Parametric Representation of Lines

[NoahsArk]

TL;DR

Help understanding the concept of representing a line in this way.

I am having trouble with the concept that the equation L = {x + tv} is the more general form of the more familiar y = mx + b (In the first equation there should be a vector sign above the x and the v). It's hard for me to see the similarities between these two equations.
1: Even if we are dealing with a 2d space like y = mx + b lies in, in the equation L = {x + tv} we are dealing with the addition of two vectors (which I visualize as two line segments), whereas in y = mx + b we are only dealing with one line.
2: In y = mx + b, we are dealing with a relationship between y and x whereas in L = {x + tv} we are defining a line.
3: The comparison becomes even harder to grasp when we go beyond 3d space. How can we even call something a line that has more than three dimensions? Any conceptual explanation about what L = {x + tv} means would be helpful. I understand that y = mx + b represents a relationship between two variables and how one variable changes as the other changes, but I am not sure what L = {x + tv} is trying to show.
Thanks!

 

[PeroK]

First $y = mx + b$ is a straight line in a 2D plane, described using a relationship between the two coordinates. Alternatively, we could write $y - y_0 = m(x - x_0)$. Here we have one point on the line $(x_0, y_0)$ and then a gradient defined by $m$. You should see that this also describes a straight line in the 2D x-y plane.

We can replace the point $(x_0, y_0)$ with a vector and describe the direction of the line by the vector $(1, m)$. I'll leave it to you to check that if we take the set of points $(x, y)$ where $(x, y) = (x_0, y_0) + t(1, m)$, then these points satisfy the equation $y - y_0 = m(x - x_0)$. Where $t$ is a (real number) parameter.

Finally, we can define $\vec x = (x_0, y_0)$ and $\vec v = (1, m)$ and we have the equation of the line in vector format: $\vec x + t\vec v$.

This form of the equation then generalises to a line in any number of dimensions. Note that the line so defined is one dimensional, no matter how many dimensions the vector space has.

 

[fresh_42]

TL;DR Summary: Help understanding the concept of representing a line in this way.

I am having trouble with the concept that the equation L = {x + tv} is the more general form of the more familiar y = mx + b (In the first equation there should be a vector sign above the x and the v). It's hard for me to see the similarities between these two equations.
1: Even if we are dealing with a 2d space like y = mx + b lies in, in the equation L = {x + tv} we are dealing with the addition of two vectors (which I visualize as two line segments), whereas in y = mx + b we are only dealing with one line.
2: In y = mx + b, we are dealing with a relationship between y and x whereas in L = {x + tv} we are defining a line.
3: The comparison becomes even harder to grasp when we go beyond 3d space. How can we even call something a line that has more than three dimensions? Any conceptual explanation about what L = {x + tv} means would be helpful. I understand that y = mx + b represents a relationship between two variables and how one variable changes as the other changes, but I am not sure what L = {x + tv} is trying to show.
Thanks!

\begin{align*}
L&=\left\{\left.\vec{x}(t)\,\right|\,\vec{x}(t)=(x_1(t),x_2(t))=\vec{b}+ t\vec{v} \text{ for }t\in \mathbb{R}\right\}\\
&=\left\{\left.\vec{x}(t)\,\right|\,x_1(t)=b_1+tv_1\wedge x_2(t)=b_2+tv_2 \text{ for }t\in \mathbb{R}\right\}\\
&=\left\{\vec{x}(t)\,\left|\,t=\dfrac{x_1(t)-b_1}{v_1}\wedge x_2(t)=b_2+tv_2 \text{ for }t\in \mathbb{R}\right.\right\}\\
&=\left\{\vec{x}(t)\, \left| \,x_2(t)=b_2+\dfrac{x_1(t)-b_1}{v_1} v_2 \text{ for }t\in \mathbb{R}\right. \right\}\\
&=\left\{\vec{x}(t)\, \left| \,x_2=\underbrace{\dfrac{v_2}{v_1}}_{=:m}x_1+\underbrace{\dfrac{b_2v_1-b_1v_2}{v_1}}_{=:b}\text{ for }x_1\in \mathbb{R}\right.\right\}\\
&=\left\{\vec{x}\, \left| \,\vec{x}=(x,y)\wedge y=mx+b\text{ for }x\in \mathbb{R}\right.\right\}
\end{align*}

The set is the same, but the parameterizations of its elements are different, one by all $t\in \mathbb{R}$ and the other by all $x\in \mathbb{R}.$ They both run over all real numbers. The parameterization with $t$ is better because we do not have to separately deal with vertical lines. The equations above require $v_1\neq 0.$ Hence, the case $v_1=0$ requires a different calculation (parameterization by $y\in \mathbb{R}$).

 

[NoahsArk]

We can replace the point (x0,y0) with a vector and describe the direction of the line by the vector (1,m). I'll leave it to you to check that if we take the set of points (x,y) where (x,y)=(x0,y0)+t(1,m), then these points satisfy the equation y−y0=m(x−x0). Where t is a (real number) parameter.

Thanks. I just started studying linear algebra, so I am not sure what (1,m) means. I thought direction of a vector was measured by an angle. Can you please give an example where:

(x,y)=(x0,y0)+t(1,m)

I am also a bit unclear on what the t means- my understanding is that it's a factor that you can use to scale up or down the size of a vector, but I'm not sure what it means in the context of t(1,m).

 

[NoahsArk]

\begin{align*}
L&=\left\{\left.\vec{x}(t)\,\right|\,\vec{x}(t)=(x_1(t),x_2(t))=\vec{b}+ t\vec{v} \text{ for }t\in \mathbb{R}\right\}\\
&=\left\{\left.\vec{x}(t)\,\right|\,x_1(t)=b_1+tv_1\wedge x_2(t)=b_2+tv_2 \text{ for }t\in \mathbb{R}\right\}\\
&=\left\{\vec{x}(t)\,\left|\,t=\dfrac{x_1(t)-b_1}{v_1}\wedge x_2(t)=b_2+tv_2 \text{ for }t\in \mathbb{R}\right.\right\}\\
&=\left\{\vec{x}(t)\, \left| \,x_2(t)=b_2+\dfrac{x_1(t)-b_1}{v_1} v_2 \text{ for }t\in \mathbb{R}\right. \right\}\\
&=\left\{\vec{x}(t)\, \left| \,x_2=\underbrace{\dfrac{v_2}{v_1}}_{=:m}x_1+\underbrace{\dfrac{b_2v_1-b_1v_2}{v_1}}_{=:b}\text{ for }x_1\in \mathbb{R}\right.\right\}\\
&=\left\{\vec{x}\, \left| \,\vec{x}=(x,y)\wedge y=mx+b\text{ for }x\in \mathbb{R}\right.\right\}
\end{align*}

The set is the same, but the parameterizations of its elements are different, one by all $t\in \mathbb{R}$ and the other by all $x\in \mathbb{R}.$ They both run over all real numbers. The parameterization with $t$ is better because we do not have to separately deal with vertical lines. The equations above require $v_1\neq 0.$ Hence, the case $v_1=0$ requires a different calculation (parameterization by $y\in \mathbb{R}$).

Thanks Fresh. Sorry I should've indicated beginner level instead of intermediate for my question since I am just starting to learn linear algebra. I was not able to follow all the equations you wrote.

 

[PeroK]

I am not sure what (1,m) means.

Those are the components of a vector.

I thought direction of a vector was measured by an angle.

A vector defines a direction.

Can you please give an example where:

$(x, y) = (x_0, y_0) + t(1, m)$

I'm not sure what you mean by an example. That's an equation. You can put in any values for $x_0, y_0$ and $m$. That gives the points on a line $(x, y)$ defined in terms of the parameter $t$.

I am also a bit unclear on what the t means- my understanding is that it's a factor that you can use to scale up or down the size of a vector, but I'm not sure what it means in the context of t(1,m).

In this context, $t$ is a variable that can take any real value ($t \in (-\infty, +\infty)$). This is often called a parameter.

You need to get familiar with these concepts.

 

[PeroK]

For example:

Any point $P$ in the x-y plane can be described by its x and y coordinates $(x, y)$. That can also be seen as a position vector from the origin to the point $P$. This can be written $\vec{OP} = (x, y)$. This "duality" between points and position vectors is important to grasp. Although, it should be fairly natural and is not very abstract.

The line segment from point $P$ to $Q$ can also be seen as a vector: $\vec{PQ} = (x_Q, y_Q) - (x_P, y_P) = (x_Q - x_P, y_Q - y_P)$. This is sometimes called a displacement vector. Note that the position vector is a special case of this.

Finally, there is a unique straight line through any pairs of points $P$ and $Q$. One natural way to describe this line $L$ is:
$$L = \vec{OP} + t\vec{PQ} \ \ (t \in (-\infty, +\infty))$$This seems to me perhaps the most natural way to define a line. Pick any two points on the line, choose one as your starting point and the displacement vector between the points is the direction of your line. The parameter $t$ is used to extend the natural line segment between the two points indefinitely in both directions.

If you are not happy with this, then linear algebra is going to be difficult, I would say.

 

[fresh_42]

Thanks Fresh. Sorry I should've indicated beginner level instead of intermediate for my question since I am just starting to learn linear algebra. I was not able to follow all the equations you wrote.

It's basically this:

(1) $L=\{(x,y)\,|\,y=mx+b\}$ describes the points $(x,y)$ of the plane, that are related by $y=mx+b,$ the equation of a line in that plane, i.e. all points $(x,mx+b).$

(2) $L=\{\vec{x}+t\cdot \vec{v}\,|\,t\in (-\infty ,+\infty )\}$ describes the points when you travel along the line, where $t$ is the time on your watch.

Hence, it is the perspective that changes. (1) is the outside view of the plane with a line in it, (2) is the inside view of or along a line that is located in a plane.

 

[NoahsArk]

@PeroK "I'll leave it to you to check that if we take the set of points $(x,y)$ where $(x,y) = (x_0, y_0) + t(1,m)$ then these points satisfy the equation $y - y_0 = mx (x-x_0)$ "

I think I am starting to understand this, and, if I am, then this means that whereas $y = mx + b$ is a way to take us from one coordinate (x) to another coordinate (y), the equation $\vec s = \vec x + t\vec v$ (where $\vec s$ represents (x, y) and $\vec v$ represents the slope in vector form) is a way to take us from one set of coordinates (i.e. vector) to another set of coordinates. Am I on the right track?

@fresh_42 if I'm not mistaken, going from one set of coordinates to another is similar to what you meant by "describes the points when you travel along the line, where t is the time on your watch."?

So, $\vec s = \vec x + t\vec v$ and $L = \vec x + t\vec v$ are saying something similar- the first is showing how we get $\vec s$ and the second is saying that when we plug in t with all the real numbers we get the line L.

I understand now, I think, why (1,m) is the slope vector in coordinate form. E.g. if m = 4, we'd have (1,4) which means if we go one to the right, we go four up and we have a slope of 4. If the slope vector were (2,3) then it means the slope is 2/3. Is this correct?

 

[fresh_42]

@fresh_42 if I'm not mistaken going from one set of coordinates to another is similar to what you meant by "describes the points when you travel along the line, where t is the time on your watch."?

It is actually a very good question you have posed here considering we are a physics website!

The $(x,y)$ coordinates describe the line in an absolute way. We first have the Euclidean two-dimensional plane, the Cartesian coordinate system, and then - in a second step - we identify certain points $(x,y)=(x,mx+b)$ in this plane as a line. This is the outside view of someone who is aware of the plane.

The parameter representation by the time/watch $t$ is an inside view. We are at position $\vec{x}$ when the watch shows noon $(t=0)$, and we proceed from there in direction $\vec{v}$ if time goes on. We can also see where we came from if we plug in negative time values. This description does not need the plane. We are at $\vec{x}$ and travel in direction (velocity) $\vec{v}.$ Where this all happens is irrelevant. It could be in a plane as in our case with two-dimensional vectors, or in space with three-dimensional vectors, or even in higher dimensions. $\vec{x}+t\cdot \vec{v}\;(t\in \mathbb{R})$ is always the equation of a line. And it doesn't matter in which direction $\vec{v}$ we travel whereas the "outside equation $y=mx+b$" does not cover the cases $x=\text{const.}.$

The notation $\vec{s}(t)=\vec{x}+t\cdot \vec{v}$ is independent of absolute $(x,y)$ coordinates. We only need coordinates if we want to perform calculations.

If you want to do analytical geometry then the Cartesian coordinates and the line $\{(x,y)\,|\,y=mx+b\}$ are fine. If you want to do physics and describe a uniform linear motion then $\vec{s}(t)=\vec{x}+t\cdot \vec{v}$ is fine. Even the choice of the variable names reflect this: $t$ for time, $\vec{v}$ for velocity, and $\vec{s}$ ... well, I don't what $\vec{s}$ stands for. In German, I would say "Standort" (position, location), but I don't know an appropriate English word, maybe "state".

We can differentiate according to time and we don't have to provide an environment for the equation. Starting point, time, and velocity are all that counts. Sure, if we are heading for a meeting, we better use coordinates, e.g. $(x,y,z)$ if the building has multiple floors ($z$). But to express that we have a uniform linear motion, all we need to say is
$$
\dfrac{d}{dt}\vec{s}(t)=(\vec{x}+t\cdot\vec{v})'=\vec{v}
$$
You do theoretical physics with $\vec{s}(t)$ and experimental physics with $(x,y)$ because you are in a lab and want to measure against scales that are also in the lab: a defining outside location.

So, $\vec s = \vec x + t\vec v$ and $L = \vec x + t\vec v$ are saying something similar- the first is showing how we get $\vec s$ and the second is saying that when we plug in t with all the real numbers we get the line L.

Yes.

I understand now, I think, why (1,m) is the slope vector in coordinate form. E.g. if m = 4, we'd have (1,4) which means if we go one to the right, we go four up and we have a slope of 4. If the slope vector were (2,3) then it means the slope is 2/3. Is this correct?

Let me think about it. I tend to confuse the order here ...

We have $y=mx+b.$ Comparing possible different units of $x$ and $y,$ e.g. $x$ is time (hours $[h]$) and $y$ is distance (miles $[mi]$), then $y [mi]= m\cdot x[h]+ b [mi]$ requires
$$
m=\dfrac{[mi]}{[h]}=\dfrac{\Delta y[mi]}{\Delta x [h]}=\dfrac{y_2-y_1}{x_2-x_1}
$$
So $m=4$ is $1$ in positive $x$ direction and $4$ in positive $y$ direction; it is steeper than the other way around which would be $m=0.25.$

$m=\dfrac{2}{3}$ is $3$ in positive $x$ direction and $2$ in positive $y$ direction.

(The program I used does not have the same scale on the x-axis and the y-axis.)

​

I think you confused directions, too, so I hope my considerations will be helpful to avoid this mistake. Think about the units! Units are your friend; or short: $x$ must cancel, so it is in the denominator.

$m=4=\dfrac{4}{1}=\dfrac{\Delta y}{\Delta x}\, : \,$ $1$ to the right, $4$ up

$m=\dfrac{2}{3}=\dfrac{\Delta y}{\Delta x}\, : \,$ $3$ to the right, $2$ up

 

[PeroK]

@PeroK "I'll leave it to you to check that if we take the set of points $(x,y)$ where $(x,y) = (x_0, y_0) + t(1,m)$ then these points satisfy the equation $y - y_0 = mx (x-x_0)$ "

I think I am starting to understand this ...

I was prompting you to do the work and confirm algebraically that every line (*) can be written in each of these three forms:
$$y = mx + c$$$$y - y_0 = m(x-x_0)$$$$(x, y) = (x_0, y_0) + t(1, m) \ \ (t \in (-\infty, +\infty))$$(*) note that, as I think has already been mentioned, vertical lines such as $x = 1$ cannot be written in this form.

 

[NoahsArk]

Thanks @fresh_42 and @PeroK.

@fresh_42 yes I see I had the slope coordinates backward for m =2/3

@PeroK I did confirm with a few examples, plugging in x and y and x0, y0 with different numbers, that the third line form you wrote above, the vector form, works. I was just observing what’s apparent from the form: i.e. that instead of having just a y value on the left side of the equation you have an x AND y value (which I assume together are a vector). I also assume you could’ve written the vector form as (x,y,z) = …… or (x,y,z,e) = … etc?

 

[WWGD]

Another perspective.
You're given two pieces of information that uniquely determine the line.
The x tells you a point the line goes through. The v, a vector, tells you the direction.
Try using $x=0$first. Then your line is tv, or $\{tv : t\in \mathbb R \}$. Choose , say, $v=(1,1)$. Then the line is formed by all the pairs
$(t,t):\{(0,0), (1,2), ...,(t,t), t \in \mathbb R \}$.
Above defines a line through the origin. If it doesn't go through it, it goes through some point $x$, then you add $x$ to it, as a translation of the line.

 

[PeroK]

I did confirm with a few examples, plugging in x and y and x0, y0 with different numbers, that the third line form you wrote above, the vector form, works. I was just observing what’s apparent from the form: i.e. that instead of having just a y value on the left side of the equation you have an x AND y value (which I assume together are a vector). I also assume you could’ve written the vector form as (x,y,z) = …… or (x,y,z,e) = … etc?

I'd stick with 2D and the x-y plane until you are comfortable with these concepts. There's no substitute for doing lots of algebra, so you get comfortable manipulating equations. For example:

Suppose we have a line $y = mx + c$. We can find a point on this line by choosing any value of $x$. E.g. $x = 0$ gives $y = c$ and we know that the point $(0, c)$ is on the line. This is a special point where the line crosses the y-axis. I think that's also called the y-intercept. This is our $(x_0, y_0)$ and we can rewrite the equatio of our line as: $y - c = mx$.

Altenatively, we could find a point on the line where $y = 0$. If that's our $y_0$, then $x_0 = -\frac c m$. And, we can rewrite our equation as: $y = m(x + \frac c m)$.

Going the other way, if we start with: $y - y_0 = m(x - x_0)$, then we can rewrite this as: $y = mx + (y_0 -mx_0)$. And we see that with $c = (y_0 -mx_0)$ we have an equation of the form $y = mx + c$.

If you grasp these ideas, you should be able to work on the vector-based definition of a line.