# What's the physical meaning of the standard base vectors?

I can find 3 displacement $$\vec d_1 ,\vec d_2, \vec d_3$$ and use them as basis so that a displacement is $$\vec d=a\vec d_1 +b\vec d_2+c \vec d_3$$. I can find 3 forces $$\vec F_1 ,\vec F_2, \vec F_3$$ and use them as basis so that a force is $$\vec F=a\vec F_1 +b\vec F_2+c \vec F_3$$. So the basis are vectors with the meaning of displacement or force or others... the physical meaning is clear in these examples. But standard base vectors $$\vec i, \vec j, \vec k$$ are different because they are good for all the kind of vectors for example they can be used to write forces as $$\vec F=a\vec I+b \vec j+c \vec k$$ or displacements as $$\vec r=d\vec I+e \vec j+f \vec k$$ and so on... But what is their physical meaning? I thought they were just direction but then it doesn't make sense the fact that they can be multiplied by a scalar and change the intensity. Furthermore it looks strange that a family of vectors like forces can be obtained by linear combinations of other objects that aren't the same kind of vectors.

• I assume that the basis vectors are orthonormal, so the component a is \begin{aligned}a=\left( \overrightarrow{F}_{1}\right) ^{T}\cdot \overrightarrow{F}, a=\left( \overrightarrow{I}\right) ^{T}\cdot \overrightarrow{F}\end{aligned} this mean that the basis I and F1 must be equal – Eli Oct 20 '19 at 21:48

Your basis vectors are unit vectors that are (ironically) unitless. So for all of the examples of basis vectors you propose, specifying what the vector represents is somewhat misleading.

In other words, you can think of all of your basis vectors ($$\hat d_1$$, $$\hat d_2$$, $$\hat d_3$$, $$\hat F_1$$, $$\hat F_2$$, $$\hat F_3$$, $$\hat i$$, $$\hat j$$, $$\hat k$$) as just directions in space. These sets of basis vectors do not represent positions, forces, etc. Your actual component magnitudes ($$a$$, $$b$$, $$c$$, etc.) have units, and this is what determines what the vector physically represents.

This is why you can explain many vector quantities with the same set of basis vectors. Position, force, etc. all have some direction in the space you are looking at. Therefore, you can express each vector using the same basis vectors.

So, when you say, "I can find three displacement vectors and use them as basis vectors," what you really mean is, "I have three displacement vectors, and I can take their directions and define basis vectors." (assuming they actually form a valid basis). With your example, these are $$\hat d_1=\frac{\mathbf d_1}{|\mathbf d_1|}$$ $$\hat d_2=\frac{\mathbf d_2}{|\mathbf d_2|}$$ $$\hat d_3=\frac{\mathbf d_3}{|\mathbf d_3|}$$

These basis vectors are just directions, they have no units associated with them, and you can use them to explain any other vector. For example, a force could be $$\mathbf F=F_1\,\hat d_1 +F_2\,\hat d_2+F_3\,\hat d_3$$ In other words, you have broken your force vector into components where each component is along the direction of each of displacement vectors you used to define your basis. The values $$F_i$$ have units of force, and this is what makes the vector a force vector. Also, note, for example, that the term $$F_1\,\hat d_1$$ is not "changing the intensity" of $$\hat d_1$$. It is just a scalar multiplication of the unit vector $$\hat d_1$$. This is analogous to how computing $$3\cdot 4=12$$ does not change what $$4$$ actually is.

1. If you are specifically considering some $$\mathbf{\hat{F}}_i$$, $$\mathbf{\hat{F}}_j$$, and $$\mathbf{\hat{F}}_k$$ as a 'force basis', you would express an arbitrary force $$\mathbf{F}$$ as $$\mathbf{F}=F_i\mathbf{\hat{F}}_i+F_j\mathbf{\hat{F}}_j+F_k\mathbf{\hat{F}}_k$$ where $$F_i$$, $$F_j$$, and $$F_k$$ are dimensionless scalars. This is not generally done because it's impractical in multiple ways. Linear independence of the basis vectors is obviously easy to work out if you're considering something like $$\mathbf{\hat{F}}_i=\langle1~\mathrm{N},0~\mathrm{N},0~\mathrm{N}\rangle$$. You claim that the basis is a set of unit vectors, so the modulus of $$\mathbf{\hat{F}}_a$$ (where $$a$$ represents $$i$$, $$j$$, or $$k$$) must be $$1$$. But the dimensions get nasty here, and we get $$||\mathbf{\hat{F}}_i||=\sqrt{1~\mathrm{N}^2}=1~\mathrm{N}\approx0.225~\mathrm{lbf}$$. This means that your unit vectors favor Newtons for some reason, since the modulus in other units is not $$1$$. That's not something you'd want. You can still call them a basis, of course, because they do work as a basisby allowing you to express any other force vector in a unique manner of the form $$\mathbf{F}=F_i\mathbf{\hat{F}}_i+F_j\mathbf{\hat{F}}_j+F_k\mathbf{\hat{F}}_k$$.

2. Let us consider a mathematical formula for instantaneous power, $$P=\mathbf{F}\cdot\mathbf{v}$$. Using the proposed set of velocity unit vectors $$\mathbf{\hat{v}}_a$$ and force unit vectors $$\mathbf{\hat{F}}_i$$, we have a highly unhelpful $$P=(F_i\mathbf{\hat{F}}_i+F_j\mathbf{\hat{F}}_j+F_k\mathbf{\hat{F}}_k)\cdot(v_i\mathbf{\hat{v}}_i+v_j\mathbf{\hat{v}}_j+v_k\mathbf{\hat{v}}_k),$$ the expansion of which isn't even clearly a scalar, since $$\mathbf{\hat{F}}_a\cdot\mathbf{\hat{v}}_a$$ and $$\mathbf{\hat{F}}_a\cdot\mathbf{\hat{v}}_b$$ haven't been defined (and defining them will give you far from elegant results).

3. The solution is to use a 'universal' set of basis vectors $$\mathbf{\hat{i}}$$, $$\mathbf{\hat{j}}$$, and $$\mathbf{\hat{k}}$$, which can be multiplied by dimensionful scalars to yield desired vectors like displacement and force. These unit vectors don't make you struggle with different sets of units (like centimeters to meters and such conversions), and they're simply indications of mutually orthogonal directions in three-dimensional space $$\mathbb{R}^3$$. The critical point is that the coefficients of these unit vectors are generally not dimensionless, though they're scalars, and that's how the vector eventually acquires an intuitive physical significance.