# Can I find out magnitude and direction (+starting point) of a vectror using only one method? (graphical or mathematical)

Let's say I need to find out the magnitude and direction (by the direction i mean "position of a vector"/"starting point + direction") in uniform circular motion.

When I draw it into a diagram and take 2 vectors of displacement and bring them closer to each other (by that i mean till Δt is infinitely small), I can graphically see, that the instantaneous velocity vector is going to begin in the point itself and its direction will be a tangent line to the circle. But I cannot see its magnitude from the graphical display, because the magnitude looks infinitely small.

On the other hand, if I do it using calculus and algebra i get to some vector described as $$\vec v = (x\vec i, y\vec j)$$ and now it is easy to count the magnitude from this, but if i want to draw the vector into the diagram, how do I know its starting point is in the point doing the circular motion itself?

Or another example, in the same situation (uniform circular motion) if you get the vector of position described as $$\vec p = (a\vec i, b\vec j)$$ and vector of velocity decribed as $$\vec v = (x\vec i, y\vec j)$$ and you have to draw these into the diagram, how do you know the position vector begins in $$[0,0]$$ but the velocity vector begins in the point doing the motion?

What I am trying to ask is I know i can find the magnitude and direction using only mathematical approach but from this I will not know the starting point of the vector, is that correct? So do I need to find out the starting point separately and apply this knowledge when drawing the vector into a diagram?

• what you have here is the difference between an affine space (where your unit circle lives), which have an origin, and a plain old Euclidean space like the difference in positions of 2 points (hence velocity too).
– JEB
Jan 19, 2019 at 14:23

do I need to find out the starting point separately and apply this knowledge when drawing the vector into a diagram?

The short answer is: yes, if you want to draw the velocity vector as an arrow with its tail starting at the point mass position, you have to know where it is (its position vector) and use this information. It is not part of the information contained in the velocity vector.

Let me elaborate a little to explain more in detail the reason for this answer. In physics we know that some quantities have combination rules which allow a mapping between physical quantities and mathematical structures. Positions with respect to a given starting point are an example. In this case, there no problem of "origin". We are speaking about positions wrt a fixed point and that's all. There is only one "zero position" (i.e. the position of the chosen reference point) which is perfectly in agreement with the uniqueness of the null vector in a mathematical vector space. Notice, that in this case, all the vectors we are allowed to draw have to share a common origin. There is no room for arrows with a tail not starting from the reference point.

Although kinematics could be done by using only these vectors belonging to the same vector space (all the vectors "starting" from the same point), it turns out that in some cases, either it might be useful, or it is necessary, to introduce vectors with a different origin.

An example where it might be useful is again the example of the displacements. It is true that, since they are differences of position vectors with respect to the same reference points, they are in the same mathematical vector space as positions, and than could be drawn as arrows starting from the reference point. But it is also useful to draw a displacement vector as an arrow with the tail at the starting position and the head at the final position. Ore, more in general (see below) even starting form an arbitrary point.

In other cases, it is necessary to introduce vectors carrying information about where they start. That is the case of the so called vector fields. A couple of examples could be the field of velocities at the same time in a fluid or the electric field at each point in the space.

A graphical representation of such physical quantities would imply to draw many arrows, each starting from a different point. Those arrows cannot be modeled by a mathematical vector space. They require a different, more complex mathematical structure which is the affine vector space. Basically, it corresponds to look at the arrows with different origin as an ordered pair made by a space point and a vector, with suitable combination rules. This mathematical model mirrors the graphical manipulations we can do with arrows, and both models are a faithful representation of the corresponding physical quantities.

So, going back to your starting example. We can describe any motion as the evolution of a position vector which provides information about the relative position between the moving point and a reference point chosen as origin. In a uniform circular motion, all positions corresponding to equal interval of time are on the circumference of a circle at equally spaced points.

A displacement, i.e. a difference of position vectors may be drawn as an arrow starting from the origin, or as an arrow staring from the initial position and arriving to the final position and even other choices may be used. For example, such a freedom in the starting position could be exploited to make more clear the fact that the instantaneous velocity is tangent to the circumference. One only needs to notice that if a function $$f(t)$$ has a first derivative $$f^{\prime}(t)$$, it can be proved that $$f^{\prime}(t) ~{\overset{def}{=}}~\lim_{\Delta t \rightarrow 0} \frac{f(t+\Delta t)-f(t)}{\Delta t} = ~\lim_{\Delta t \rightarrow 0} \frac{f(t+\Delta t)-f(t-\Delta t)}{2\Delta t}$$ where the second formula is the limit of the so-called symmetric finite difference. The nice thing about using the symmetric difference formula is that, if used in connection with the position vector, for all $$\Delta t$$ it provides a difference vector with the same direction of the tangent at the point $${\bf r}(t)$$. The limit is only required to get the size (the modulus) of the velocity. However, a key point to allow to exploit such a simplification, is to be able to see the complete decoupling between vector and space point we want to associate to the vector. In a similar way, working with velocity vectors, it is also trivial to see that acceleration is orthogonal to velocity.