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Just like we have the equation $y=mx+b$ for $\mathbb{R}^{2}$, what would be a equation for $\mathbb{R}^{3}$? Thanks.

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6 Answers 6

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Here are three ways to describe the formula of a line in $3$ dimensions. Let's assume the line $L$ passes through the point $(x_0,y_0,z_0)$ and is traveling in the direction $(a,b,c)$.

Vector Form

$$(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)$$

Here $t$ is a parameter describing a particular point on the line $L$.

Parametric Form

$$x=x_0+ta\\y=y_0+tb\\z=z_0+tc$$

These are basically the equations that result from the three components of vector form.

Symmetric Form

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$

Here we assume $a,b,$ and $c$ are all nonzero. All we've done is solve the parametric equations for $t$ and set them all equal.

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    $\begingroup$ In my opinion, the symmetric form is the most useless one. $\endgroup$
    – Lemon
    May 28, 2013 at 4:05
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    $\begingroup$ @Hawk, Can you please explain your point? $\endgroup$ Nov 5, 2017 at 5:24
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    $\begingroup$ t is an index of step from xo,yo,zo to xn,yn,zn ... as this is an typical stepping function. If you calculate t you will find at which fraction of the line (a,b,c) -> (x0,y0,z0) is point with coordinates (x,y,z) $\endgroup$
    – Danilo
    Dec 2, 2018 at 20:21
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    $\begingroup$ @MrReality I'm programming a line intersection with a z=z_0 plane. For my case, Hawk is wrong. Symmetric form immediately gives me the x and y values I wanted $\endgroup$
    – Nathan
    Mar 11, 2019 at 13:45
  • $\begingroup$ @Jared, I have misunderstanding with normal form of a line. Is symmetric form == normal $\endgroup$ May 13, 2020 at 10:20
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You can describe a line in space as the intersection of two planes. Thus, $$\{(x,y,z)\in{\mathbb R}^3: a_1x+b_1y+c_1z=d_1 \text{ and } a_2x+b_2y+c_2z=d_2\}.$$ Alternatively, you can use vector notation to describe it as $$\vec{p}(t) = \vec{p}_0 + \vec{d}t.$$

I used this relationship to generate this picture:

enter image description here

This is largely a topic that you will learn about in a third semester calculus course, at least in the states.

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    $\begingroup$ One representation uses 8 numbers and the other uses 6. Is there a representation that uses fewer than 6? $\endgroup$ Jul 9, 2016 at 3:15
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    $\begingroup$ @SamuelDanielson Spherical Coordinates: theta, phi, x0, y0, z0. $\endgroup$
    – User3910
    May 3, 2017 at 14:37
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    $\begingroup$ Could you say what program you used to draw this graph? $\endgroup$ Nov 29, 2017 at 18:18
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    $\begingroup$ @TurkhanBadalov I used Mathematica. $\endgroup$ Nov 29, 2017 at 19:47
  • $\begingroup$ @SamuelDanielson A representations with 4 coordinates is the minimum, e.g. based on the nearest point to the origin and an angle about that axis in which it points, or based on a line direction in theta/phi sphericals and a cylindrical offset from the origin. $\endgroup$ Mar 24 at 23:49
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When I originally asked this question, I was not expecting these seemingly indirect ways of describing a line, such as an intersection of two planes, or vector equations. Just like how $y=mx+b$ is the equation of a line in $2$D, I was expecting some sort of equation $z = f(x, y)$, where $f$ is some nice elementary function. I am writing this answer for anyone who has this same idea that I did. I want to quickly explain why the equation of a line cannot be $z = f(x, y)$, where $f$ is a nice function.

The problem is that if $f$ is a nice function, then it is probably defined for all pairs $(x, y)$, or almost all. That means that if you try to graph it, there will be a point of the graph of $f$ above almost every point of the floor, so the graph of $f$ cannot be a line.

Another way to say it is this: imagine the graph of the line. If we want an equation $f(x, y)$ for the line, the domain of $f$ can only be the shadow of the line on the $xy$ plane. But any nice function $f$ will have as a domain either all pairs $(x, y)$, or almost all of them.

With all of that being said, it is possible to cook up a function $f(x, y)$ whose graph is a line. We know that if we could take a plane, for example $g(x, y) = x+y$, and somehow restrict its domain to a line on the $xy$ plane, that would give us a line in $xyz$ space. Here is one way to do it: $$f(x, y) = x+y+\sqrt{-(y-x)^2}$$

The expression $-(y-x)^2$ is $\le 0$ for any $x, y$ and it equals zero precisely when $y = x$. Therefore $\sqrt{-(y-x)^2}$ will only be defined precisely when $y = x$, and when $y$ does equal $x$, $f(x, y) = x+y$. Thus the graph of $f$ is a line in 3D space.

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  • $\begingroup$ For a line, the number of coordinates that can be freely determined is one. However, in this function, f(x,y), there are two. So it is impossible to write it in this form. $\endgroup$
    – ANuo
    Aug 21, 2020 at 6:56
  • $\begingroup$ @ANuo It's inconvenient, but no impossible. Please see the updated answer. $\endgroup$
    – Ovi
    Feb 15, 2021 at 15:29
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I am giving you an example. Let $A(-2,0,1),~~B(4,5,3)$ be two points in $\mathbb R^3$. And let $C$ be the end point for the vector which is drawn from the origin. In addition, we assume that this vector has the same direction as the vector $AB$. So we have its coordinates is $(4,5,3)-(-2,0,1)=(6,5,2)$. Therefore the equation of the line passing through $A$ and $B$ is $$L_{AB}: x=(-2,0,1)+t(6,5,2)$$

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  • $\begingroup$ Please advise my friend if you have the time.Thank you. math.stackexchange.com/questions/404862/… $\endgroup$
    – Software
    May 28, 2013 at 16:58
  • $\begingroup$ @BabakS.: Nice answer + 1, and congratulations on doing 1000 edit reviews - I know how hard those are to do my friend! $\endgroup$
    – Amzoti
    May 28, 2013 at 19:45
  • $\begingroup$ @Amzoti: Thanks so much. Yes indeed it was. Huuuh :-) $\endgroup$
    – Mikasa
    May 28, 2013 at 19:46
  • $\begingroup$ Hello, dear friend! I hope your students did well on the exam you gave them! $\endgroup$
    – amWhy
    May 29, 2013 at 0:28
  • $\begingroup$ @amWhy: I am red-penciling their jobs. They were not so bad. $\endgroup$
    – Mikasa
    May 29, 2013 at 4:26
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I love your answer for a line equation in the form of z = f(x, y)... Unfortunately calculating square roots can be impractical from the calculational standpoint and hence I really doubt any plotting software will be able to graph it correctly.

I was thinking of a little bit different approach in order to achieve this. Based on a symmetric equation of a line. However it requires some support for if ... else logic.

$$z=if\:\left(\frac{x}{2}+1=\frac{y}{3}+2\right)\:then\left(\frac{x}{5}+3\right)\:else\left(undefined\right)$$

Of course the advantage here is that it's easy to read and you can potentially define any line in $R^3$ by changing the constant parameters. In practice though, even when conditional statements are supported, for example in GeoGebra - it still fails to actually draw the line.

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Besides the parametric form, another equation of a line in 3D to get it in the form $f(x,y,z)=0$ could be written as: $\frac{\mathbf{r}-\mathbf{r_0}}{|\mathbf{r}-\mathbf{r_0}|} \cdot \mathbf{n} = 1$

Here $\mathbf{r}=(x,y,z)$ is a vector representing any general point on the line. $\mathbf{r_0}=(x_0,y_0,z_0)$ is a given point that lies on the line. $\mathbf{n}=(n_x,n_y,n_z)$ is a given unit vector (that has a magnitude of unity) that is parallel to the line. If two separate points $\mathbf{r_1}$ and $\mathbf{r_2}$ are given through which the line passes, then we could write $\mathbf{n} = \frac{\mathbf{r_2}-\mathbf{r_1}}{|\mathbf{r_2}-\mathbf{r_1}|}$.

Written differently, the equation would read: $(x-x_0)n_x + (y-y_0)n_y + (z-z_0)n_z - \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2} = 0$

This is a quadratic equation in, say, $z$, in that a closed form solution could be written for $z$ in terms of $x, y$ and other given parameters. However, it is not easy to see that this equation should lead to a real solution for $z$ for only a specific subset of $x, y$ pairs on the $x-y$ plane.

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