Confusion about the equation of a line in space (vector form)

[AfterSunShine]

Homework Statement

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Relevant Equations

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Hi,
In my calculus book,I found this vector form of line equation in space (bold means vector):
Given point (x1,x2,x3) lies on line L & v=<a,b,c>, then equation of line is :
r = <x1,x2,x3> + t <a,b,c> with t any number.
Now, my question if I plug any number for t, then result will be vector <x1,x2,x3> + <at,bt,ct> which is a vector not the whole line.!
In otherwords, what am supposed to get when I plug in a number for t? Am getting a vector which partial of the straight line not the line itself. is this right?

 

[jedishrfu]

If you plug in t=0, what answer do you get?

 

[FactChecker]

When you plug in a single number for t, it gives you a single vector and the associated single position in space at the tip of the vector. That is one point on the line. When you look at all the possible values of t, you get all the associated possible points on the line.

 

[AfterSunShine]

If you plug in t=0, what answer do you get?

I will get the vector <x1,x2,x3>

 

[AfterSunShine]

When you plug in a single number for t, it gives you a single vector and the associated single position in space at the tip of the vector. That is one point on the line. When you look at all the possible values of t, you get all the associated possible points on the line.

If I plug a number for t i will get vector not point! Still confuses me.
Put t=k, then i will get <x1,x2,x3> + <ka,kb,kc>=< x1+ka , x2+kb , x3+kc > which a vector going from origin to the line. Vector not a point.

 

[jedishrfu]

But it's a position vector, the equation describes the position (vector) of a spaceship at any time t.

https://www.math.usm.edu/lambers/mat169/fall09/lecture25.pdf

 

[FactChecker]

If I plug a number for t i will get vector not point! Still confuses me.

Given a single value, $t_0$, for $t$, the vector goes from the origin $(0,0,0)$ to a point $(x1+at_0, x2+bt_0, x3+ct_0)$ at the tip. That tip is at one point on the line with coordinates $(x1+at_0, x2+bt_0, x3+ct_0)$. When you consider all possible values for $t_0$, you get all possible points on the line.

 

[mathwonk]

This stuff is confusing becuse sometimes three coordinates stands for a point and sometimes for a vector. I personally like to think of it this way: you can add two vectors and get a vector, but you can also add a point to a vector and you get a point, namely the point obtained by translating the original point in the direction of the given vector. In this sense, p+tv is a point. Similarly although you cannot add two points, you can subtract two points, the result is vector, namely q-p = the vector v that would translate p to q. Hence q = p+v if and only if q-p = v.

Thus the equation for the (points on the) line through p and in the direction v, is {p+tv, for all t}. I.e. each choice of t makes p+tv a point on that line. And the set of all points on that line is the set of all points of form p+tv for all choices of t.

This of course is essentially only a tiny twist on what factchecker just said.

 

[robphy]

Some texts (e.g. Bamberg & Sternberg A course in mathematics for students of physics)
will use different brackets to distinguish a point from a vector
$\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{c} x_0 \\ y_0 \end{array}\right] + t \left(\begin{array}{c} u \\ v \end{array}\right)$

 

[mathwonk]

thank you robphy, that is absolutely brilliant.

 

[Mark44]

This stuff is confusing becuse sometimes three coordinates stands for a point and sometimes for a vector.

Yes, there is some ambiguity in how points in space and vectors in space are represented. For example, (1, 2, 3) could be both the terminal point of a vector that originates from the origin, as well as the vector itself. When I write vectors/points, I usually will put angle brackets around the vector coordinates (e.g., $\vec v = <1, 2, 3>$) and parentheses around coordinates of a point (e.g. $A = (1, 2, 3)$). For $\vec v$ in my example, it is implied that this vector starts at the origin.

 

[MidgetDwarf]

Some texts (e.g. Bamberg & Sternberg A course in mathematics for students of physics)
will use different brackets to distinguish a point from a vector
$\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{c} x_0 \\ y_0 \end{array}\right] + t \left(\begin{array}{c} u \\ v \end{array}\right)$

Yes, or some books use the two interchangeably. Causes more confusion when brackets/parenthesis are not used.

@ OP.

We can think of points and vectors as the same thing in R^n.

A line in space contains two parts:
(1) a point it passes through.
(2)direction we want parallel to a vector.

ie., r = a + tb , where a is (1) and b is (2) from above.

Now, why do we have t there? Recall that vectors are parallel to one another if they are scalar multiples.
Consider <1,1,1>. Are <1,1,1> and <2,2,2> parallel? Yes, since <2,2,2> = 2<1,1,1>.

Can you see what t, the parameter is doing now?

I would advice that you go read and go through the derivation of the parameterization of a line in R^n a few times, and convince yourself why it works...

Hmm. Read Mathwonk's reply over and over again. He summarized what the derivation explains succinctly.

 

[malawi_glenn]

A line can be defined using a vector. However, a line is set of points.
$L = \lbrace \left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{c} x_0 \\ y_0 \end{array}\right] + t \left(\begin{array}{c} u \\ v \end{array}\right) \, : \, t \in \mathbb{R}\rbrace$

I totally agree that it can be confusing to things like "adding a point to a vector", and having the same notation (a,b,c) for both a point and a vector. I never write (a,b,c) alone but always "Q=# or "$\vec q =$" infront. There are so much notation in math that is not there to make it easy to learn. They are what they are and at some point you have to be able to deal with different notations.

You can think of it like this then. You want a line that goes through the point $(x_0, y_0)$ and has $(u,v)$ as directional vector. Consider the vector that starts in the origin O = (0,0) and ends at the point $(x_0, y_0)$. This vector will also be represented by $(x_0, y_0)$. So now you can add this with $t(u,v)$

 

[mathwonk]

some semi philosophical thoughts, which may be flawed:

a vector is really determined by 2 points. intuitively a vector is something dynamic, whereas a point is something static. the coordinates of a vector are things to be added to the coordinates of a point. i.e. a vector acts on a point. the velocity vector of a particle acts on the position of that particle and takes it to a new position. the position is a point, the velocity is a vector.

unfortunately we sometimes speak of the "position vector" which is another source of confusion. we do this because one can in practice only specify the location of something in reference to the location of something else. so we speak of the velocity vector that would take the origin to our point as the "position vector" of our point. but this has no physical meaning independent of the choice of origin, whereas a velocity vector does have.

 

[robphy]

some semi philosophical thoughts, which may be flawed:

a vector is really determined by 2 points. intuitively a vector is something dynamic, whereas a point is something static. the coordinates of a vector are things to be added to the coordinates of a point. i.e. a vector acts on a point. the velocity vector of a particle acts on the position of that particle and takes it to a new position. the position is a point, the velocity is a vector.

unfortunately we sometimes speak of the "position vector" which is another source of confusion. we do this because one can in practice only specify the location of something in reference to the location of something else. so we speak of the velocity vector that would take the origin to our point as the "position vector" of our point. but this has no physical meaning independent of the choice of origin, whereas a velocity vector does have.

I sometimes like to use "location" instead of "position" and
I try to suggest the idea of convenient "labeling" and
suggest how the origin and orientation of axes might be arbitrary.

The term "position vector" could be troubling because
what would it mean physically to add two position vectors or scalar multiply a position vector
(whose result depends on the choice of origin)?

We could do an affine sum of positions to get a sort of "average position".
We could subtract two positions to get a true vector--the displacement vector.

Maybe many really want to say that the "position" requires 3 coordinates (like a vector does).

It's likely that many users of the word "vector" think of it primarily as an array of 3 coordinates,
and not as an element of a "vector space".
(Indeed, in physics classes, the usual definition of vector makes reference to a "magnitude",
which makes it a little difficult when other vectors in more advanced classes may have
no assignment of "magnitude" or have a non-Euclidean magnitude.)

 

[mathwonk]

"We could subtract two positions to get a true vector--the displacement vector."

In this vein, I like a coordinate free conception of vectors and points. Namely we have a flat space of points, without coordinates, and on this space there acts a family of translations, which are the vectors. I.e.a vector is by definition a translation of the space of points. The vectors thus act on the points, and vectors can be added by composing the translations, and scaled by stretching or contracting the translations. Then given any point and any "vector", i.e. any translation, we obtain a second point by translating the first one. These two points represent the translation, which you can picture by an arrow joining the points, and all such arrows represent the same translation, i.e. the same vector. Of course you can't calculate anything until you introduce coordinates. But this gives a mathematically rigorous way to consider all arrows with the same length and direction as "equal" vectors, since they represent the same translation. It also makes clear the dynamic nature of a vector as opposed to the static nature of a point.

 

[vanhees71]

I think one problem leading to this confusion is that, at least in physics, we often do not clear distinguish between a (Euclidean) vector space and an affine (Euclidean) manifold, and analytical geometry deals with the latter. An affine manifold consists of a stet with elements called "points" and a (Euclidean) vector space. The points and vectors are related in the sense that for two points there's a vector $\overrightarrow{AB}$. Its physical meaning in our usual space is that it's the directed straight connection of point $A$ with point $B$. All such "arrows" are identified as the same if they can be parallel shifted into the other. Now the vector operations (addition of two vectors and multiplication with a real number) have an intuitive meaning. E.g., for any points $A$, $B$, and $C$ we have $\overrightarrow{AB} + \overrightarrow{BC}=\overrightarrow{AC}$ etc. To write down all the rules is lengthy but pretty straight forward, and they build the axiomatics of what an affine space is. If you endow the vector space with a scalar product you have automatically also a Euclidean metric and you can thus define distances between points and angles between straight lines by this metric. What you get is indeed equivalent to the usual Euclidean geometry but now with the possibility to use vector algebra and vector calculus to deal with it in a much more convenient way than without these algebraic and analytic tools.

Now a position vector indicates uniquely any point $P$ of the affine manifold by defining an arbitrary point $O$ ("the origin") and then defining the positive vector as $\vec{r}=\overrightarrow{OP}$. Then you can describe arbitrary curves by defining functions $\vec{r}(t)$ with $t \in \mathbb{R}$. Then it should be pretty obvious that with
$$\overrightarrow{OP(t)}=\vec{r}(t)=\overrightarrow{OP_0} + t \vec{n}$$
the points $P(t)$ define a straight line which goes through $P_0$.

 

[FactChecker]

"We could subtract two positions to get a true vector--the displacement vector."

In this vein, I like a coordinate free conception of vectors and points. Namely we have a flat space of points, without coordinates, and on this space there acts a family of translations, which are the vectors. I.e.a vector is by definition a translation of the space of points.

A coordinate-agnostic vector would be a tensor. That implies much more than the simple mathematical vector does.

 

[vanhees71]

Vectors are always independent of coordinates and are of course tensors of rank 1. In the axiomatics of affine manifolds there's no definition of subtraction of addition of two points but, of course, of vectors.

 

[FactChecker]

Vectors are always independent of coordinates and are of course tensors of rank 1. In the axiomatics of affine manifolds there's no definition of subtraction of addition of two points but, of course, of vectors.

A general mathematical definition and use of the term "vector" does not always imply that it is a tensor. A general vector space can be multiplied by the scalars of a field. It's more general and abstract than tensors.

EDIT: In the context of this thread, I don't know if the abstract mathematical definition of a vector space is appropriate. I may have hijacked the thread. Sorry.

 

[vanhees71]

A vector is always a tensor of first rank, because you can canonically identify the bidual of a vector space with this vector space, and that's usually done for convenience.

 

[FactChecker]

A vector is always a tensor of first rank, because you can canonically identify the bidual of a vector space with this vector space, and that's usually done for convenience.

In the abstract mathematical definition of a vector space, that is simply not true. The coordinate transformation properties of tensors are not part of the definition.

 

[vanhees71]

Of course the coordinate transformation properties are deduced. I don't have any coordinates in mind to begin with.

You have a finite-dimesnional vector space $V$ over a field $K$ with the usual axioms of vectors (addition of two vectors as an abelian group and multiplication of a vector with a scalar). Then you can define tensors of 1st rank (dual vectors) as linear forms, i.e., a mapping $L:V \rightarrow K$ with $L(\lambda_1 \vec{v}_1 + \lambda_2 \vec{v}_2)=\lambda_1 L(\vec{v}_1) + \lambda_2 L(\vec{v}_2)$ for all $\lambda_1,\lambda_2 \in K$ and all $\vec{v}_1,\vec{v}_2 \in V$. These linear forms build themselves a vector space when defining addition and multiplication of a scalar in the obvious "pointwise" way. That's called the dual space $V^*$. Now you can build the "double dual", i.e., the linear forms defined on $V^*$, forming the dual of the dual space, $V^{**}$ but these can be "canonically maped" to the vectors. You define $A:V \rightarrow V^{**}$, $\vec{v} \mapsto A_{\vec{v}}$ by defining $A_{\vec{v}}$ by $A_{\vec{v}}(L)=L(\vec{v})$ for any $L \in V^*$. It's easy to see that $A$ is linear and an isomorphism.

 

[Mark44]

Of course the coordinate transformation properties are deduced. I don't have any coordinates in mind to begin with.

You have a finite-dimesnional vector space $V$ over a field $K$ with the usual axioms of vectors (addition of two vectors as an abelian group and multiplication of a vector with a scalar). Then you can define tensors of 1st rank (dual vectors) as linear forms, i.e., a mapping $L:V \rightarrow K$ with $L(\lambda_1 \vec{v}_1 + \lambda_2 \vec{v}_2)=\lambda_1 L(\vec{v}_1) + \lambda_2 L(\vec{v}_2)$ for all $\lambda_1,\lambda_2 \in K$ and all $\vec{v}_1,\vec{v}_2 \in V$. These linear forms build themselves a vector space when defining addition and multiplication of a scalar in the obvious "pointwise" way. That's called the dual space $V^*$. Now you can build the "double dual", i.e., the linear forms defined on $V^*$, forming the dual of the dual space, $V^{**}$ but these can be "canonically maped" to the vectors. You define $A:V \rightarrow V^{**}$, $\vec{v} \mapsto A_{\vec{v}}$ by defining $A_{\vec{v}}$ by $A_{\vec{v}}(L)=L(\vec{v})$ for any $L \in V^*$. It's easy to see that $A$ is linear and an isomorphism.

I'm going to go out on a limb and speculate that any talk about tensors and finite-dimensional vector spaces is well beyond the scope of the OP's question.

Let's see whether the OP comes back with more questions before venturing further down this rabbit hole.