3.14 Vector Equation of a Line in 3D (HL)

1. 3D Line Equations

The vector logic defining 2D lines expands identically into 3-dimensional space by merely introducing the third $z$-axis parameter.

For a line anchored by spatial coordinate $A(a_1, a_2, a_3)$ tracking parallel to the 3D direction vector $\vec{b} = \binom{b_1}{b_2}{b_3}$:

  • Vector Equation: $\vec{r} = \vec{a} + \lambda\vec{b}$
  • Parametric Equations: $x = a_1 + \lambda b_1$, $\quad y = a_2 + \lambda b_2$, $\quad z = a_3 + \lambda b_3$
  • Cartesian Equations: $\frac{x - a_1}{b_1} = \frac{y - a_2}{b_2} = \frac{z - a_3}{b_3}$

Any point dynamically traversing this mathematical line adopts the unified positional structure $P(a_1 + \lambda b_1, a_2 + \lambda b_2, a_3 + \lambda b_3)$.

EXAMPLE 1

(a) Construct the line operating through spatial nodes $A(1,2,3)$ and $B(5,2,-1)$.
(b) Determine whether the isolated point $C(21,2,-17)$ mathematically rests on this specific line.

(a) Establishing the anchor $\vec{a} = \binom{1}{2}{3}$. The direction vector evaluates via subtraction $\vec{b} = \vec{AB} = \binom{5-1}{2-2}{-1-3} = \binom{4}{0}{-4}$.
The vector structure forms as $\mathbf{\vec{r} = \binom{1}{2}{3} + \lambda\binom{4}{0}{-4}}$.
(b) Point $C$ rests geometrically on the line only if a singular, identical scalar $\lambda$ satisfies all three coordinate axes concurrently.
$21 = 1 + 4\lambda \Rightarrow \lambda = 5$
$2 = 2 + 0\lambda \Rightarrow 2 = 2$ (Valid universally)
$-17 = 3 - 4\lambda \Rightarrow 4\lambda = 20 \Rightarrow \lambda = 5$
Because $\lambda=5$ aligns continuously across all conditions, point $C$ logically lies exactly on the line.

2. Relative Configurations of 3D Lines

In bounded 2D planes, non-parallel lines inevitably collide. Within expansive 3D space, lines can bypass each other entirely without ever intersecting, creating three categorized spatial conditions.

Spatial Condition Evaluation Methodology
Parallel (or Coincident) Direction vectors act as direct scalar multiples ($\vec{b}_1 \parallel \vec{b}_2$). If they also share an intersecting coordinate point, they logically coincide perfectly into one line.
Intersecting Set positional vectors equal ($\vec{r}_1 = \vec{r}_2$). This yields a system of 3 equations utilizing 2 variables ($\lambda, \mu$). Solving the first two equations establishes values for $\lambda$ and $\mu$. If these outputs perfectly satisfy the third equation, the lines physically intersect.
Skew Direction vectors are not parallel, yet the computed $\lambda$ and $\mu$ from the first two equations mathematically fail the third equation constraint. The lines traverse distinctly separate spatial planes.

EXAMPLE 2 (Analyzing Intersection Cases)

Intersecting Lines:
$L_1: \vec{r}_1 = \binom{3}{2}{1} + \lambda\binom{5}{4}{3}$ and $L_2: \vec{r}_2 = \binom{4}{4}{1} + \mu\binom{3}{2}{2}$.
System equates to: $3+5\lambda = 4+3\mu \Rightarrow 5\lambda - 3\mu = 1$
$2+4\lambda = 4+2\mu \Rightarrow 4\lambda - 2\mu = 2$
$1+3\lambda = 1+2\mu \Rightarrow 3\lambda - 2\mu = 0$.
Solving the second and third equations outputs $\lambda=2, \mu=3$. Checking against the first equation validates the solution ($5(2) - 3(3) = 1$). Lines physically intersect at spatial coordinate $\mathbf{(13, 10, 7)}$.
Skew Lines:
$L_1: \vec{r}_1 = \binom{3}{2}{1} + \lambda\binom{5}{4}{3}$ and $L_2: \vec{r}_2 = \binom{4}{4}{1} + \mu\binom{2}{2}{2}$.
System evaluates: $5\lambda - 2\mu = 1$ | $4\lambda - 2\mu = 2$ | $3\lambda - 2\mu = 0$.
Solving the latter two gives $\lambda=2, \mu=3$. Verifying this within the first equation yields $5(2) - 2(3) = 4$, which fatally contradicts the required output of $1$. The lines are conclusively skew.