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 $\vec{b} = \binom{b_1}{b_2}$ defining the trajectory.

The position vector $\vec{r} = \binom{x}{y}$ 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 point A ($\vec{OA}$), $\vec{b}$ is the parallel direction vector, and $\lambda$ acts as an unbounded scalar parameter ($\lambda \in \mathbb{R}$).

Derivation: The pathway to point $P$ follows $\vec{OP} = \vec{OA} + \vec{AP}$. Since $\vec{AP}$ is perfectly parallel to direction $\vec{b}$, it must equal $\lambda\vec{b}$.

2. Parametric and Cartesian Forms

Breaking the singular vector equation $\binom{x}{y} = \binom{a_1}{a_2} + \lambda\binom{b_1}{b_2}$ 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:

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

EXAMPLE 1

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

Vector Form: $\vec{r} = \binom{1}{2} + \lambda\binom{3}{4}$.
Parametric Form: $x = 1 + 3\lambda$ and $y = 2 + 4\lambda$.
Cartesian Form: Equating $\lambda$ yields $\frac{x-1}{3} = \frac{y-2}{4}$.
Algebraic rearrangement standardizes this to $4x - 4 = 3y - 6 \Rightarrow \mathbf{4x - 3y = -2}$.

Notice on Non-Unique Representations

A singular geometric line can be legitimately described by infinitely many valid vector equations. The anchor 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., $\binom{6}{8}$ or $\binom{-3}{-4}$).

3. Formulating Lines from Two Points

EXAMPLE 2

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

Solution: Any coordinate acts as the anchor (e.g., $\vec{a} = \binom{1}{2}$). The direction vector is formulated precisely by extracting the relative difference $\vec{b} = \vec{AB} = \binom{4-1}{7-2} = \binom{3}{5}$.
The final equation merges to $\mathbf{\vec{r} = \binom{1}{2} + \lambda\binom{3}{5}}$.

4. Intersecting Lines and Angular Geometry

To detect the specific 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 (Intersection)

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

Equating structural forms builds the linear system:
$1 + 3\lambda = 2 + \mu \Rightarrow 3\lambda - \mu = 1$
$2 + 4\lambda = -2 + 4\mu \Rightarrow 4\lambda - 4\mu = -4 \Rightarrow \lambda - \mu = -1$.
Solving the simultaneous system identifies $\lambda = 1$ and $\mu = 2$.
Injecting $\lambda = 1$ back into $L_1$ calculates the absolute coordinate point: $\binom{1+3}{2+4} = \mathbf{\binom{4}{6}}$.

EXAMPLE 6 (Angle Between Lines)

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

The intersection angle relies exclusively upon evaluating the dot product of the direction vectors: $\vec{b}_1 = \binom{3}{4}$ and $\vec{b}_2 = \binom{1}{4}$.
$\cos\theta = \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|} = \frac{3(1) + 4(4)}{5\sqrt{17}} = \frac{19}{5\sqrt{17}} \approx 0.922$.
Inverse computation extracts the acute geometric angle $\theta \approx \mathbf{22.8^\circ}$.