Equations of Lines

Main idea

In space a line has no slope, so you describe it the way you'd give directions: start here, and go this way. That's $\vec r(t) = \vec r_0 + t\,\vec v$ , a point plus every multiple of a direction vector.

\vec r({\htmlData{sym=2}{t}}) = {\htmlData{sym=0}{\vec r_0}} + {\htmlData{sym=2}{t}}\,{\htmlData{sym=1}{\vec v}} \quad\iff\quad x = x_0 + a{\htmlData{sym=2}{t}},\;\; y = y_0 + b{\htmlData{sym=2}{t}},\;\; z = z_0 + c{\htmlData{sym=2}{t}}

What the symbols mean

$\vec r_0$
A point on the line
Where you are at t = 0.
$\vec v$
Direction vector
⟨a, b, c⟩: which way the line runs.
$t$
Parameter
Any real number; each value gives one point.

Symmetric form

\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}

Explanation

A line in space is simply what you get when you start at one point and keep walking in one fixed direction, forward and backward forever. It's how you describe a laser beam, a flight path or the edge where two walls meet. In $\vec r(t) = \vec r_0 + t\,\vec v$ , the piece $\vec r_0$ is where you start, $\vec v$ is which way you walk (and how big a step is), and $t$ is how many steps you've taken. Plug in any $t$ and you get a point on the line. It looks more abstract than $y = mx + b$ , but it's actually simpler.

In the plane you'd write a line as $y = mx + b$ . That doesn't carry over to 3D: there's no single "slope" when the line can tilt in two different ways at once, and we just saw that one equation in $x$ , $y$ , $z$ gives a surface anyway. So we go back to what a line really is: a point and a direction.

Start at $\vec r_0$ and add $t$ copies of the direction $\vec v$ . At $t = 0$ you're at the starting point, at $t = 1$ you've gone one $\vec v$ along, and negative $t$ goes backward. Letting $t$ run over all real numbers sweeps out the whole line. Writing out each component gives the parametric equations, and solving each one for $t$ and setting them equal gives the symmetric form.

Through two points $P$ and $Q$ : use $\vec v = \overrightarrow{PQ}$ , the "end minus start" vector from the first unit.

Two lines in space can be

parallel, when their direction vectors are multiples of each other,
intersecting, when some $t$ and $s$ give the same point, or
skew: not parallel and never meeting. This can't happen in a plane, where non-parallel lines always cross. In space, one line can pass over the other, like a highway overpass.

To tell intersecting from skew, give the two lines different parameters, $t$ and $s$ , set the components equal, and solve. If the equations are inconsistent, the lines never meet.

See it

A drone at (2, 1, 6) fires a laser along ⟨1, 2, −3⟩, so the beam is r(t) = ⟨2 + t, 1 + 2t, 6 − 3t⟩. As t runs, the dot slides down the beam one point at a time. It reaches the ground when the height 6 − 3t hits 0, at t = 2, landing at (4, 5, 0).

Worked examples

Example 1

Find the line through $(2, 0, -1)$ and $(-1, 3, 4)$ , and where it crosses the $xy$ -plane.

1
The direction is end minus start, $\vec v = \langle -3, 3, 5\rangle$ . Starting from the first point, the line is $x = 2 - 3t$ , $y = 3t$ , $z = -1 + 5t$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find a direction vector for the line through $(1, 2, 3)$ and $(4, 6, 8)$ .

Problem 2+5 XP

Find the point at $t = 2$ on $\vec r(t) = \langle 1, -1, 0\rangle + t\langle 2, 3, 1\rangle$ .

Problem 3+5 XP

Where does $x = 1 + t$ , $y = 2 - t$ , $z = 3t$ meet the plane $z = 6$ ?

Problem 4+5 XP

Are $\vec r_1 = \langle 1, 0, 0\rangle + t\langle 1, 1, 1\rangle$ and $\vec r_2 = \langle 0, 1, 0\rangle + s\langle 2, 2, 2\rangle$ parallel?

Problem 5+5 XP

How are $\langle t, t, 0\rangle$ and $\langle 1, s, s\rangle$ related?

Problem 6+5 XP

In a pinhole camera model, the camera sits at the origin and the image plane is $z = 1$ . A point in the scene at $P(4, 2, 10)$ shows up where the line from the origin to $P$ crosses that plane. Find the image point.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Two drone flight paths are $\vec r_1(t) = \langle 1, 0, 0\rangle + t\langle 1, 1, 0\rangle$ and $\vec r_2(s) = \langle 0, 0, 2\rangle + s\langle 0, 1, 1\rangle$ . They never meet (the lines are skew). How close do they come?

Where it's used

Ray tracing, the technique behind film-quality graphics, fires a line $\vec r_0 + t\vec v$ from the camera through each pixel and solves for where it first hits an object.

How a Camera SeesLin Alg · Computer Science

The 3-D Coordinate System Equations of Planes