What is the equation for a 3D line?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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.

Answer 5

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.

Answer 6

Ovi, I too like your answer about having initially set out looking for something of the form $z=f(x,y)$. I think maybe I have something that tends in a similar direction, and that hopefully feels like a more generalizable extension from $\mathbb{R}^2$ to $\mathbb{R}^3$ (and beyond).

TLDR: In $\mathbb{R}^2$, points on a line have coordinates $(x, f(x))$. In $\mathbb{R}^3$, points on a line have coordinates $(x, f(x), g(x))$.

The above wasn't obvious to me at all though, so for those who care to walk through it as I did, my thinking is as follows:

Let's put aside $y=mx+b$ for a minute, and think instead of a line in $\mathbb{R}^2$ as a collection of points whose ($x$, $y$) coordinates are of the form $(x,f(x))$. This just amounts to saying that the points' $y$ values are a function of $x$. Simply, $y = f(x)$.

That might seem tautologically obvious, but let's consider that if $x$ and $y$ weren't in this constrained relationship to one another where $y$ is determined by $f(x)$ -- in other words, if they were free to vary independently of one another, and every $x$ value could be combined with any $y$ value in a coordinate $(x,y)$ -- then you wouldn't have an object describing a line, you would have an object that describes the whole $xy$ plane itself!

Moving into $\mathbb{R}^3$, it's true, exactly as you said, that if we construct an object whose $z$ value is determined by a function $f(x,y)$ we'll have two values, $x$ and $y$, that are allowed to vary independently of one another, and will thus end up with a plane. (The only difference between this plane and the plane described in the paragraph above is that in $\mathbb{R}^3$ we have the added $z$-dimension in which the plane can be tilted -- provided $f$ and $g$ are both linear functions -- whereas in $\mathbb{R}^2$ the only possible plane is the $xy$ plane by definition, but this is just an aside).

What if, instead of a function $f(x,y)$ which has two variables, all the points in our new 3-dimensional line object's coordinates are functions of one variable, say, $x$? In other words, if we have an object in $3$-space whose coordinates are of the form: $$(x,y,z) = (x, f(x), g(x))$$

Then every $x$ value can only have one corresponding $y$ value and one corresponding $z$ value (if $f$ and $g$ are functions, they must have a unique output for any input value of $x$).

And if there is only one $y$ and $z$ value for every $x$ value, you know this is a line. As a counterexample, consider a linear equation in 3 independent variables: $ax + by + cz = d$. This describes a plane in $3$-space. For every $x$ value, $y$ and $z$ can still vary in a dependent relationship with each other, i.e. at each value of $x$ you have a slice through a plane instead of a point, the latter being what you'd need for the object described to a line.

You can arrive at this by looking at the intersection of two planes, as shown in Mark's answer above. If you set two planes' equations equal to one another, and try to solve for the intersection, you end up with a solution (in linear algebra, a general solution or parametric solution) where two out of the three unknowns are given as functions of the third.