3.13 Vector Equation of a Line in 2DHL ONLY

1. The Vector Equation Structure

A line is determined by a point $A(a_1, a_2)$ acting as an anchor and a direction vector $\vec{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$.

The position vector $\vec{r} = \begin{pmatrix} x \\ y \end{pmatrix}$ of any point $P(x,y)$ on the line is given by the vector equation:

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

Where:

$\vec{a}$ is the position vector of point A ($\overrightarrow{OA}$).
$\vec{b}$ is the direction vector parallel to the line.
$\lambda$ is a scalar parameter ($\lambda \in \mathbb{R}$).

Derivation: The path from the origin to point $P$ is $\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}$. Since $\overrightarrow{AP}$ is parallel to $\vec{b}$, $\overrightarrow{AP} = \lambda\vec{b}$.

2. Parametric, Cartesian, and Normal Forms

Separating the 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 individual coordinates gives the Parametric Equations:

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

Isolating $\lambda$ in both equations and equating them yields the Cartesian Equation:

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

Note on Normal Vectors: Rearranging the Cartesian equation into standard form $Ax + By + C = 0$ gives the perpendicular normal vector: $\mathbf{\vec{n} = \begin{pmatrix} A \\ B \end{pmatrix}}$.

EXAMPLE 1 (Formulating the Line Equations)

Given point $A(1,2)$ and direction vector $\vec{b} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, find all forms of the line equation.

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 $\lambda$ gives $\dfrac{x-1}{3} = \dfrac{y-2}{4}$.

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

Notice on Non-Unique Representations

A line can be described by infinitely many vector equations. The position vector $\vec{a}$ can be any point on the line, and the direction vector $\vec{b}$ can be any scalar multiple of itself (e.g., replacing $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ with $\begin{pmatrix} 6 \\ 8 \end{pmatrix}$ or $\begin{pmatrix} -3 \\ -4 \end{pmatrix}$).

3. Finding Lines and Checking Points

EXAMPLE 2 (Line Between Two Points)

Find the vector equation of the line passing through points $A(1,2)$ and $B(4,7)$.

Step 1: Choose a point as the anchor, such as $A \implies \vec{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.

Step 2: Find the direction vector by subtraction.

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

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

EXAMPLE 3 (Testing Point Incidence)

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

Set the line equation equal to the position vector of point C:

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

Since $\lambda = 3$ for both components, point $C$ lies on the line.

EXAMPLE 4 (Parallel and Perpendicular Lines)

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

Since $L_1$ is perpendicular to $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, its direction vector $\vec{b}$ must have a dot product of $0$. Swap the components and negate one to find the perpendicular vector:

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

Combine the point and direction vector: $\mathbf{\vec{r} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 4 \end{pmatrix}}$.

4. Intersecting Lines and Angular Geometry

To find the intersection of lines $L_1$ and $L_2$, set their position vectors equal ($\vec{r}_1 = \vec{r}_2$). Use different parameters (e.g., $\lambda$ and $\mu$) for each line before solving the system.

EXAMPLE 5 (Calculating Intersection Coordinates)

Find the intersection point of $\vec{r}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{r}_2 = \begin{pmatrix} 2 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 4 \end{pmatrix}$.

Step 1: Set the components equal to form a system of equations:

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

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

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

Step 3: Substitute $\lambda = 1$ into $L_1$ to find the intersection 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)

Find the angle between the lines in Example 5.

The angle depends on the dot product of the direction vectors $\vec{b}_1 = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{b}_2 = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$.

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

Calculate the acute angle: $\theta = \arccos(0.9216) \approx \mathbf{22.8^\circ}$.