3.13 Vector Equation of a Line in 2D (HL)

1. The Vector Equation Structure

A straight line is completely determined spatially by a known coordinate point $A(a_1, a_2)$ functioning as an anchor, and a designated direction vector $\vec{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$ defining the trajectory.

The position vector $\vec{r} = \begin{pmatrix} x \\ y \end{pmatrix}$ describing any random point $P(x,y)$ residing on this continuous line is governed by the vector equation:

$\vec{r} = \vec{a} + \lambda\vec{b}$

Where:

$\vec{a}$ is the position vector of the known anchor point A ($\overrightarrow{OA}$).
$\vec{b}$ is the parallel direction vector guiding the line.
$\lambda$ acts as an unbounded scalar parameter ($\lambda \in \mathbb{R}$).

Derivation: The pathway from the origin to point $P$ follows $\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}$. Since $\overrightarrow{AP}$ is perfectly parallel to direction $\vec{b}$, it must be exactly equivalent to $\lambda\vec{b}$.

2. Parametric, Cartesian, and Normal Forms

Breaking the singular vector equation $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \lambda\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$ into isolated coordinates produces the Parametric Equations:

$x = a_1 + \lambda b_1$
$y = a_2 + \lambda b_2$

Isolating the parameter $\lambda$ in both sequences and equating the results yields the Cartesian Equation:

$\dfrac{x - a_1}{b_1} = \dfrac{y - a_2}{b_2}$

Note on Normal Vectors: When the Cartesian format is rearranged into the standard form $Ax + By + C = 0$, the coefficients organically yield the line's perpendicular Normal Vector: $\mathbf{\vec{n} = \begin{pmatrix} A \\ B \end{pmatrix}}$.

EXAMPLE 1 (Formulating the Line Equations)

Given the anchor point $A(1,2)$ and directional trajectory $\vec{b} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, construct all line equation forms.

Vector Form: $\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 4 \end{pmatrix}$

Parametric Form: $x = 1 + 3\lambda$ and $y = 2 + 4\lambda$

Cartesian Form: Equating $\lambda$ yields $\dfrac{x-1}{3} = \dfrac{y-2}{4}$.

$$\begin{aligned} 4(x - 1) &= 3(y - 2) \\ 4x - 4 &= 3y - 6 \\ \mathbf{4x - 3y} &= \mathbf{-2} \end{aligned}$$

Notice on Non-Unique Representations

A singular geometric line can be legitimately described by infinitely many valid vector equations. The anchor position vector $\vec{a}$ can target any coordinate sitting on the line (by evaluating different $\lambda$ values), and the direction vector $\vec{b}$ can be swapped for any scalar multiple of itself (e.g., scaling $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ to $\begin{pmatrix} 6 \\ 8 \end{pmatrix}$ or $\begin{pmatrix} -3 \\ -4 \end{pmatrix}$).

3. Formulating Lines and Analyzing Points

EXAMPLE 2 (Line Between Two Points)

Determine the vector equation mapping the line extending through points $A(1,2)$ and $B(4,7)$.

Step 1: Select any coordinate as the anchor. Let's use $A \implies \vec{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.

Step 2: Formulate the direction vector exactly by extracting the relative difference.

$\vec{b} = \overrightarrow{AB} = \begin{pmatrix} 4 - 1 \\ 7 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$

Step 3: The final vector equation seamlessly merges to $\mathbf{\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 5 \end{pmatrix}}$.

EXAMPLE 3 (Testing Point Incidence)

Determine strictly if the point $C(10, 17)$ lies geometrically on the line $L: \vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 5 \end{pmatrix}$.

Equate the line equation to the target point's position vector:

$x\text{-component: } 10 = 1 + 3\lambda \implies 3\lambda = 9 \implies \lambda = 3$
$y\text{-component: } 17 = 2 + 5\lambda \implies 5\lambda = 15 \implies \lambda = 3$

Because the scalar parameter $\lambda$ produces identical values across all components, the point $C$ does securely lie on the line.

EXAMPLE 4 (Parallel and Perpendicular Lines)

A straight line $L_1$ passes through $P(2, -1)$ and is strictly perpendicular to the direction vector $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$. Find the vector equation of $L_1$.

If $L_1$ is perpendicular to $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, its own direction vector $\vec{b}$ must yield a dot product of exactly $0$. Swapping the components and student-negating one provides the orthogonal vector:

$\vec{b} = \begin{pmatrix} -3 \\ 4 \end{pmatrix} \quad \text{since} \quad \begin{pmatrix} 4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 4 \end{pmatrix} = -12 + 12 = 0$

Merging the anchor point and the derived direction constructs $\mathbf{\vec{r} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 4 \end{pmatrix}}$.

4. Intersecting Lines and Angular Geometry

To precisely detect the intersection coordinate between line $L_1$ and line $L_2$, equate their respective position vectors ($\vec{r}_1 = \vec{r}_2$). It is strictly mandatory to designate different scalar parameters (e.g., $\lambda$ and $\mu$) for each independent line prior to evaluating the resulting linear system.

EXAMPLE 5 (Calculating Intersection Coordinates)

Locate the exact intersection junction for $\vec{r}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{r}_2 = \begin{pmatrix} 2 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 4 \end{pmatrix}$.

Step 1: Equating structural forms independently builds the linear system:

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

Step 2: Subtract (Eq. 2) strictly from (Eq. 1) to solve the simultaneous system:

$2\lambda = 2 \implies \mathbf{\lambda = 1}$
Substitute back: $1 - \mu = -1 \implies \mathbf{\mu = 2}$

Step 3: Injecting $\lambda = 1$ back into $L_1$ exclusively calculates the absolute coordinate point:
$\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + 1\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \mathbf{\begin{pmatrix} 4 \\ 6 \end{pmatrix}}$

EXAMPLE 6 (Angle Between Intersecting Lines)

Compute the exact geometric angle intersecting the lines utilized in Example 5.

The intersection angle relies exclusively upon evaluating the dot product of the respective direction vectors: $\vec{b}_1 = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{b}_2 = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$.

$$\begin{aligned} \cos\theta &= \dfrac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|} \\ &= \dfrac{3(1) + 4(4)}{\sqrt{3^2 + 4^2}\sqrt{1^2 + 4^2}} \\ &= \dfrac{3 + 16}{5\sqrt{17}} = \dfrac{19}{5\sqrt{17}} \approx 0.9216 \end{aligned}$$

Inverse computation analytically extracts the acute geometric angle $\theta = \arccos(0.9216) \approx \mathbf{22.8^\circ}$.