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I was reviewing notes of physics, and i realized that something about the mathematics of vectors was wrong in my head. The problem is the following:

Suppose a vector is $A=5\textbf{i} + 3\textbf{j}$, and other $B=7\textbf{i}+3\textbf{j}$. Then $A-B=C=-2\textbf{i}$.

The question is: Why C is not placed at the origin, from $x=0$ to $x=-2$? Is not it what $C=-2\textbf{i}$ indicates? The general question is: does a vector has an origin?

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    $\begingroup$ The essential information about vectors are direction and length. If you move the tail of a vector you do not alter its direction or length. $\endgroup$ – Tucker May 19 '17 at 2:05
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    $\begingroup$ This is the difference between vector space and affine space. Take a look here. $\endgroup$ – Masacroso May 19 '17 at 2:49
  • $\begingroup$ @Tucker Your claim is wrong! A vector in a normed vector space has a length but this is an extra feature and is not intrinsic to the vector. $\endgroup$ – Sebastian Bechtel May 19 '17 at 7:15
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    $\begingroup$ @SebastianBechtel, while your comment is technically correct, it may not be germane to this discussion. If one means by "a vector" just "an element of a vector space", then vectors have nothing in common. I think that this problem carries the context that it is asking about vectors in a Euclidean space … $\endgroup$ – LSpice May 19 '17 at 8:59
  • $\begingroup$ … or rather, as @Masacroso nicely points out (which I think is really the best perspective on the answer, rather than just declaring it by fiat), in the vector space underlying an affine Euclidean space. $\endgroup$ – LSpice May 19 '17 at 8:59
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Notice the calculation for $B-A$ is not correct in the question.

$$B-A = (7i -3j) - (5i + 3j) = (7-5)i + (-3 -3)j = 2i - 6j$$

So $i$ represents a unit vector in $x$ direction, and $j$ represents a unit vector in $y$ direction.

So what $B-A$ means is that it has component in $x$ with $2$ of unit vectors, and component in $y$ with $-6$ unit vectors.

EDIT

Now $B$ has been edited to $B=7i+3j$, so we have $A-B=C=-2i$, so how to draw that?

So a vector is a line-segment with a direction, and any translation to it won't change it. e.g. a line segment starting from $(0,0)$, ending at $(3,4)$ is the same vector as line segment starting from $(2,1)$, ending at $(5,5)$; and they are all referring to vector $3i + 4j$. Having discussed this, let's assume we always draw vector starting from $(0,0)$, and you could move them accordingly if you want, but all of them represents the same vector.

$C=-2i$ meaning C has component of $-2$ on $x$ axis, and component of $0$ on $y$ axis. Thus it is a vector starting from $(0,0)$ and ending at $(-2,0)$.

To extend this, let $C = ai + bj$, where $a,b \in \mathbb R$, then $C$ has component of $a$ on $x$ axis, and component of $b$ on $y$ axis, and the vector thus starts at $(0,0)$ and ends at $(a,b)$.

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  • $\begingroup$ yes, thanks, i edited that $\endgroup$ – santimirandarp May 19 '17 at 2:13
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    $\begingroup$ @santimirandarp For Coulomb vector, it is to represent the force, it actually do not need to be drawn starting in the charge (but we usually do so). Notice that the force vector itself only contains two pieces of information: how large is the force (length of vector), direction of the force (direction of the vector) - thus it does not contain information about where the force is applied. But we know that the force is always applied to the charge, though this piece of information is not contained in vector itself. So you do not need to, but if you do, it will be very intuitive and clear. $\endgroup$ – Yujie Zha May 19 '17 at 3:06
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    $\begingroup$ @santimirandarp Plus, if you do so, it will be very easy to visualize the addition of several forces, which is a very usual case. $\endgroup$ – Yujie Zha May 19 '17 at 3:07
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    $\begingroup$ @santimirandarp No problem! Hope this helps! $\endgroup$ – Yujie Zha May 19 '17 at 3:14
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    $\begingroup$ @santimirandarp It could form a force field, but for each point, the value is a vector. So the vector is the value at a point, given a point, we have a vector, but the value/vector itself does not information about the point. And if we want to analyze the force of a charge, we know where the charge is, and given the position of the charge, we get a vector back. e.g. $f(0,1) = v_1$ $\endgroup$ – Yujie Zha May 19 '17 at 14:03
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For two vectors to be equivalent, they need the same magnitude and direction. The 'starting point' for a vector is undefined.

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  • $\begingroup$ great, that is the thing i do not understand. If starting point is not defined, does not it produce trouble in physical problems where the starting point would have meaning? $\endgroup$ – santimirandarp May 19 '17 at 2:22
  • $\begingroup$ the SP is a co-ordinate, you have a pair $\{(x_1,\dots,x_n),\overline v\}$ $\endgroup$ – JonMark Perry May 19 '17 at 2:28
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    $\begingroup$ Perhaps it might be helpful think about velocity (which is something that a vector often represents in the physical sense). Two people can be going the same exact speed in the same exact direction even though they are in completely different locations. In that case, we'd say that their velocity is the same, despite the different specific positions. $\endgroup$ – rnrstopstraffic May 19 '17 at 2:29
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    $\begingroup$ @santimirandarp One way to reconcile this is through the idea of a tangent space, which is a separate vector space attached to each point. In a Euclidean space, each of these is an identical copy of $\mathbb R^n$ and you’re able to move the “tails” of vectors around from point to point without worrying very much about the fact that they belong to different spaces. When working with more complicated structures, the distinction between tangent spaces of different points becomes important. $\endgroup$ – amd May 19 '17 at 2:32
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    $\begingroup$ @santimirandarp In physics we usually have vector fields: each position in space is associated with a vector. The vectors alone do not give all the information needed to describe the physics of a system, so we need to attach them to a specific point. $\endgroup$ – Joren May 19 '17 at 8:54
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C=-2i indicates that vector C is parallel to the vector joining origin$(0,0)$ and $(-2,0)$, and obviously, having a length I.e.magnitude $2$.

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A vector can be placed anywhere, or rather, it does not have location.

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You're confusing vectors, which model direction and magnitude, with points, which model a location in space. It's easy to confuse the two because we use lists of real numbers to model both, but they're different concepts.

If you know calculus, the derivative can help you keep them straight. Consider a mapping from time to a point in space. The derivative of that is a mapping from time to a vector, the velocity of the point at the given time. The vector doesn't have any starting point, but if you place it on the original curve at the corresponding time, you'll see that it's tangent to that curve.

Here's an example. Suppose you have the position of an object as a function of time as so:

$$f(t) = (a + vt, b - \frac{1}{2} mgt^2)$$

The derivative of this would be:

$$f'(t) = (v, -mgt)$$

The first function deals with locations, namely the object is at $(a,b)$ at t = 0. The derivative deals with velocity, a vector. At time = 0, the object's velocity is $(v, 0)$, in other words, it's moving at a speed of $v$ in the direction of the positive x-axis.

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  • $\begingroup$ Mapping? Sorry but I did not understand $\endgroup$ – santimirandarp May 19 '17 at 12:55
  • $\begingroup$ Mapping is just another way of saying function. $\endgroup$ – David Stanley May 19 '17 at 14:30
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The resultant vector C = B + (-A). As mentioned before, a vector does not represent location.

For this case, consider the start of both the vectors A and B at origin. If you draw B and -A, you get C by parallelogram law. In this way, you can visualise the result obtained (where C.j = 0 and C.k = 0, i.e. C is parallel to x-axis)

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