# What is the equation for a 3D line?

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

• $$(a_0 + a_1 t, b_0 + b_1 t,c_0 + c_1 t)$$ where $a_1, b_1, c_1$ are not all $0.$ – Will Jagy May 28 '13 at 3:39
• Spherical Coordinates are generally simpler. – User3910 May 3 '17 at 14:25

## 4 Answers

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

• One representation uses 8 numbers and the other uses 6. Is there a representation that uses fewer than 6? – Samuel Danielson Jul 9 '16 at 3:15
• @SamuelDanielson Spherical Coordinates: theta, phi, x0, y0, z0. – User3910 May 3 '17 at 14:37
• Could you say what program you used to draw this graph? – Turkhan Badalov Nov 29 '17 at 18:18
• @TurkhanBadalov I used Mathematica. – Mark McClure Nov 29 '17 at 19:47

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.

• In my opinion, the symmetric form is the most useless one. – Hawk May 28 '13 at 4:05
• @Hawk, Can you please explain your point? – HeWhoMustBeNamed Nov 5 '17 at 5:24
• 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) – Danilo Dec 2 '18 at 20:21
• @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 – Nathan Mar 11 '19 at 13:45
• @Jared, I have misunderstanding with normal form of a line. Is symmetric form == normal – Dan Klymenko May 13 '20 at 10:20

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.

• 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. – ANuo Aug 21 '20 at 6:56
• @ANuo It's inconvenient, but no impossible. Please see the updated answer. – Ovi Feb 15 at 15:29

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 $$L_{AB}: x=(-2,0,1)+t(6,5,2)$$

• Please advise my friend if you have the time.Thank you. math.stackexchange.com/questions/404862/… – Software May 28 '13 at 16:58
• @BabakS.: Nice answer + 1, and congratulations on doing 1000 edit reviews - I know how hard those are to do my friend! – Amzoti May 28 '13 at 19:45
• @Amzoti: Thanks so much. Yes indeed it was. Huuuh :-) – Mikasa May 28 '13 at 19:46
• Hello, dear friend! I hope your students did well on the exam you gave them! – amWhy May 29 '13 at 0:28
• @amWhy: I am red-penciling their jobs. They were not so bad. – Mikasa May 29 '13 at 4:26