Lines in 3D
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.5: “Equations of Lines and Planes in Space” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-5-equations-of-lines-and-planes-in-space |
| Textbook used in class | Stewart, Calculus, Section 12.5: “Equations of Lines and Planes” (Examples 1, 2, 3) |
Opening Scenario
A line in 2D needs one equation. A line in 3D cannot be described by a single equation (a single equation gives a plane). Instead, a line in 3D is described parametrically: a starting point plus a direction vector scaled by a parameter $t$.
Quick Reference
Vector equation of a line through $P_0 = (x_0, y_0, z_0)$ with direction $\mathbf{d} = \langle a, b, c\rangle$: \[ \mathbf{r} = \mathbf{r}_0 + t\,\mathbf{d}, \quad t \in \mathbb{R}. \]
Parametric equations: \[ x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct. \]
Symmetric equations (when $a, b, c \neq 0$): \[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}. \]
Key Concepts
1. Writing the Equations of a Line
Example 1. Find parametric and symmetric equations of the line through $A = (2, 4, -3)$ and $B = (3, -1, 1)$. (Stewart 12.5, Example 1.)
Direction vector: $\overrightarrow{AB} = \langle 1, -5, 4\rangle$.
Parametric: $x = 2 + t$, $y = 4 - 5t$, $z = -3 + 4t$.
Symmetric: $\dfrac{x-2}{1} = \dfrac{y-4}{-5} = \dfrac{z+3}{4}$.
2. Line Through a Point Parallel to Another Line
Two lines are parallel if their direction vectors are scalar multiples of each other.
Example 2. Write equations for the line through $(1, 1, 0)$ parallel to $\langle 2, -1, 3\rangle$.
$x = 1 + 2t$, $y = 1 - t$, $z = 3t$.
3. Do Two Lines Intersect?
Set the parametric equations equal (using different parameters $s$ and $t$) and solve. If a consistent solution exists, the lines intersect; if not, they are parallel or skew.
Example 3. Do the lines $\mathbf{r}_1 = \langle 1, 1, 0\rangle + t\langle 1, -1, 2\rangle$ and $\mathbf{r}_2 = \langle 2, 0, -1\rangle + s\langle -1, 1, 0\rangle$ intersect?
Set equal: $1 + t = 2 - s$, $1 - t = s$, $2t = -1$.
From the third: $t = -1/2$, then $s = 1 - (-1/2) = 3/2$. Check first: $1 + (-1/2) = 1/2$ and $2 - 3/2 = 1/2$. They agree, so the lines intersect at $t = -1/2$: point $\langle 1/2, 3/2, -1\rangle$.
using a single equation to describe a line in 3D. In 3D, one equation like $ax + by + cz = d$ describes a plane, not a line. A line in 3D requires either parametric equations (one for each coordinate) or two simultaneous plane equations. This contrasts with 2D, where one equation is enough for a line.
Common Errors Summary
| Error | Correction |
|---|---|
| Setting $t = 0$ as the only point on the line | $t$ ranges over all reals; the line is infinite; $t = 0$ gives only the point $P_0$ |
| Using the same parameter name for two different lines when testing intersection | Use different letters ($t$ and $s$); the same parameter would force you to stay on one line |
Common Misconceptions
one linear equation in three variables describes a line in 3D.
This is the iconic-graph error carried over from two dimensions. In two dimensions, one equation such as $ax + by = c$ describes a line. In three dimensions, one equation $ax + by + cz = d$ describes a plane, because all points in the plane satisfy the single constraint and the third coordinate is free. A line in 3D requires parametric equations (one per coordinate variable) or the intersection of two planes. Specifying only one equation leaves too much freedom.
Leveled Practice
Problem 1. Find parametric equations for the line through $(0, 1, 2)$ with direction $\langle 3, 0, -1\rangle$.
Show answer
$x = 3t$, $y = 1$, $z = 2 - t$.
Problem 2. Find the symmetric equations of the line through $(1, -2, 3)$ and $(4, 0, -1)$.
Show answer
Direction: $\langle 3, 2, -4\rangle$. Symmetric: $\dfrac{x-1}{3} = \dfrac{y+2}{2} = \dfrac{z-3}{-4}$.