sqrt{v_1^2[m^2]+\ldots+v_n^2[m^2]}= \ldots [m].$

I officially declare post #3 wrong.

 

[etotheipi]

Then the solution of the paradox is that it only regains its dimension if drawn, i.e. we actually draw $\dfrac{\vec{v}}{|\vec{v}|}\cdot |\vec{1}|$ where $\vec{1}$ carries the units of our drawing.

I see! That makes sense. Thanks!

 

[Mark44]

These notes suggested to me that a unit vector is dimensionless, and of magnitude 1, whilst the vector components and the magnitude of the vector have dimensions. Specifically,

A unit vector is a dimensionless vector one unit in length used only to specify a given direction

Doesn't it make sense to say $|\vec{v}| = 10\text{ms}^{-1}$?

In the linked-to notes, the examples are either vectors with no dimensions given or one about the coordinates of a park in Salt Lake City, in which he was careful to exclude any information about units.

None of the examples deals with vector quantities such as velocity or force, that do have dimensions. If you have a vector that represents the wind's velocity, a unit vector in the same direction would have the same units: miles per hour, or km per hour, or whatever.

 

[jedishrfu]

I think a unit vector is considered dimensionless and only indicates a direction.

However when paired with a scalar quantity such as speed it becomes a vector quantity velocity.

 

[Mark44]

I think a unit vector is considered dimensionless and only indicates a direction.

I disagree. Forces are vectors, as are velocities, accelerations, and many other physical entities.
If you multiply a vector by a scalar, you get a new vector that points in the same or opposite direction, but still having the same units (e.g. Newtons, pounds, whatever). If you multiply a vector by the reciprocal of its magnitude/norm, the new vector still has the same units, but has a magnitude of 1.

However when paired with a scalar quantity such as speed it becomes a vector quantity velocity.

... which has the same units.

 

[PeroK]

I disagree. Forces are vectors, as are velocities, accelerations, and many other physical entities.
If you multiply a vector by a scalar, you get a new vector that points in the same or opposite direction, but still having the same units (e.g. Newtons, pounds, whatever). If you multiply a vector by the reciprocal of its magnitude/norm, the new vector still has the same units, but has a magnitude of 1.
... which has the same units.

A unit vector is, by definition, dimensionless and independent of the units. In both SI and Imperial units, for example, a unit vector has magnitude $1$ and not $1m$ or $1ft$ respectively.

The units come from the quantity you multiply the unit vector by.

 

[Delta2]

I agree that unit vectors are dimensionless.

When we draw them we probably do silently a convention that they are in the units of the diagram in which we draw them at.

 

[DrClaude]

In an expression like
$$
\vec{v} = v_x \hat{\imath} + v_y \hat{\jmath} + v_z \hat{k}
$$
I expect the $v_i$'s to have units of velocity, not $\{ \hat{\imath}, \hat{\jmath}, \hat{k} \}$.

 

[jedishrfu]

We’re conflating two concepts into one here. A vector may represent a physical quantity such a position, velocity, acceleration ... with units of measure but it may also represent simply a direction without any units.

In physics, you get used to the notion that in some instances a vector has units and in other instances it doesn’t. When a vector is divided by its length, you get a unitless unit vector. This vector is useful for determining direction. It’s components are known as direction cosines.

When describing vector quantities, we often use unit vectors like i, j, and k in vector expressions like

$$x \hat {i} + y \hat {j}+z \hat {k} $$

where it is understood that x,y and z have units of measure but the i,j,k unit vectors do not.

 

[hutchphd]

However, if the unit vectors are dimensionless, why does it make sense to draw them on the diagram? That is to say, how do you judge how long a length of a "vector of magnitude 1 (dimensionless)" is?

The appropriate length of the unit vector is determined not by the vector nor by the pencil but by the paper...

 

[fresh_42]

The appropriate length of the unit vector is determined not by the vector nor by the pencil but by the paper...

Or short: You cannot draw pure direction.