7
$\begingroup$

I have four points as shown in this figure:

enter image description here

I want to calculate one vector for all these points. So, what would be the correct way:

1) I take the vector between $A-B, B-C, C-D$ and add them $(A-B + B-C + C-D)$ — for example: $$A-B = (x_2-x_1)i + (y_2-y_1)j + (z_2-z_1)k$$ 2) Directly take the vector between $(A-D)$, would it be the same?

Or is there any other possible solution?

$\endgroup$
  • 3
    $\begingroup$ “… one vector for all these points” There is no such thing as one vector for multiple points. In the Euclidian plane, a vector can i.e. be used to describe the relative position of two points to each other, but this does not relate to four or more points. Do you mean that you want to obtain a vector from $A$ to $D$ by “going over” the points $B$ and $C$, or do you just want some distance? To the downvoter, please leave something constructive. The matter here is obviously an uncertainty regarding mathematical terms and correct formulation, I do not think this should get -1 just for this. $\endgroup$ – Luke Jan 29 '15 at 5:35
  • $\begingroup$ @lukas Actually I want to find the single vector representation for all the four point (A, B, C, and D) that going over points B and C $\endgroup$ – Frq Khan Jan 29 '15 at 5:40
  • $\begingroup$ by the way, use Dollar signs to format LaTeX math code: $(x_1, x_2)$ gets $(x_1, x_2)$ — cf. here $\endgroup$ – Luke Jan 29 '15 at 5:44
  • $\begingroup$ @lukas Is it possible to obtain a vector which is going over all the points? $\endgroup$ – Frq Khan Jan 29 '15 at 6:02
10
$\begingroup$

First of all, welcome to Math Stackexchange!

There is no such thing as a single vector representation over all of these four points.

Mathematically, a vector in the Euclidean $\mathbb{R}^n$ is a tuple of $n$ numbers — for example $$\begin{pmatrix}a\\b\end{pmatrix},\quad a,b\in\mathbb{R} $$ If you consider the 2-dimensional case, just as in your question.

So, what are these vectors used for?

First, you can describe points with it: $\begin{pmatrix}1\\2\end{pmatrix}$ refers to the point that is located by $1$ on the X-Axis and $2$ on the Y-Axis.

If you want to think in pictures, think of the vector as an arrow pointing 1 unit to the right and 2 units up. It has no certain starting point, by this definition, the vector just tells you about the direction! But if you use this vector to describe the location of a certain point, you have to imagine this arrow to be drawn from the origin. In this situation, the vector “points” to this point. If a vector describes the location of a point $A$, we usually write $\vec{OA}$ for it.

Another thing to do with vectors is describing the relative position of two points to each other. In this case, the arrow would start at point $A$ and then point to point $B$. For such a usage of a vector, we would write $\vec{AB}$.

Geometrically, one can “chain” tho vectors — mathematically, you just add them: $$\vec{AC}=\vec{AB}+\vec{BC}$$ Flipping the direction makes it negative in each component: $$\vec{AB}=-\vec{BA}$$

By that, one can conclude i.e. that $\vec{AB}=\vec{OB}-\vec{OA}$

By the pythagorean theorem, we can even find the length of such a vector: $|\vec{a}|=\sqrt{a_x^2+a_y^2+a_z^2}$.

Back to your original question:

I want to calculate one vector for all these points

Apparently, it is just not possible to use one single vector for that purpose.

You will need to describe each point by itself, so you will end up with $\vec{OA}, \vec{OB}, \vec{OC}, \vec{OD}$ — or you pick let's say $A$ to be your first point and describe all the other points relative to the previous one, which will give you $\vec{OA}, \vec{AB}, \vec{BC}, \vec{CD}$.

But either way, you will end up with four vectors.

EDIT: As I think this might be your actual conceptual problem: You can obtain a vector from $A$ to $D$ by “going over” $B$ and $C$ ($\vec{AD}=\vec{AB}+\vec{BC}+\vec{CD}$), but this would in the end result in one vector that just describes the relative position from $A$ to $D$ — but the information with the help of which vectors you calculated at is lost.

LATER EDIT: Let me add a bit about the component or basis notation:

There are multiple ways to describe a vector. in the $\mathbb{R}_n$, one way is to write it directly, just as I did above all the time — for $n=2$, this looks like $$\vec{x}:=\begin{pmatrix}a\\b\end{pmatrix},\quad a,b\in\mathbb{R}$$.

But one can look at a vector as being a sum of each component:

Let (consider $\mathbb{R}_3$ now) $$ \vec{i}:=\begin{pmatrix}1\\0\\0\end{pmatrix}\\ \vec{j}:=\begin{pmatrix}0\\1\\0\end{pmatrix}\\ \vec{k}:=\begin{pmatrix}0\\0\\1\end{pmatrix}\\ $$ Be the “unit vectors of $\mathbb{R}^3$”

Now, using scalar multiplication, you can say $$ \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}x_1\\0\\0\end{pmatrix} + \begin{pmatrix}0\\x_2\\0\end{pmatrix} + \begin{pmatrix}0\\0\\x_3\end{pmatrix} = x_1\begin{pmatrix}1\\0\\0\end{pmatrix} + x_2\begin{pmatrix}0\\1\\0\end{pmatrix} + x_3\begin{pmatrix}0\\0\\1\end{pmatrix}\\ = x_1\vec{i} + x_2\vec{j} + x_3\vec{k} $$

Therefore, another representation of $$\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}-\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$$ would be $$(a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$$

And do not be confused by how I named the components: Because one might use vectors with more than three components some time, mathematicians tend not to use $x$, $y$ and $z$, but rather use for a vector $\vec{a}$ the components $a_1, a_2, \cdots, a_n$.

$\endgroup$
  • 1
    $\begingroup$ Can you please explain what do you mean by but the information with the help of which vectors you calculated at is lost. $\endgroup$ – Frq Khan Jan 29 '15 at 6:42
  • $\begingroup$ I mean that no matter over which vectors you “go” to obtain the resuling vector, you will always get the same result. Therefore, this resulting vector does not represent or describe those vectors you “went over” anymore. $\endgroup$ – Luke Jan 29 '15 at 6:45
  • $\begingroup$ While adding three vectors, I found that there is a different way to add three vectors (i.e. using component method). So is it same to add three vectors as you state above and using component method? $\endgroup$ – Frq Khan Jan 31 '15 at 9:37
  • $\begingroup$ Also To calculate the vector $\vec{AB}$ we will use this formula? $\vec{AB} = (x_2-x_1)i + (y_2-y_1)j + (z_2-z_1)k$? $\endgroup$ – Frq Khan Jan 31 '15 at 11:27
  • $\begingroup$ I added this to my answer, see my edit $\endgroup$ – Luke Jan 31 '15 at 19:48
0
$\begingroup$

The second option would be upper-bounded by the first. See the triangle inequality.

$\endgroup$
  • $\begingroup$ Sorry I just edited my question, can you check and tell me which way is correct? $\endgroup$ – Frq Khan Jan 29 '15 at 5:30
  • $\begingroup$ It seems to be the same. What did you change? $\endgroup$ – Kevin Sheng Jan 29 '15 at 5:53
  • $\begingroup$ I mean I want to single vector that represent all these four points.. So which way I should adopt? $\endgroup$ – Frq Khan Jan 29 '15 at 5:54
  • $\begingroup$ If you're looking for the distance between A and B then option 2 $\endgroup$ – Kevin Sheng Jan 29 '15 at 5:56
  • $\begingroup$ and if I want to find the single vector representation for all four points? how can I do that? $\endgroup$ – Frq Khan Jan 29 '15 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.