# Why we use vectors?

When we say that the position of an object is +5m on x axis why we need to use vectors? I mean could we don't use vectors and just say +5m on x or y or z axis instead of writing 5*unit vector either $$i,j,k$$?

• Sure but then how do you do any kind of math, e.g. adding two displacements? – DanielSank Nov 1 '18 at 22:44
• @DanielSank if you want numerical answers, you end up doing exactly what the OP wants to do - choose a coordinate system and work in it. But the advantage of vectors is that vector equations do not depend on a particular coordinate system - and that matches how physics works, because a physical object doesn't "know" anything about the coordinate system that a human decides to use, to describe how it behaves. – alephzero Nov 1 '18 at 22:55
• Because quaternions are a bigger pain? – dmckee --- ex-moderator kitten Nov 2 '18 at 1:28
• Because doing cross products by hand is tedious, versus just writing down $\vec{\omega} \times \vec{r}$ and deferring the actual calculation for later, or letting computers do the calculating. – John Alexiou Nov 2 '18 at 11:36
• Define "vector". Hint: position is NOT a vector quantity. When you understand why that is so, you will have a much better understanding of why vectors are used. Also be aware that there are competing definitions of "vector". To a physicist, a vector is something he carries around in his pocket and takes it out, when and where he needs it. A mathematician keeps all of his vectors locked in a "vector space". – Steven Thomas Hatton Nov 2 '18 at 11:55

You can do it that way, but... how is that different from vectors? If you think about it, it is the same, after all.

I get you, the formalism is complicated. Why so much complication? Well, despite some methods, I keep thinking that everybody should start separating components.

$$x=5m; y=2m$$.

That's much easier to work.

For example, when you throw an object, the $$x$$-axis is a uniform straight movement $$x=x_0+v_x\cdot t$$ and the vertical movement is a free fall: $$y=y_0+v_y\cdot t+\frac{at^2}{2}$$

It is really easier that way, I agree. I also support that method, because then it is just about putting them together. You calculate all things like that, and then, if the question asks you to give them in form of vectors, you just have to "put them together", and that doesn't need much effort:

$$\vec{r}=(x,y); \qquad \vec{v}=(v_x,v_y), \dots$$

So, the first point is that

You are already using them, altough you don't really notice.

In fact, you'd sometimes want to calculate the distance from the origin, and that's $$\sqrt{x^2+y^2}$$. The modulus of the vector, in fact.

But there are more reasons: vectors are really useful to

• Do calculations (like adding up distances and velocities). And we do know their properties very well.
• Work in different frames.

And let me explain this last point. If you say $$x=5m$$, that is attached to the reference frame you have chosen. IF you wanted to switch to another reference frame, you'd have to re-calculate everything. That's tedious.

However, vectors are a powerful tool to do this. You can use a matrix to transform ALL vectors. So you only have to calculate one matrix, and that will work to transform all positions, all velocities, and all accelerations, with one simple calculation.

It also works for forces, fields, and so on.

So yes, vectors are indeed useful:

• They appear naturally. In 1 dimension, you can work with numbers. But as soon as you go to 2 dimensions, the best way to locate points is using coordinates in a plane, $$P=(x,y)$$. But that's absolutely linked to a vector $$\vec{r}=(x,y)$$. So they appear in a natural way. A Cartesian reference frame invites us to do so. There's a small step from points to vectors.
• They allow easy visualizations: how to add up quantities, even if they are perpendicular. Calculating moduli and directions...
• They are easily transformed when switching to another reference frame.

I read an article once that said that the whole of mechanics could have been formulated in terms of quaternions rather than vectors ... but that vectors just wan!

It also showed how the elliptical orbit of a body under gravitational attraction to a point is formulated in terms of quaternions. A gorgeous piece of mathematics, it was!

• Good point, but it would be nice if you could ellaborate it a bit more: do you have a link to the article? any example? or anything else? – FGSUZ Nov 2 '18 at 11:21
• It's a long time since I read that article! It's on some ancient memory stick, or possibly on the internal storage of tablet №1 - if the latter it'll be relatively easy to find. But it was one of those articles that did make a strong impression. I do have doubts about it's being so that mathematics was 'poised between' going down the vector route & going down the £quaternion_ route: I think there might have been a bit of fancy on the part of the author there. Bi – AmbretteOrrisey Nov 2 '18 at 11:29
• but having said that I remember that the elementary celestial mechanics calculation just 'slotted in' to the quaternion machinery so very naturally & beautifully. – AmbretteOrrisey Nov 2 '18 at 11:32
• – AmbretteOrrisey Nov 2 '18 at 11:45
• – AmbretteOrrisey Nov 2 '18 at 11:46

Other people have already given some great answers, but I just wanted to give another really great reason to use vectors: dot and cross products:

Let me ask this question: How do I calculate Angular momentum of a particle moving with momentum perpendicular to its radial vector?

Non-Vectorial Picture: $$L_x = r_y p_z - r_z p_y$$ $$L_y = r_z p_x - r_x p_z$$ $$L_z = r_x p_y - r_y p_z$$

Vectorial Picture: $$\vec{L} = \vec{r} \times \vec{p} = |\vec{r}||\vec{p}| \sin(\theta)$$

Not only is the vectorial picture much easier to think about, it is also the source of the equations I wrote in the Non-vectorial picture. Furthermore, it is much easier to adjust the vectorial picture model to changes in initial conditions. For example, if I said that the momentum was no longer perpendicular to the motion but at $$45^o$$ to it, the vectorial picture would only need a small angle adjustment, whereas the non-vectorial picture would require a complete recalculation.

Hope that helps!

When you say "5 meters along the x-axis" then two things must be defined first:

• In which direction the x-axis points, and
• how long one unit is on this axis.

If you don't know this, your statement is meaningless.

And knowing those two things is the same as knowing the basis vector $$\vec i$$. The basis vector is the mathematical version of those two things written as one.

Vectors like other mathematical constructs make life easier. A vector is a finite list of numbers (called components or elements) that are grouped together and their location in the list is important.

Their power comes from the fact that they have some specific behaviors, which means that anything that can be represented by a vector also has the same behavior. These are like "badges of merit" that tells students that read textbooks and scientists and engineers that work with vectors how these things behave.

## Linear Algebra

For example: Let us explore the behaviors of vectors under the umbrella of linearity. Below are some basic rules of all vectors:

1. You add two vectors together by adding each component to the same location in the list together. So the 1st component of two vectors gets added up to produce the 1st component of the resulting vector \begin{aligned} \boldsymbol{c} & = \boldsymbol{a} + \boldsymbol{b} \\ \pmatrix{c_1 \\ c_2 \\ \vdots \\ c_n} & = \pmatrix{a_1+b_1 \\ a_2+b_2 \\ \vdots \\ a_n+b_n} \end{aligned} As a result, all vectors have the following properties:

• vector addition is commutative: $$\boldsymbol{a}+\boldsymbol{b} = \boldsymbol{c}+\boldsymbol{a}$$
• vector addition is associative: $$\boldsymbol{a}+(\boldsymbol{b}+\boldsymbol{c}) = (\boldsymbol{a}+\boldsymbol{b})+\boldsymbol{c}$$
2. You multiply any scalar value with a vector by multiplying each component with the same value \begin{aligned} \boldsymbol{c} & = \lambda \boldsymbol{a} \\ \pmatrix{c_1 \\ c_2 \\ \vdots \\ c_n} & = \pmatrix{\lambda\, a_1 \\ \lambda\, a_2 \\ \vdots \\ \lambda\, a_n} \end{aligned}

As a result, all vectors have the following properties:

• vectors have the distributive property (with respect to scalar values): $$\lambda (\boldsymbol{a}+\boldsymbol{b}) = \lambda\,\boldsymbol{a}+ \lambda\,\boldsymbol{b}$$
• you can constuct vectors from a linear combination of other vectors: $$\boldsymbol{c} = 3\, \boldsymbol{a} - 2\, \boldsymbol{b} + \ldots$$
3. Each vector has a magnitude that is defined by the square root of the sum of the squares of each component. $$\| \boldsymbol{a} \| = \sqrt{ \sum_{i=1}^n a_i^2}$$

## A Practical Example

An object is moving at a constant velocity $$\boldsymbol{v}=\pmatrix{37 \\ -5.2 \\ -22.4}$$ what is its displacement after $$t=7$$ seconds.

• The displacement vector is calculated by the equation $$\boldsymbol{r} = \boldsymbol{v}\,t$$.

$$\pmatrix{x \\ y \\ z} =7\,\pmatrix{37 \\ -5.2 \\ -22.4} = \pmatrix{259 \\ -36.4 \\ -156.8}$$

• The displacement $$d$$ is the magnitude of the displacement vector $$d = \| \boldsymbol{r} \|$$

$$d = \sqrt{ 259^2 +(-36.4)^2 + (-156.8)^2 } = \sqrt{92992.2} = 304.94622$$

So what have we learned. That although the results are equally calculatable working with components vs. with vectors, it is a lot easier to write $$d = \| \boldsymbol{v}\,t \|$$ that the expressions this expands out to.

Also we all understand that the distance if the object was moving with $$3 \boldsymbol{v}$$ would be three times as much, because of the linearity properties above. On the other hand if we specify the velocity by component, it is not obvious what the relationship is to the original velocity.

Finally, given a vector equation like $$\boldsymbol{r} = \frac{1}{2} \boldsymbol{a} t^2$$ or $$\boldsymbol{F} = m \boldsymbol{a}$$ or $$v = \| \boldsymbol{v} \|$$ we all understand what the properties of this relationship are. Double the mass $$m$$ and the force vector $$\boldsymbol{F}$$ doubles also, or the force magnitude of a combination of two forces should be $$F = \| \boldsymbol{F}_1 + \boldsymbol{F}_2 \|$$. We don't have to proove the above behavior every time we use it, because we have prooved the properies of linear algebra once already and that is enough.

Final Note I did all of the above without one mention of what coordinate frame I am using or what each component means (like x, y, z). As long I am consistent with my vector components (common coordinate system) I can perform all of the actions that linear algebra allows me and I know the results are correct. Only when you want to interpret the results you have think about things like, oh the 2nd component is along the "up" direction, or whatever convention you are using.

• yes but we must define a coordinate system..i cant say what is the unit vector if i dont define a coordinate system... – Antonios Sarikas Nov 5 '18 at 18:05
• @adosar - As I said, only when you want to interpret the results you need to be aware of the coordinate system. There is nothing stoping you from modeling a system using vectors assuming some coordinate system, without being specific until it is time to get numerical results that are meaningful. – John Alexiou Nov 5 '18 at 20:19