Vectors—Arrows in Space
Discover the mathematical objects that have both magnitude and direction
Think about giving someone directions. “Walk 500 feet” is not particularly helpful on its own. Walk 500 feet in which direction? North? East? Toward the big oak tree? The distance alone does not tell you where you will end up. But “walk 500 feet north” tells you everything you need to know. This simple observation, that sometimes direction matters just as much as size, is the foundation of one of the most powerful ideas in mathematics: the vector.
You have been working with vectors your entire life, even if you did not call them that. When you describe the wind as “20 miles per hour from the west,” you are using a vector. When you push a shopping cart and it veers to the left, you are experiencing the result of force vectors. When your GPS shows an arrow pointing toward your destination, that is a vector. Linear algebra takes this intuitive idea and gives you the tools to work with it precisely, opening up applications from physics to computer graphics to machine learning.
In this lesson, we will build your understanding of vectors from the ground up. We will start with the intuitive picture (arrows) and develop the algebraic representation (lists of numbers) that makes vectors so powerful for computation. By the end, you will see how these two perspectives, geometric and algebraic, work together beautifully.
Core Concepts
What Is a Vector?
A vector is a mathematical object that has two essential properties: magnitude (how big it is) and direction (which way it points). You can visualize a vector as an arrow. The length of the arrow represents the magnitude, and the way the arrow points represents the direction.
This is fundamentally different from a regular number like 5 or -3.7, which has magnitude but no direction. In mathematics, we call these regular numbers scalars to distinguish them from vectors.
Here is the key distinction:
- Scalar: Temperature is 72 degrees. There is no direction associated with temperature (read: temperature does not point anywhere).
- Vector: The wind blows at 15 mph toward the east. Both the speed (15 mph) and the direction (east) matter.
Some real-world quantities that are vectors:
- Displacement: “Move 3 meters forward and 2 meters to the right”
- Velocity: “Traveling 60 km/h northeast”
- Force: “Push with 50 newtons to the left”
- Acceleration: “Speeding up at 9.8 m/s² downward” (gravity)
Some real-world quantities that are scalars:
- Temperature
- Mass
- Time
- Speed (just the number, without direction)
- Distance (total path length, without considering direction)
Scalars vs. Vectors
The distinction between scalars and vectors is not just academic. It changes how we do mathematics.
A scalar is simply a number, a single value that tells you “how much.” Scalars can be positive, negative, or zero. They can be integers, fractions, or irrational numbers like $\pi$. When you add two scalars, you get another scalar. When you multiply two scalars, you get another scalar. Nothing surprising there.
A vector, on the other hand, packages multiple pieces of information together. It tells you both “how much” and “which way.” You cannot fully describe a vector with just one number. In two-dimensional space, you need two numbers. In three-dimensional space, you need three. And the rules for adding and multiplying vectors are different from scalars (as we will see in future lessons).
Here is a helpful way to think about it: if you can represent something with a single number and that completely describes it, you have a scalar. If you need multiple numbers to capture the full picture, you probably have a vector.
Representing Vectors: The Geometric View
Geometrically, we represent a vector as an arrow. The arrow has two key features:
- Length: The length of the arrow represents the magnitude of the vector (how big it is).
- Direction: The way the arrow points represents the direction of the vector.
Here is something important: a vector is defined only by its length and direction, not by where it is located. Two arrows with the same length pointing in the same direction represent the same vector, even if they are in different locations. You can pick up a vector arrow and move it anywhere in space, and as long as you do not rotate it or stretch it, it remains the same vector.
This is like saying “move 3 steps forward.” It does not matter where you start. The instruction “3 steps forward” is the same instruction regardless of your starting position. The starting position is separate information.
Representing Vectors: The Algebraic View
While the arrow picture is great for intuition, we need a way to work with vectors using numbers and algebra. This is where component form comes in.
In two dimensions, we can describe any vector by two numbers: how far it goes in the horizontal direction (the $x$-component) and how far it goes in the vertical direction (the $y$-component).
We write a 2D vector as a column of numbers:
$$\vec{v} = \begin{pmatrix} x \ y \end{pmatrix}$$
The notation $\vec{v}$ (with the arrow on top) tells us that $v$ is a vector, not a scalar. The column format with numbers stacked vertically is called a column vector.
For example, the vector $\begin{pmatrix} 3 \ 4 \end{pmatrix}$ represents an arrow that goes 3 units to the right and 4 units up.
Vectors in 2D
In two-dimensional space (think: a flat piece of paper, a computer screen, a map), every vector can be written as:
$$\vec{v} = \begin{pmatrix} x \ y \end{pmatrix}$$
where:
- $x$ is the horizontal component (positive means right, negative means left)
- $y$ is the vertical component (positive means up, negative means down)
Some examples:
- $\begin{pmatrix} 3 \ 0 \end{pmatrix}$ points directly to the right
- $\begin{pmatrix} 0 \ -2 \end{pmatrix}$ points straight down
- $\begin{pmatrix} 1 \ 1 \end{pmatrix}$ points diagonally up and to the right (at a 45-degree angle)
- $\begin{pmatrix} -4 \ 3 \end{pmatrix}$ points up and to the left
The beauty of this representation is that we can now do arithmetic with vectors. We have converted a geometric object (an arrow) into an algebraic object (a list of numbers).
Vectors in 3D
When we move into three-dimensional space, we simply add a third component. A 3D vector looks like:
$$\vec{v} = \begin{pmatrix} x \ y \ z \end{pmatrix}$$
where:
- $x$ is the component in the first direction (often left-right)
- $y$ is the component in the second direction (often forward-backward or up-down, depending on convention)
- $z$ is the component in the third direction (often up-down or forward-backward)
For example, $\begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix}$ is a vector in 3D space with components 2, -1, and 3.
The same principles apply: this vector goes 2 units in the $x$-direction, -1 unit in the $y$-direction (so 1 unit in the negative $y$-direction), and 3 units in the $z$-direction.
Three-dimensional vectors are essential for describing the physical world we live in. Every position in space, every movement, every force can be described with a 3D vector.
Position Vectors and Displacement Vectors
There are two common ways to interpret what a vector represents:
A position vector describes a location relative to some reference point (called the origin). If you place the tail of the vector at the origin, the tip of the arrow points to the location being described. The vector $\begin{pmatrix} 3 \ 4 \end{pmatrix}$ as a position vector describes the point that is 3 units right and 4 units up from the origin.
A displacement vector describes a change in position, a movement from one place to another. The vector $\begin{pmatrix} 3 \ 4 \end{pmatrix}$ as a displacement vector means “move 3 units right and 4 units up,” regardless of where you start.
Mathematically, these are the same object. The difference is in interpretation. Here is the connection: if you start at point $A$ and move by displacement vector $\vec{d}$, you end up at point $B$, and the displacement vector equals the difference between the position vectors:
$$\vec{d} = \text{(position of } B\text{)} - \text{(position of } A\text{)}$$
To find the displacement vector from point $(x_1, y_1)$ to point $(x_2, y_2)$:
$$\vec{d} = \begin{pmatrix} x_2 - x_1 \ y_2 - y_1 \end{pmatrix}$$
You subtract the starting coordinates from the ending coordinates.
The Zero Vector
The zero vector is the vector with all components equal to zero:
$$\vec{0} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \quad \text{(in 2D)} \qquad \vec{0} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \quad \text{(in 3D)}$$
The zero vector is special. It has magnitude zero (no length) and no defined direction (since an arrow with no length cannot point anywhere). Geometrically, the zero vector represents “no displacement” or “staying in place.”
The zero vector plays the same role for vectors that the number 0 plays for regular numbers. Adding the zero vector to any vector leaves it unchanged:
$$\vec{v} + \vec{0} = \vec{v}$$
Magnitude (Length) of a Vector
The magnitude of a vector, also called its length or norm, tells you how long the arrow is. We write the magnitude of vector $\vec{v}$ as $|\vec{v}|$ (double vertical bars, like an “enhanced” absolute value).
For a 2D vector $\vec{v} = \begin{pmatrix} x \ y \end{pmatrix}$, the magnitude is:
$$|\vec{v}| = \sqrt{x^2 + y^2}$$
This formula comes directly from the Pythagorean theorem. If you draw the vector as an arrow, the horizontal and vertical components form the two legs of a right triangle, and the vector itself is the hypotenuse.
For a 3D vector $\vec{v} = \begin{pmatrix} x \ y \ z \end{pmatrix}$, the formula extends naturally:
$$|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$$
This is the 3D version of the Pythagorean theorem, also called the distance formula in three dimensions.
Important properties of magnitude:
- Magnitude is always non-negative: $|\vec{v}| \geq 0$
- The only vector with magnitude zero is the zero vector
- Magnitude gives you the “size” without any direction information
Unit Vectors
A unit vector is a vector with magnitude exactly 1. Unit vectors are useful because they capture pure direction without any scaling. Think of a unit vector as answering the question “which way?” without answering “how far?”
The most important unit vectors are the standard basis vectors:
In 2D: $$\hat{i} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \quad \text{(points in the positive } x \text{-direction)}$$ $$\hat{j} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \quad \text{(points in the positive } y \text{-direction)}$$
In 3D: $$\hat{i} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}, \quad \hat{j} = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}, \quad \hat{k} = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}$$
The “hat” notation ($\hat{i}$, read as “i-hat”) indicates a unit vector.
To find a unit vector in the same direction as any non-zero vector $\vec{v}$, divide the vector by its magnitude:
$$\hat{v} = \frac{\vec{v}}{|\vec{v}|}$$
This process is called normalizing a vector.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| Vector | Quantity with magnitude and direction | $\vec{v} = \begin{pmatrix} 3 \ 4 \end{pmatrix}$ |
| Scalar | A single number (no direction) | $5$, $-2.7$, $\pi$ |
| Component | Individual entry in a vector | In $\begin{pmatrix} 3 \ 4 \end{pmatrix}$, components are 3 and 4 |
| Magnitude | Length of the vector | $|\vec{v}|$ |
| $\vec{0}$ | Zero vector (no displacement) | $\begin{pmatrix} 0 \ 0 \end{pmatrix}$ |
| Unit vector | Vector with magnitude 1 | $\hat{i} = \begin{pmatrix} 1 \ 0 \end{pmatrix}$ |
| $\vec{v}$ | Arrow notation for a vector | Indicates $v$ is a vector, not a scalar |
| $|\vec{v}|$ | Magnitude (length) of $\vec{v}$ | $\left|\begin{pmatrix} 3 \ 4 \end{pmatrix}\right| = 5$ |
| Column vector | Vector written as a vertical column | $\begin{pmatrix} x \ y \end{pmatrix}$ |
| Position vector | Vector from origin to a point | Describes location |
| Displacement vector | Vector representing change in position | Describes movement |
Examples
Represent the vector from point $(1, 2)$ to $(4, 6)$.
Solution:
To find the displacement vector from one point to another, subtract the starting coordinates from the ending coordinates.
Step 1: Identify the starting and ending points.
- Starting point: $(1, 2)$
- Ending point: $(4, 6)$
Step 2: Calculate each component by subtracting.
- $x$-component: $4 - 1 = 3$
- $y$-component: $6 - 2 = 4$
Step 3: Write the vector.
$$\vec{v} = \begin{pmatrix} 3 \ 4 \end{pmatrix}$$
Interpretation: To get from $(1, 2)$ to $(4, 6)$, you move 3 units to the right and 4 units up. This makes sense if you visualize the two points on a coordinate plane.
Find the magnitude of $\vec{v} = \begin{pmatrix} 3 \ 4 \end{pmatrix}$.
Solution:
Use the magnitude formula for 2D vectors: $|\vec{v}| = \sqrt{x^2 + y^2}$.
Step 1: Identify the components.
- $x = 3$
- $y = 4$
Step 2: Substitute into the formula.
$$|\vec{v}| = \sqrt{3^2 + 4^2}$$
Step 3: Calculate.
$$|\vec{v}| = \sqrt{9 + 16} = \sqrt{25} = 5$$
Answer: $|\vec{v}| = 5$
Note: If you recognized this as a 3-4-5 right triangle, you could have gotten the answer immediately. The 3-4-5 triangle is one of the most common Pythagorean triples.
Find the magnitude of $\vec{u} = \begin{pmatrix} 1 \ 2 \ 2 \end{pmatrix}$.
Solution:
Use the magnitude formula for 3D vectors: $|\vec{u}| = \sqrt{x^2 + y^2 + z^2}$.
Step 1: Identify the components.
- $x = 1$
- $y = 2$
- $z = 2$
Step 2: Substitute into the formula.
$$|\vec{u}| = \sqrt{1^2 + 2^2 + 2^2}$$
Step 3: Calculate.
$$|\vec{u}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$
Answer: $|\vec{u}| = 3$
Note: This is a nice example where the magnitude works out to a whole number. In general, magnitudes are often irrational numbers involving square roots.
Find a unit vector in the direction of $\begin{pmatrix} 6 \ 8 \end{pmatrix}$.
Solution:
To find a unit vector in the same direction as a given vector, divide the vector by its magnitude.
Step 1: Find the magnitude of the original vector.
$$\left|\begin{pmatrix} 6 \ 8 \end{pmatrix}\right| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
Step 2: Divide each component by the magnitude.
$$\hat{v} = \frac{1}{10}\begin{pmatrix} 6 \ 8 \end{pmatrix} = \begin{pmatrix} 6/10 \ 8/10 \end{pmatrix} = \begin{pmatrix} 0.6 \ 0.8 \end{pmatrix}$$
Or, in fraction form:
$$\hat{v} = \begin{pmatrix} 3/5 \ 4/5 \end{pmatrix}$$
Step 3: Verify the magnitude is 1 (always a good check).
$$\left|\begin{pmatrix} 3/5 \ 4/5 \end{pmatrix}\right| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1 \quad \checkmark$$
Answer: The unit vector is $\begin{pmatrix} 3/5 \ 4/5 \end{pmatrix}$ or equivalently $\begin{pmatrix} 0.6 \ 0.8 \end{pmatrix}$.
Interpretation: This unit vector points in the exact same direction as $\begin{pmatrix} 6 \ 8 \end{pmatrix}$, but has been scaled down to have length exactly 1.
A plane flies 300 km east, then 400 km north. Represent each leg of the journey as a vector, then find the total displacement magnitude.
Solution:
Let us set up a coordinate system where east is the positive $x$-direction and north is the positive $y$-direction.
Step 1: Represent the first leg (300 km east) as a vector.
$$\vec{v}_1 = \begin{pmatrix} 300 \ 0 \end{pmatrix}$$
The plane moves 300 km in the $x$-direction (east) and 0 km in the $y$-direction (no north-south movement).
Step 2: Represent the second leg (400 km north) as a vector.
$$\vec{v}_2 = \begin{pmatrix} 0 \ 400 \end{pmatrix}$$
The plane moves 0 km in the $x$-direction and 400 km in the $y$-direction (north).
Step 3: Find the total displacement vector.
The total displacement is the sum of the two displacement vectors (we will cover vector addition formally in the next lesson, but the idea is straightforward: add corresponding components).
$$\vec{d} = \vec{v}_1 + \vec{v}_2 = \begin{pmatrix} 300 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 400 \end{pmatrix} = \begin{pmatrix} 300 \ 400 \end{pmatrix}$$
Step 4: Find the magnitude of the total displacement.
$$|\vec{d}| = \sqrt{300^2 + 400^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \text{ km}$$
Answer: The total displacement magnitude is 500 km.
Interpretation: Even though the plane flew a total distance of $300 + 400 = 700$ km along its path, it ended up only 500 km from its starting point. This is the difference between distance (total path length, a scalar) and displacement (net change in position, a vector). The displacement magnitude is always less than or equal to the distance traveled, with equality only when you travel in a straight line.
Notice that this is another 3-4-5 right triangle, just scaled up by a factor of 100.
Key Properties and Rules
Magnitude Properties
- Non-negativity: $|\vec{v}| \geq 0$ for all vectors
- Zero vector: $|\vec{v}| = 0$ if and only if $\vec{v} = \vec{0}$
- Pythagorean formula (2D): $|\vec{v}| = \sqrt{x^2 + y^2}$
- Pythagorean formula (3D): $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$
The Zero Vector
- The zero vector $\vec{0}$ has magnitude 0
- The zero vector has no defined direction
- Adding $\vec{0}$ to any vector leaves it unchanged: $\vec{v} + \vec{0} = \vec{v}$
Unit Vectors
- A unit vector has magnitude exactly 1
- To create a unit vector from $\vec{v}$: divide by the magnitude, $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$
- The standard basis vectors $\hat{i}$, $\hat{j}$, $\hat{k}$ are unit vectors pointing along the coordinate axes
Displacement Vectors
- From point $A = (x_1, y_1)$ to point $B = (x_2, y_2)$: $$\overrightarrow{AB} = \begin{pmatrix} x_2 - x_1 \ y_2 - y_1 \end{pmatrix}$$
- The displacement from $A$ to $B$ is the negative of the displacement from $B$ to $A$: $\overrightarrow{AB} = -\overrightarrow{BA}$
Scalars vs. Vectors
- Scalars have magnitude only; vectors have magnitude and direction
- Scalars are single numbers; vectors are lists of numbers (components)
- Temperature, mass, and speed are scalars; force, velocity, and displacement are vectors
Real-World Applications
GPS Navigation and Displacement
Your smartphone’s GPS constantly works with vectors. When the app calculates how to get from your current location to a destination, it is computing a displacement vector. The “as the crow flies” distance shown on the map is the magnitude of this displacement vector. Turn-by-turn directions break this overall displacement into a sequence of smaller displacement vectors (read: turn left, go 500 feet, turn right, etc.).
When you are hiking and want to know how far you are from camp, the GPS computes your position vector relative to camp and finds its magnitude. The bearing (direction) to camp is the direction of that position vector.
Physics: Force, Velocity, and Acceleration
In physics, vectors are everywhere. Force is a vector: it matters not just how hard you push, but in which direction. When multiple forces act on an object (gravity pulling down, friction pushing back, you pushing forward), the object accelerates in the direction of the total force, which is the vector sum of all individual forces.
Velocity is a vector: speed with a direction attached. A car going 60 mph north has a different velocity than a car going 60 mph south, even though they have the same speed.
Acceleration is a vector: the rate of change of velocity. Gravity causes a constant acceleration of about 9.8 m/s² downward everywhere on Earth’s surface, represented as the vector $\begin{pmatrix} 0 \ -9.8 \end{pmatrix}$ in a coordinate system where up is positive.
Computer Graphics: Positions and Directions
Video games and animated movies represent every object’s location as a position vector. When a character moves, the game updates the position vector by adding a velocity vector (scaled by the time elapsed). Lighting calculations use the angle between surface normal vectors (which way a surface faces) and light direction vectors to determine how bright each point appears.
Even 2D games and user interfaces use vectors. When you drag an icon across your screen, the computer tracks the displacement vector of your mouse movement and moves the icon accordingly.
Wind Speed and Direction in Weather Reports
Weather forecasts describe wind as a vector quantity: “winds from the northwest at 15 mph.” The direction tells you where the wind is coming from, and the speed gives the magnitude. Meteorologists represent wind fields as collections of vectors at different locations, showing how air moves across a region.
Pilots and sailors must account for wind vectors when planning routes. If you want to fly due north but there is a wind from the west, you must aim your plane slightly into the wind (toward the west) so that your actual path (the vector sum of your airspeed vector and the wind vector) takes you north.
Self-Test Problems
Problem 1: Write the vector from point $(-2, 5)$ to point $(4, 1)$ in component form.
Show Answer
Subtract the starting coordinates from the ending coordinates:
$$\vec{v} = \begin{pmatrix} 4 - (-2) \ 1 - 5 \end{pmatrix} = \begin{pmatrix} 6 \ -4 \end{pmatrix}$$
The displacement is 6 units to the right and 4 units down.
Problem 2: Find the magnitude of $\vec{w} = \begin{pmatrix} -5 \ 12 \end{pmatrix}$.
Show Answer
$$|\vec{w}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$
The magnitude is 13. (This is a 5-12-13 right triangle, another common Pythagorean triple.)
Problem 3: Find the magnitude of the 3D vector $\vec{u} = \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix}$.
Show Answer
$$|\vec{u}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$
The magnitude is 3.
Problem 4: Find the unit vector in the direction of $\vec{v} = \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix}$.
Show Answer
Step 1: Find the magnitude. $$|\vec{v}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$
Step 2: Divide each component by the magnitude. $$\hat{v} = \frac{1}{3}\begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix} = \begin{pmatrix} 1/3 \ -2/3 \ 2/3 \end{pmatrix}$$
The unit vector is $\begin{pmatrix} 1/3 \ -2/3 \ 2/3 \end{pmatrix}$.
Problem 5: A hiker walks 5 km south, then 12 km east. What is the magnitude of their total displacement?
Show Answer
Set up coordinates with east as positive $x$ and north as positive $y$.
First leg (5 km south): $\vec{v}_1 = \begin{pmatrix} 0 \ -5 \end{pmatrix}$
Second leg (12 km east): $\vec{v}_2 = \begin{pmatrix} 12 \ 0 \end{pmatrix}$
Total displacement: $\vec{d} = \begin{pmatrix} 12 \ -5 \end{pmatrix}$
Magnitude: $|\vec{d}| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ km
The hiker is 13 km from their starting point. (Another 5-12-13 triangle!)
Problem 6: Is $\begin{pmatrix} 0.6 \ 0.8 \end{pmatrix}$ a unit vector? Verify your answer.
Show Answer
Calculate the magnitude: $$\left|\begin{pmatrix} 0.6 \ 0.8 \end{pmatrix}\right| = \sqrt{0.6^2 + 0.8^2} = \sqrt{0.36 + 0.64} = \sqrt{1} = 1$$
Yes, it is a unit vector because its magnitude is exactly 1.
Problem 7: The zero vector is sometimes written as $\vec{0}$. What is its magnitude? Does it have a direction?
Show Answer
The zero vector in 2D is $\begin{pmatrix} 0 \ 0 \end{pmatrix}$.
Magnitude: $|\vec{0}| = \sqrt{0^2 + 0^2} = \sqrt{0} = 0$
The zero vector has magnitude 0 and no defined direction. A zero-length arrow cannot point in any particular direction, so direction is undefined for the zero vector.
Summary
- A vector is a mathematical quantity with both magnitude (size) and direction, visualized as an arrow.
- A scalar is just a number with magnitude only, no direction attached.
- Vectors in 2D are written as $\vec{v} = \begin{pmatrix} x \ y \end{pmatrix}$, where $x$ and $y$ are the horizontal and vertical components.
- Vectors in 3D are written as $\vec{v} = \begin{pmatrix} x \ y \ z \end{pmatrix}$ with three components.
- The magnitude (length) of a 2D vector is $|\vec{v}| = \sqrt{x^2 + y^2}$, coming directly from the Pythagorean theorem.
- The magnitude of a 3D vector is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
- A displacement vector from point $(x_1, y_1)$ to $(x_2, y_2)$ is found by subtracting: $\begin{pmatrix} x_2 - x_1 \ y_2 - y_1 \end{pmatrix}$.
- The zero vector $\vec{0}$ has magnitude zero and no defined direction.
- A unit vector has magnitude exactly 1 and represents pure direction.
- To find a unit vector in the direction of $\vec{v}$, divide by the magnitude: $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$.
- Vectors appear throughout physics (force, velocity, acceleration), computer graphics (positions, directions), navigation (GPS, wind), and countless other applications where direction matters.