Ultimate Guide to Vector Equations of Lines

Introduction

In junior mathematics, you described straight lines using the Cartesian gradient-intercept formula: $y = mx + c$ . However, this formula completely breaks down in 3-dimensional space. How do you describe a line's gradient when there are three axes? In HSC Mathematics Extension 2, we use Vector Equations of Lines. This method elegantly describes lines in both 2D and 3D space using just two vectors: a starting position and a direction.

Executive Summary

This guide covers the foundation of vector line equations:

The Core Equation: Understanding $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ .
Position and Direction: Distinguishing between the position vector ( $\mathbf{a}$ ) that anchors the line, and the direction vector ( $\mathbf{b}$ ) that determines its angle.
Forming Equations from Points: Creating a line equation given two coordinates in space.
Converting Formats: Translating between 2D Cartesian ( $y = mx + c$ ) and 2D vector forms.

What is this about?

Imagine standing at a specific point on a map. To tell someone how to draw a line through your location, you need to tell them two things: where you are standing (the position), and which way you are facing (the direction).

In vector mathematics:

The place you are standing is the Position Vector ( $\mathbf{a}$ ). It points from the origin $(0,0,0)$ to a specific point on the line.
The way you are facing is the Direction Vector ( $\mathbf{b}$ ). It describes the parallel movement along the line.
To reach any point on that line (the generic vector $\mathbf{r}$ ), you start at $\mathbf{a}$ and take a certain number of steps ( $\lambda$ , a scalar parameter) in the direction of $\mathbf{b}$ .

Main Content

1. The Vector Equation of a Line

The universal formula for a straight line in any dimension is: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$

Where:

$\mathbf{r}$ is the position vector of any generic point $(x, y, z)$ on the line.
$\mathbf{a}$ is the position vector of a known point on the line.
$\mathbf{b}$ is the direction vector of the line.
$\lambda$ (lambda) is a real scalar parameter. As $\lambda$ changes from $-\infty$ to $\infty$ , $\mathbf{r}$ traces out every single point on the line.

Note: A line has infinitely many valid vector equations! You can choose any point on the line to be $\mathbf{a}$ , and you can scale the direction vector $\mathbf{b}$ by any non-zero constant, and the equation still describes the exact same line.

2. Finding the Equation from Two Points

If you are given two points on a line, let's say $A$ and $B$ , you need to find $\mathbf{a}$ and $\mathbf{b}$ .

The Position Vector ( $\mathbf{a}$ ): You can just use the position vector of $A$ (denoted $\vec{OA}$ ) or the position vector of $B$ (denoted $\vec{OB}$ ).
The Direction Vector ( $\mathbf{b}$ ): To find the direction between the two points, calculate the vector connecting them: $\vec{AB} = \vec{OB} - \vec{OA}$ .

So, the equation of the line passing through $A$ and $B$ can be written as: $\mathbf{r} = \vec{OA} + \lambda (\vec{OB} - \vec{OA})$

3. Direction Vectors and 2D Gradients

In 2D Cartesian geometry, a line has a gradient $m = \frac{\text{rise}}{\text{run}}$ . If a line has a gradient of $m = \frac{3}{2}$ , it means for every 2 units you move in the $x$ -direction (run), you move 3 units in the $y$ -direction (rise).

This translates perfectly into a 2D direction vector! If gradient $m = \frac{y_{step}}{x_{step}}$ , then the direction vector is: $\mathbf{b} = \begin{pmatrix} x_{step} \\ y_{step} \end{pmatrix}$ For $m = \frac{3}{2}$ , the direction vector is $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ .

4. Converting Cartesian to Vector Form (2D)

To convert $y = 3x - 5$ into a vector equation:

Find a point ( $\mathbf{a}$ ): Let $x = 0$ , then $y = -5$ . The point is $(0, -5)$ , so $\mathbf{a} = \begin{pmatrix} 0 \\ -5 \end{pmatrix}$ .
Find the direction ( $\mathbf{b}$ ): The gradient is $m = 3 = \frac{3}{1}$ . So the direction is $\mathbf{b} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ .
Write the equation: $\mathbf{r} = \begin{pmatrix} 0 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ .

Simple Worked Example

Question: Find a vector equation for the straight line passing through the points $A(1, -2, 4)$ and $B(3, 1, 5)$ in 3D space.

Solution: Step 1: Write down the position vectors of the points. $\vec{OA} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}$ $\vec{OB} = \begin{pmatrix} 3 \\ 1 \\ 5 \end{pmatrix}$

Step 2: Calculate the direction vector. The line travels from $A$ to $B$ . The direction vector is $\vec{AB}$ : $\mathbf{b} = \vec{AB} = \vec{OB} - \vec{OA}$ $\mathbf{b} = \begin{pmatrix} 3 \\ 1 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 - 1 \\ 1 - (-2) \\ 5 - 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$

Step 3: Construct the final equation. Using the formula $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ . We will use $\vec{OA}$ as our anchor point $\mathbf{a}$ . $\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$

(Note: $\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ -3 \\ -1 \end{pmatrix}$ would also be a completely correct and valid answer!)

mini-FAQ page

Q: Does it matter if I write vectors in column format or using $\mathbf{i}, \mathbf{j}, \mathbf{k}$ ? A: No, both are mathematically identical. $\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$ is exactly the same as $2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$ . Column vectors are generally faster to write and easier to read during long calculations.

Q: Can a 3D line be written in Cartesian form like $y = mx+c$ ? A: Not with a single equation! To describe a 1D line in 3D space using Cartesian algebra, you actually need a system of two plane equations intersecting. This is why the vector parameter form is vastly superior for 3D geometry.

Common mistakes to avoid

Confusing position and direction vectors: If a question asks for a line through $(1,2,3)$ parallel to another line with equation $\mathbf{r} = \begin{pmatrix} 4\\5\\6 \end{pmatrix} + \lambda \begin{pmatrix} 7\\8\\9 \end{pmatrix}$ , do not use the $(4,5,6)$ part! The direction of the parallel line is only the $(7,8,9)$ part.
Forgetting the $\mathbf{r} =$ part: Writing just $\begin{pmatrix} 1\\2\\3 \end{pmatrix} + \lambda \begin{pmatrix} 4\\5\\6 \end{pmatrix}$ is an expression, not an equation. You must write $\mathbf{r} = \dots$ or $\begin{pmatrix} x\\y\\z \end{pmatrix} = \dots$ to get full marks.

Practice on Vu's Maths Hub

Converting fluidly between geometric lines and vectors is a core skill.

Practice deriving 3D line equations in the HSC Vectors Booklet.
Test your 2D Cartesian-to-Vector conversions at Vu's Maths Hub.

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