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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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    $\begingroup$ $$ (a_0 + a_1 t, b_0 + b_1 t,c_0 + c_1 t) $$ where $a_1, b_1, c_1$ are not all $0.$ $\endgroup$ – Will Jagy May 28 '13 at 3:39
  • $\begingroup$ Spherical Coordinates are generally simpler. $\endgroup$ – User3910 May 3 '17 at 14:25
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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? $\endgroup$ – Samuel Danielson Jul 9 '16 at 3:15
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    $\begingroup$ @SamuelDanielson Spherical Coordinates: theta, phi, x0, y0, z0. $\endgroup$ – User3910 May 3 '17 at 14:37
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    $\begingroup$ Could you say what program you used to draw this graph? $\endgroup$ – Turkhan Badalov Nov 29 '17 at 18:18
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    $\begingroup$ @TurkhanBadalov I used Mathematica. $\endgroup$ – Mark McClure Nov 29 '17 at 19:47
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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. $\endgroup$ – Hawk May 28 '13 at 4:05
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    $\begingroup$ @Hawk, Can you please explain your point? $\endgroup$ – HeWhoMustBeNamed Nov 5 '17 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 '18 at 20:21
  • $\begingroup$ @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 '19 at 13:45
  • $\begingroup$ @Jared, I have misunderstanding with normal form of a line. Is symmetric form == normal $\endgroup$ – Dan Klymenko May 13 '20 at 10:20
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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 '20 at 6:56
  • $\begingroup$ @ANuo It's inconvenient, but no impossible. Please see the updated answer. $\endgroup$ – Ovi Feb 15 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 '13 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 '13 at 19:45
  • $\begingroup$ @Amzoti: Thanks so much. Yes indeed it was. Huuuh :-) $\endgroup$ – Mikasa May 28 '13 at 19:46
  • $\begingroup$ Hello, dear friend! I hope your students did well on the exam you gave them! $\endgroup$ – amWhy May 29 '13 at 0:28
  • $\begingroup$ @amWhy: I am red-penciling their jobs. They were not so bad. $\endgroup$ – Mikasa May 29 '13 at 4:26

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