Equations of Lines in 3D

Core Definition

In three-dimensional space ( $\mathbb{R}^3$ ), a line cannot be defined by a single slope value ( $y = mx + b$ ). Instead, a line is uniquely determined by a known initial point $P_0(x_0, y_0, z_0)$ with position vector $\vec{r}_0 = \langle x_0, y_0, z_0 \rangle$ and a non-zero direction vector $\vec{v} = \langle a, b, c \rangle$ :

Vector Equation (Vector-Valued Function): $\vec{r}(t) = \vec{r}_0 + t\vec{v} = \langle x_0 + at, y_0 + bt, z_0 + ct \rangle, \quad t \in \mathbb{R}$
Parametric Equations: $x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct$
Symmetric Equations (assuming $a, b, c \neq 0$ ): $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

Concept Explanation

Geometric & Vector Interpretation

A line in space is generated by tracing the terminal point of a position vector $\vec{r}(t)$ as the real scalar parameter $t$ varies from $-\infty$ to $+\infty$ .

When $t = 0$ , $\vec{r}(0) = \vec{r}_0$ , placing us directly at the initial reference point $P_0$ .
When $t = 1$ , $\vec{r}(1) = \vec{r}_0 + \vec{v}$ , advancing one full direction vector along the trajectory.
Changing the scalar parameter $t$ moves a point forward or backward along the directed path.

            ^ z
            |           . Q (t = 1)
            |          /
            |   v     /  r(t) = r_0 + t*v
            |  -->   /
            |       . P_0 (t = 0)
            |      /
            |     / r_0
            |    /
            +------------------> y
           /
          /
         v x

From Parametric to Symmetric Equations

Eliminating the parameter $t$ transforms time-dependent motion into a purely static geometric path in $\mathbb{R}^3$ :

Solve each coordinate equation for $t$ : $t = \frac{x - x_0}{a}, \quad t = \frac{y - y_0}{b}, \quad t = \frac{z - z_0}{c}$
Equate the expressions for $t$ to form two simultaneous linear equations defining the intersection of two planes: $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

Edge Case: Zero Components in Direction Vector (

a, b, \text{ or } c = 0

)

If one of the direction vector components is zero (e.g., $b = 0$ ), division by zero is undefined. We isolate the constant coordinate and equate the remaining terms:

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

Geometrically, this restricts the line to lie entirely within the vertical plane

y = y_0

Bounded Line Segments

To define a finite line segment between two endpoints $P(x_0, y_0, z_0)$ and $Q(x_1, y_1, z_1)$ , we set the direction vector to $\vec{v} = \vec{PQ} = \langle x_1 - x_0, y_1 - y_0, z_1 - z_0 \rangle$ and restrict the parameter domain:

\vec{r}(t) = \vec{r}_0 + t\vec{v}, \quad 0 \le t \le 1

Engineering Applications

Aerospace Trajectory Mapping: Describing spacecraft launch corridors, satellite orbit insertions, and commercial flight waypoints as time-dependent parametric curves $\vec{r}(t)$ .
Robotics & CNC Machining: Generating linear G-code toolpaths where robotic cutting tools or laser heads translate between Cartesian coordinates.
Structural Engineering: Defining geometric centerlines for steel cables, truss members, and utility conduits in 3D Building Information Modeling (BIM).

Example Problems

Example 1: Three Standard Forms from Two Points

Find the vector, parametric, and symmetric equations of the line passing through $P(2, -1, 4)$ and $Q(5, 3, -2)$ .

Solution

Compute Direction Vector: $\vec{v} = \vec{PQ} = \langle 5 - 2, 3 - (-1), -2 - 4 \rangle = \langle 3, 4, -6 \rangle$
Vector Equation: $\vec{r}(t) = \langle 2, -1, 4 \rangle + t\langle 3, 4, -6 \rangle = \langle 2 + 3t, -1 + 4t, 4 - 6t \rangle$
Parametric Equations: $x = 2 + 3t, \quad y = -1 + 4t, \quad z = 4 - 6t$
Symmetric Equations: $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 4}{-6}$

Example 2: Direction Vector with a Zero Component

Find the symmetric equations for the line through $P_0(1, 4, -3)$ parallel to $\vec{v} = \langle 2, 0, 5 \rangle$ .