Background
I was reviewing notes of physics, and i realized that something about the mathematics of vectors was wrong in my head.
Example-problem
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 have an origin?
$\endgroup$ 76 Answers
$\begingroup$For two vectors to be equivalent, they need the same magnitude and direction. The 'starting point' for a vector is undefined.
$\endgroup$ 6 $\begingroup$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 $(a,b)$.
$\endgroup$ 17 $\begingroup$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 $2$.
$\endgroup$ $\begingroup$A vector can be placed anywhere, or rather, it does not have location.
$\endgroup$ $\begingroup$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.
$\endgroup$ 2 $\begingroup$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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