Equation of a line in 3D.

Question

I'm about halfway through a highschool class in vectors and calculus, and we are about to finish up the vectors portion. I've asked pretty much every time it's brought up, but why is there no Cartesian equation for a line in 3D? My teacher never answered it or I don't remember the answer. What is so fundamentally different about 3D than 2D or is it super subtle?

Answer 1

It’s not possible to represent a line in $\mathbb R^n$ with a single implcit Cartesian linear equation when $n\ge3$. A consequence of the rank-nullity theorem from linear algebra is that the solution set to the equation $a_1x_1+\cdots+a_nx_n=b$ is an $(n-1)$-dimensional subset of the space—a “hyperplane”—but a line is one-dimensional.

That same theorem tells us how many independent linear equations are required to specify a line, namely $n-1$. This makes geometric sense. If two distinct $(n-1)$-dimensional hyperplanes intersect, their intersection will be $(n-2)$ dimensional. In 3D, this says that two planes intersect in a line (or not at all). If you then throw in another hyperplane you can reduce the dimension of the common intersection to $n-3$, although it could also be empty or the same subset that you had with the first two hyperplanes, and so on. Relating this to 3D again, the common intersection of three planes is either a point, a line, or empty.

In sum, if you’re going to stick to linear equations, you need two of them to specify a line in 3D.

Now, you might have seen what’s called “the symmetric equation” of a line in 3D: $${x-x_0\over\alpha} = {y-y_0\over\beta} = {z-z_0\over\gamma}.\tag1$$ Here, $(x_0,y_0,z_0)$ is some known point on the line and $(\alpha,\beta,\gamma)$ is a vector parallel to it—a direction vector. If in addition $\alpha^2+\beta^2+\gamma^2=1$, then these numbers are the line’s direction cosines. Notice, however, that there are two equal signs in expression (1), so it’s not really a single equation. It’s a compact way of writing the system of equations $${x-x_0\over\alpha} = {y-y_0\over\beta} \\ {y-y_0\over\beta} = {z-z_0\over\gamma}$$ so we’re right back to using two linear equations to describe a line in 3D.

It is possible to describe a line in 3D with a single Cartesian equation of some other kind (e.g., the solution set to $y^2+z^2=0$ is the $x$-axis) but that’s going to obscure the nature of the object being described and is generally harder to work with. On the other hand, there are contexts in which you might want to think of a line as a cylinder with zero radius, which is how I came up with my example.

There are also various parametric equations that can describe a line, but now we’re leaving the realm of implicit Cartesian equations. Using the same known point and direction vector as in (1), we can write the parameterization $(x_0,y_0,z_0)+t(\alpha,\beta,\gamma)$, $t\in\mathbb R$. This is an abbreviated version of the set-builder expression $$\{(x,y,z) \mid (x,y,z) = (x_0,y_0,z_0)+t(\alpha,\beta,\gamma) \land t\in\mathbb R\}.$$ Since a line can always be described by a known point $\mathbf p_0$ and a direction vector $\mathbf v$, the parameterization $\mathbf p=\mathbf p_0+t\mathbf d$ works in 2D and in higher-dimensional spaces as well. This parameterization is often called the “vector equation” of a line.

A line can also be defined by two points $\mathbf p_1$ and $\mathbf p_2$. A direction vector for the line is $\mathbf p_2-\mathbf p_1$, so we get the parameterization $\mathbf p = \mathbf p_1+t(\mathbf p_2-\mathbf p_1) = (1-t)\mathbf p_1+t\mathbf p_2$. Not coincidentally, this looks like a linear interpolation formula.

There’s one other option specific to $\mathbb R^3$ that’s not as widely known as the ones above. If we take the cross product of both sides of the vector equation of a line with $\mathbf d$, since $\mathbf d\times\mathbf d=0$, we have the equation $$\mathbf p\times\mathbf d = \mathbf p_0\times\mathbf d.$$ This is a single implicit vector equation, but if you expand it by coordinates, you get a system of three linear equations, one of which is redundant, so this is really another compact way to write the system of linear equations that we need to describe a line.

Answer 2

There is.

Suppose the vector equation of a line passing through $A(\vec a)$ and parallel to $\vec b$ is $(\vec r-\vec a)=\lambda\vec b$

Now let $\vec r=x\hat i+y\hat j+z\hat k$,$\space\space\vec a=x_1\hat i+y_1\hat j+z_1\hat k$ and $\vec b=a\hat i+b\hat j+c\hat k$ $$(x-x_1)\hat i+(y-y_1)\hat j+(z-z_1)\hat k=\lambda a\hat i+\lambda b\hat j+\lambda c\hat k$$ $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}=\lambda$$ This is the equation of the line in cartesian form with $a,b$ and $c$ being the direction ratios for the line.

Suppose the vector equation for a line passing through points $A(\vec a)$ and $B(\vec b)$ is $(\vec r-\vec a)=\lambda (\vec b-\vec a)$

Now let $\vec r=x\hat i+y\hat j+z\hat k$,$\space\space\vec a=x_1\hat i+y_1\hat j+z_1\hat k$,$\space\space \vec b=x_2\hat i+y_2\hat j+z_2\hat k\space\space$ $$(x-x_1)\hat i+(y-y_1)\hat j+(z-z_1)\hat k=\lambda (x_2-x_1)\hat i+\lambda (y_2-y_1)\hat j+\lambda (z_2-z_1)\hat k$$ $$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}=\lambda$$ This is the equation of the line in cartesian form with $(x_2-x_1),(y_2-y_1)$ and $(z_2-z_1)$ being the direction ratios for the line.

In general, whenever you want to represent any vector equation of a line or plane in cartesian form, you write the vector in the form of combination of the basic unit vectors $\hat i,\hat j\space$ and $\space\hat k$ and then continue with the vector equation as shown in the above methods.