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Equations of Lines in 3D

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Core Definition

In three-dimensional space (R3\mathbb{R}^3), a line cannot be defined by a single slope value (y=mx+by = mx + b). Instead, a line is uniquely determined by a known initial point P0(x0,y0,z0)P_0(x_0, y_0, z_0) with position vector r⃗0=⟨x0,y0,z0⟩\vec{r}_0 = \langle x_0, y_0, z_0 \rangle and a non-zero direction vector v⃗=⟨a,b,c⟩\vec{v} = \langle a, b, c \rangle:

  1. Vector Equation (Vector-Valued Function):
    r⃗(t)=r⃗0+tv⃗=⟨x0+at,y0+bt,z0+ct⟩,t∈R\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}
  2. Parametric Equations:
    x=x0+at,y=y0+bt,z=z0+ctx = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct
  3. Symmetric Equations (assuming a,b,c≠0a, b, c \neq 0):
    x−x0a=y−y0b=z−z0c\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 r⃗(t)\vec{r}(t) as the real scalar parameter tt varies from −∞-\infty to +∞+\infty.

  • When t=0t = 0, r⃗(0)=r⃗0\vec{r}(0) = \vec{r}_0, placing us directly at the initial reference point P0P_0.
  • When t=1t = 1, r⃗(1)=r⃗0+v⃗\vec{r}(1) = \vec{r}_0 + \vec{v}, advancing one full direction vector along the trajectory.
  • Changing the scalar parameter tt 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 tt transforms time-dependent motion into a purely static geometric path in R3\mathbb{R}^3:

  1. Solve each coordinate equation for tt:
    t=x−x0a,t=y−y0b,t=z−z0ct = \frac{x - x_0}{a}, \quad t = \frac{y - y_0}{b}, \quad t = \frac{z - z_0}{c}
  2. Equate the expressions for tt to form two simultaneous linear equations defining the intersection of two planes:
    x−x0a=y−y0b=z−z0c\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}
Edge Case: Zero Components in Direction Vector (a,b, or c=0a, b, \text{ or } c = 0)

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

y=y0,x−x0a=z−z0cy = 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=y0y = y_0.

Bounded Line Segments

To define a finite line segment between two endpoints P(x0,y0,z0)P(x_0, y_0, z_0) and Q(x1,y1,z1)Q(x_1, y_1, z_1), we set the direction vector to v⃗=PQ⃗=⟨x1−x0,y1−y0,z1−z0⟩\vec{v} = \vec{PQ} = \langle x_1 - x_0, y_1 - y_0, z_1 - z_0 \rangle and restrict the parameter domain:

r⃗(t)=r⃗0+tv⃗,0≤t≤1\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 r⃗(t)\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)P(2, -1, 4) and Q(5,3,−2)Q(5, 3, -2).

Solution
  1. Compute Direction Vector:
    v⃗=PQ⃗=⟨5−2,3−(−1),−2−4⟩=⟨3,4,−6⟩\vec{v} = \vec{PQ} = \langle 5 - 2, 3 - (-1), -2 - 4 \rangle = \langle 3, 4, -6 \rangle
  2. Vector Equation:
    r⃗(t)=⟨2,−1,4⟩+t⟨3,4,−6⟩=⟨2+3t,−1+4t,4−6t⟩\vec{r}(t) = \langle 2, -1, 4 \rangle + t\langle 3, 4, -6 \rangle = \langle 2 + 3t, -1 + 4t, 4 - 6t \rangle
  3. Parametric Equations:
    x=2+3t,y=−1+4t,z=4−6tx = 2 + 3t, \quad y = -1 + 4t, \quad z = 4 - 6t
  4. Symmetric Equations:
    x−23=y+14=z−4−6\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 P0(1,4,−3)P_0(1, 4, -3) parallel to v⃗=⟨2,0,5⟩\vec{v} = \langle 2, 0, 5 \rangle.

Solution
  1. Parametric Equations:
    x=1+2t,y=4+0t=4,z=−3+5tx = 1 + 2t, \quad y = 4 + 0t = 4, \quad z = -3 + 5t
  2. Symmetric Form: Since b=0b = 0, yy remains fixed at 44:
    y=4,x−12=z+35y = 4, \quad \frac{x - 1}{2} = \frac{z + 3}{5}

Connections & References