Parameterizing a Line Using Vectors

It's pretty easy to find the equation of a line in two dimensions. We just whip out the slope-intercept formula, point-slope formula, or any of the other options available to us. But what if somebody asked you to find the equation of a line in three dimensions? How about four dimensions?

If you're anything like me, you'd be a little thrown off! But have no fear, we can describe any line in any dimension using our favorite mathematical construct: vectors!

Let's take this point in 3D:

For the purposes of making this a little easier to understand, let's say this point is (1, 2, 3).

Now how do we usually find lines in two dimensions? Well, we normally have two points and we want to find the equation of a line that passes through both of them. So let's say (1, 2, 3) is our first point and (3, 1, 2) is our second point.

What do we do here? We can't really use a slope-intercept formula, can we? However, let's think about the vector that goes from the first point to the second. Using our knowledge of vectors, we would know what that is: <(3-1), (1-2), (2-3)> = <2, -1, -1>.

You might be thinking, "But this still doesn't help us write an equation for a line..." Hang on a second! Let's do a quick mental exercise first.

Bob the Bug is currently standing on point (1, 2, 3). In one second, he can travel to point (3, 1, 2). If Bob is traveling in a straight line, where will be the next point he goes? Well, following the pattern from the previous jump, he would go to point (5, 0, 1). In other words, he is using a vector to travel in a straight line from one point to another. If he keeps traveling using that same vector to determine his next location, he'll be going in a straight line!

Now let's make some equations for Bob's jumping.

In one second, Bob went from (1, 2, 3) to (3, 1, 2). Let's use a variable t to describe Bob's time. For the first second, t = 1.

$x_1 = x_0 + 2*t$

$y_1 = y_0 + -1*t$

$z_1 = z_0 + -1*t$

If you plug in t = 1, that gets you the point you want! Now try plugging in t = 2. What point do you get? It should be (5, 0, 1)!

Now let's use a more general definition to describe these vector-lines:

For two points $(x_0, y_0, z_0$ and $(x_1, y_1, z_1)$ and the vector <a, b, c> that travels between them, we can describe any point on the same line as:

$x_1 = x_0 + at$

$y_1 = y_0 + bt$

$z_1 = z_0 + ct$

where t is any real number.

(Please note: You can do this in any dimension you like! Can you see why?)

A line may be defined by any two distinct points on the line, but it is often more convenient to describe a line by only one point on the line and the line's direction. With the new tool of vectors, we may describe lines in any number of dimensions more easily and effectively.

Suppose that the point $P=(p_1,p_2,...,p_n)$ in n-dimensional space lies on the line that we are trying to describe, and let us say that the direction vector of the line is the vector $\vec{a}=<a_1,a_2,...,a_n>$.

Our goal is to describe all points $Q$ on the line in terms of $P$ and $\vec{a}$.

We know that the vector $\vec{PQ}$ is parallel to $\vec{a}$ if and only if $\vec{PQ}=k\vec{a}$ for some real number k.

Thus, $Q$ lies on the line if and only if $\vec{Q}=\vec{P}+k\vec{a}$, where $\vec{P}$ and $\vec{Q}$ are the position vectors of $P$ and $Q$, respectively.

Thus, all points on the line can be described as follows: $\vec{Q}=<p_1,p_2,...,p_n>+k<a_1,a_2,...,a_n>\\=<ka_1+p_1,ka_2+p_2,...,ka_n+p_n>$.

We have successfully parameterized the line! Here, $k$ is called the parameter, as all of the coordinates of $Q$ are described as functions of $k$.

Be warned that a given line (or more generally, any curve at all) has infinitely many parameterizations. This may be seen by multiplying the direction vector $\vec{a}$ by some new nonzero scalar to produce a vector parallel to $\vec{a}$; this new vector is equally valid as a direction vector for the line.

A Different Definition of a Straight Line

What is a straight line? Simple question. What is a straight line in terms of vectors?

Consider a line $AP$, parallel to vector $\vec{b}$. Take another vector $\vec{a}$, which starts from the origin,$O$. Vector $\vec{r}$ is such that it starts from the origin and ends at point $P$. The diagram is shown in below (Figure 1).

In this diagram, line ${AP}$ can now be written as:

$\vec{r}$ - $\vec{a}$

Also, $AP$ is parallel to $\vec{b}$. Thus,

$AP$ = $t$$\vec{b}$ (As $AP$ is a linear multiple of $\vec{b}$ )

Combining both the equations, we get:

$\vec{r}$ - $\vec{a}$ = $t$$\vec{b}$

$\therefore \vec{r} = \vec{a} + t\vec{b}$ for some multiple or (more correctly) parameter $t$.

The equation above is the Vector equation of a straight line.