3.14 Vector Equation of a Line in 3DHL ONLY

1. 3D Line Equations

The vector logic for 2D lines extends to 3D space by introducing a third axis, the $z$-axis. The structure of anchoring a line to a point and extending it along a direction vector remains unchanged.

For a line passing through point $A(a_1, a_2, a_3)$ and parallel to the 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 on this line has the coordinates $P(a_1 + \lambda b_1, a_2 + \lambda b_2, a_3 + \lambda b_3)$.

EXAMPLE 1 (Constructing & Testing a 3D Line)

(a) Find the vector equation of the line passing through points $A(1,2,3)$ and $B(5,2,-1)$.
(b) Determine whether point $C(21,2,-17)$ lies on this line.

(a) Use the anchor vector $\vec{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$. Find the direction vector by subtraction:

$\vec{b} = \overrightarrow{AB} = \begin{pmatrix} 5-1 \\ 2-2 \\ -1-3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ -4 \end{pmatrix}$

The vector equation of the line is $\mathbf{\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 0 \\ -4 \end{pmatrix}}$.

(b) Point $C$ lies on the line if a single value of $\lambda$ satisfies the equations for all three coordinates:

$x\text{-axis: } 21 = 1 + 4\lambda \implies 4\lambda = 20 \implies \lambda = 5$ bracket
$y\text{-axis: } 2 = 2 + 0\lambda \implies 2 = 2$ (True for any value of $\lambda$)
$z\text{-axis: } -17 = 3 - 4\lambda \implies 4\lambda = 20 \implies \lambda = 5$

Since $\lambda=5$ satisfies all equations, point $C$ lies on the line.

2. Relative Configurations of 3D Lines

In 2D, non-parallel lines always intersect. In 3D space, lines can pass each other without intersecting, resulting in three possible relationships.

Spatial Condition	Evaluation Methodology
Parallel (or Coincident)	The direction vectors are scalar multiples ($\vec{b}_1 \parallel \vec{b}_2$). If the lines also share a point, they are coincident.
Intersecting	Set the vector equations equal ($\vec{r}_1 = \vec{r}_2$) to form a system of three equations with two variables ($\lambda, \mu$). Solve two equations to find $\lambda$ and $\mu$. If these values satisfy the third equation, the lines intersect.
Skew	The direction vectors are not parallel, and the values of $\lambda$ and $\mu$ found from two equations do not satisfy the third equation. The lines do not intersect.

EXAMPLE 2 (Analyzing Intersecting Lines)

Find the coordinates of the intersection point of the 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 form a system of linear equations.

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

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: Check the values in the remaining Equation 1.

$5(2) - 3(3) = 10 - 9 = 1$. The equation is satisfied.

Step 4: Substitute $\lambda=2$ into the equation for $L_1$ to find 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 that the lines are skew:

Form the system of equations:

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

Subtract Equation 3 from Equation 2 to eliminate $\mu$:

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

Substitute these values into Equation 1: $5(2) - 2(3) = 4$, which contradicts the required value of $1$. Since the system has no solution, the lines do not intersect and are skew.