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. The fundamental structure of anchoring a line to a point and projecting it along a direction vector remains completely unchanged.

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

  • Vector Equation: $\vec{r} = \vec{a} + \lambda\vec{b}$
  • Parametric Equations:
    $$ \begin{aligned} x &= a_1 + \lambda b_1 \\ y &= a_2 + \lambda b_2 \\ z &= a_3 + \lambda b_3 \end{aligned} $$
  • Cartesian Equations:
    $\dfrac{x - a_1}{b_1} = \dfrac{y - a_2}{b_2} = \dfrac{z - a_3}{b_3}$

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

O y z x A P(x,y,z) a r λb L

EXAMPLE 1 (Constructing & Testing a 3D Line)

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

(a) Establishing the anchor $\vec{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$. The direction vector evaluates via exact subtraction:
$\vec{b} = \overrightarrow{AB} = \begin{pmatrix} 5-1 \\ 2-2 \\ -1-3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ -4 \end{pmatrix}$

The final vector structure securely forms as $\mathbf{\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 0 \\ -4 \end{pmatrix}}$.
(b) Point $C$ rests geometrically on the line $\iff$ a singular, identical scalar $\lambda$ logically satisfies all three coordinate axes concurrently.
$x\text{-axis: } 21 = 1 + 4\lambda \implies 4\lambda = 20 \implies \lambda = 5$
$y\text{-axis: } 2 = 2 + 0\lambda \implies 2 = 2$ (Valid universally regardless of $\lambda$)
$z\text{-axis: } -17 = 3 - 4\lambda \implies 4\lambda = 20 \implies \lambda = 5$
Because $\lambda=5$ aligns continuously across all three 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 physically bypass each other entirely without ever intersecting, creating three distinct categorical 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 singular line.
Intersecting Equate the positional vectors ($\vec{r}_1 = \vec{r}_2$). This yields a system of $3$ equations utilizing $2$ distinct variables ($\lambda, \mu$). Solving the first two equations establishes fixed 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 $\mu$ extracted from the first two equations mathematically fail the third equation's constraint. The lines traverse distinctly separate spatial planes.

EXAMPLE 2 (Analyzing Intersecting Lines)

Determine the exact intersection coordinates for the following two lines:

$L_1: \vec{r}_1 = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 5 \\ 4 \\ 3 \end{pmatrix} \quad \text{and} \quad L_2: \vec{r}_2 = \begin{pmatrix} 4 \\ 4 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$

Step 1: Equate components to build the linear system.

$$\begin{aligned} 3 + 5\lambda &= 4 + 3\mu \implies 5\lambda - 3\mu = 1 \quad &\text{--- (Eq. 1)}\\ 2 + 4\lambda &= 4 + 2\mu \implies 4\lambda - 2\mu = 2 \implies 2\lambda - \mu = 1 \quad &\text{--- (Eq. 2)}\\ 1 + 3\lambda &= 1 + 2\mu \implies 3\lambda - 2\mu = 0 \quad &\text{--- (Eq. 3)} \end{aligned}$$

Step 2: Solve the system using Equations 2 and 3.

From (Eq. 2), $\mu = 2\lambda - 1$. Substitute this into (Eq. 3):
$3\lambda - 2(2\lambda - 1) = 0 \implies 3\lambda - 4\lambda + 2 = 0 \implies \mathbf{\lambda = 2}$
Substitute back: $\mu = 2(2) - 1 \implies \mathbf{\mu = 3}$

Step 3: Validate strictly against the unused Equation 1.

$5(2) - 3(3) = 10 - 9 = 1$. The equality holds perfectly.

Step 4: Inject $\lambda=2$ into $L_1$ to locate the point:
$\vec{r} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 5 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 13 \\ 10 \\ 7 \end{pmatrix}$. The lines intersect at $\mathbf{(13, 10, 7)}$.

EXAMPLE 3 (Identifying Skew Lines)

Show algebraically that the following lines are skew:

$L_1: \vec{r}_1 = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 5 \\ 4 \\ 3 \end{pmatrix} \quad \text{and} \quad L_2: \vec{r}_2 = \begin{pmatrix} 4 \\ 4 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}$

The system evaluates to:

$$\begin{aligned} 5\lambda - 2\mu &= 1 \quad &\text{--- (Eq. 1)}\\ 4\lambda - 2\mu &= 2 \quad &\text{--- (Eq. 2)}\\ 3\lambda - 2\mu &= 0 \quad &\text{--- (Eq. 3)} \end{aligned}$$

Subtracting (Eq. 3) from (Eq. 2) directly eliminates $\mu$:

$(4\lambda - 2\mu) - (3\lambda - 2\mu) = 2 - 0 \implies \mathbf{\lambda = 2}$
Substitute into (Eq. 3): $3(2) - 2\mu = 0 \implies 2\mu = 6 \implies \mathbf{\mu = 3}$

Verifying this within (Eq. 1) yields $5(2) - 2(3) = 10 - 6 = \mathbf{4}$, which fatally contradicts the required output of $1$. Since the system is mathematically inconsistent, the lines bypass one another and are conclusively skew.