Vectors
Represent quantities with both magnitude and direction
You have probably been working with numbers your entire mathematical life, and those numbers have always just been… numbers. Five apples, ten dollars, 98.6 degrees Fahrenheit. These quantities tell you how much, but they do not tell you anything about which way. And for most everyday situations, that is perfectly fine.
But think about directions for a moment. If someone asks you how to get to the coffee shop, saying “walk 500 feet” is not very helpful without also saying which direction to walk. Walk 500 feet north? South? Diagonally across the parking lot? The magnitude (500 feet) only becomes useful when paired with a direction. This is the essence of what vectors are all about.
In physics, engineering, computer graphics, navigation, and countless other fields, direction matters just as much as size. Vectors give us a way to capture both pieces of information in a single mathematical object. And the good news is that working with vectors is not nearly as intimidating as it might first appear. If you can add, subtract, multiply, and work with right triangles, you already have all the tools you need.
Core Concepts
What Is a Vector?
A vector is a mathematical quantity that has both magnitude (size or length) and direction. This is in contrast to a scalar, which is just a regular number that only has magnitude.
Think of the difference this way:
- Temperature is a scalar: 72 degrees. There is no “direction” to temperature.
- Velocity is a vector: 60 miles per hour heading north. Both the speed and the direction matter.
Visually, we represent a vector as an arrow. The length of the arrow represents the magnitude, and the arrow points in the direction of the vector.
Some common examples of vector quantities in the real world:
- Displacement: “Walk 3 miles east” (not just “walk 3 miles”)
- Velocity: “Moving at 25 m/s toward the goal” (speed + direction)
- Force: “Push with 50 newtons to the left”
- Wind: “15 mph from the northwest”
Vector Notation
There are several ways to write vectors, and you will encounter all of them in your mathematical journey.
Arrow notation: We often write a vector with an arrow above the letter:
$$\vec{v}$$
This tells us that $v$ is a vector, not just a regular number.
Bold notation: In textbooks, you will sometimes see vectors written in bold: v. This is common in printed materials but harder to write by hand, which is why arrow notation is more practical.
Component form: The most common way to work with vectors algebraically is to write them as an ordered pair of components:
$$\vec{v} = \langle a, b \rangle$$
Here, $a$ is the horizontal component (how far the vector goes in the $x$-direction) and $b$ is the vertical component (how far it goes in the $y$-direction). The angle brackets $\langle \rangle$ specifically indicate a vector, distinguishing it from a point $(a, b)$.
Initial and terminal points: If a vector goes from point $A = (x_1, y_1)$ to point $B = (x_2, y_2)$, we can write it as $\overrightarrow{AB}$, and its component form is:
$$\overrightarrow{AB} = \langle x_2 - x_1, y_2 - y_1 \rangle$$
You simply subtract the starting coordinates from the ending coordinates.
Find the vector from point $A = (2, 3)$ to point $B = (7, 11)$.
Step 1: Subtract the coordinates: ending minus starting.
$$\overrightarrow{AB} = \langle 7 - 2, 11 - 3 \rangle$$
Step 2: Simplify.
$$\overrightarrow{AB} = \langle 5, 8 \rangle$$
Answer: The vector is $\langle 5, 8 \rangle$. This means to get from $A$ to $B$, you move 5 units right and 8 units up.
Magnitude of a Vector
The magnitude of a vector is its length. For a vector $\vec{v} = \langle a, b \rangle$, we find the magnitude using the Pythagorean theorem:
$$|\vec{v}| = \sqrt{a^2 + b^2}$$
The vertical bars indicate magnitude (similar to absolute value for numbers). Some textbooks use $|\vec{v}|$ instead.
Why does the Pythagorean theorem apply here? If you draw the vector on a coordinate plane, the horizontal component $a$ and vertical component $b$ form the legs of a right triangle, with the vector itself as the hypotenuse.
Find the magnitude of $\vec{v} = \langle 3, 4 \rangle$.
Step 1: Apply the magnitude formula:
$$|\vec{v}| = \sqrt{3^2 + 4^2}$$
Step 2: Calculate:
$$|\vec{v}| = \sqrt{9 + 16} = \sqrt{25} = 5$$
Answer: $|\vec{v}| = 5$
(You might recognize this as a 3-4-5 right triangle.)
Direction Angle
The direction angle of a vector is the angle it makes with the positive $x$-axis, measured counterclockwise. For a vector $\vec{v} = \langle a, b \rangle$, the direction angle $\theta$ can be found using:
$$\tan \theta = \frac{b}{a}$$
Or equivalently:
$$\theta = \tan^{-1}\left(\frac{b}{a}\right)$$
Important: The inverse tangent function on your calculator only gives angles between $-90°$ and $90°$ (or $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ radians). If your vector points into the second or third quadrant, you need to add $180°$ (or $\pi$ radians) to get the correct angle.
Find the direction angle of $\vec{v} = \langle 3, 3\sqrt{3} \rangle$.
Step 1: Calculate the tangent of the angle:
$$\tan \theta = \frac{3\sqrt{3}}{3} = \sqrt{3}$$
Step 2: Find the angle whose tangent is $\sqrt{3}$:
$$\theta = \tan^{-1}(\sqrt{3}) = 60°$$
Step 3: Check the quadrant. Since both components are positive, the vector is in the first quadrant, so $60°$ is correct.
Answer: The direction angle is $60°$ (or $\frac{\pi}{3}$ radians).
Unit Vectors
A unit vector is a vector with magnitude 1. Unit vectors are extremely useful because they capture direction without any scaling. Think of a unit vector as a “pure direction” indicator.
To find the unit vector in the same direction as a given vector $\vec{v}$, divide $\vec{v}$ by its magnitude:
$$\hat{v} = \frac{\vec{v}}{|\vec{v}|}$$
The “hat” notation $\hat{v}$ (read as “v hat”) is commonly used to indicate a unit vector.
Standard unit vectors: In two dimensions, we have two special unit vectors:
$$\hat{i} = \langle 1, 0 \rangle \quad \text{(points in the positive } x \text{-direction)}$$
$$\hat{j} = \langle 0, 1 \rangle \quad \text{(points in the positive } y \text{-direction)}$$
Any vector $\vec{v} = \langle a, b \rangle$ can be written in terms of these unit vectors:
$$\vec{v} = a\hat{i} + b\hat{j}$$
This is called standard unit vector notation or $\hat{i}$-$\hat{j}$ notation.
Find the unit vector in the direction of $\vec{v} = \langle 6, 8 \rangle$. Then write the original vector in $\hat{i}$-$\hat{j}$ notation.
Step 1: Find the magnitude:
$$|\vec{v}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
Step 2: Divide the vector by its magnitude:
$$\hat{v} = \frac{\langle 6, 8 \rangle}{10} = \left\langle \frac{6}{10}, \frac{8}{10} \right\rangle = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle$$
Step 3: Verify the magnitude is 1:
$$|\hat{v}| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = 1 \quad \checkmark$$
Step 4: Write the original vector in $\hat{i}$-$\hat{j}$ notation:
$$\vec{v} = 6\hat{i} + 8\hat{j}$$
Answer: The unit vector is $\left\langle \frac{3}{5}, \frac{4}{5} \right\rangle$, and in $\hat{i}$-$\hat{j}$ notation, $\vec{v} = 6\hat{i} + 8\hat{j}$.
Vector Operations
One of the beautiful things about vectors is that we can add, subtract, and multiply them in meaningful ways.
Vector Addition: To add two vectors, add their corresponding components:
$$\vec{u} + \vec{v} = \langle u_1 + v_1, u_2 + v_2 \rangle$$
Geometrically, this is the “tip-to-tail” method: place the tail of the second vector at the tip of the first, and the sum is the vector from the start to the end.
Vector Subtraction: Subtraction works similarly:
$$\vec{u} - \vec{v} = \langle u_1 - v_1, u_2 - v_2 \rangle$$
This gives you the vector that, when added to $\vec{v}$, gives you $\vec{u}$.
Scalar Multiplication: To multiply a vector by a scalar (a regular number), multiply each component:
$$c\vec{v} = c\langle v_1, v_2 \rangle = \langle cv_1, cv_2 \rangle$$
If $c > 0$, the vector points in the same direction but is scaled by factor $c$. If $c < 0$, the vector is scaled and reversed in direction.
Let $\vec{u} = \langle 4, -2 \rangle$ and $\vec{v} = \langle -1, 5 \rangle$. Find:
- $\vec{u} + \vec{v}$
- $\vec{u} - \vec{v}$
- $3\vec{u}$
Addition: $$\vec{u} + \vec{v} = \langle 4 + (-1), -2 + 5 \rangle = \langle 3, 3 \rangle$$
Subtraction: $$\vec{u} - \vec{v} = \langle 4 - (-1), -2 - 5 \rangle = \langle 5, -7 \rangle$$
Scalar multiplication: $$3\vec{u} = 3\langle 4, -2 \rangle = \langle 12, -6 \rangle$$
Answers: $\vec{u} + \vec{v} = \langle 3, 3 \rangle$, $\vec{u} - \vec{v} = \langle 5, -7 \rangle$, $3\vec{u} = \langle 12, -6 \rangle$
The Dot Product
The dot product (also called the scalar product) is a way to multiply two vectors that produces a scalar (a single number, not a vector). For vectors $\vec{u} = \langle u_1, u_2 \rangle$ and $\vec{v} = \langle v_1, v_2 \rangle$:
$$\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2$$
Multiply corresponding components and add the results. That is all there is to it.
There is also a geometric interpretation that connects the dot product to angles:
$$\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos \theta$$
where $\theta$ is the angle between the two vectors. This formula is incredibly powerful because it lets us find angles between vectors.
Find the dot product of $\vec{u} = \langle 3, 7 \rangle$ and $\vec{v} = \langle 2, -4 \rangle$.
Step 1: Apply the formula:
$$\vec{u} \cdot \vec{v} = (3)(2) + (7)(-4)$$
Step 2: Calculate:
$$\vec{u} \cdot \vec{v} = 6 + (-28) = -22$$
Answer: $\vec{u} \cdot \vec{v} = -22$
(A negative dot product indicates that the angle between the vectors is greater than $90°$.)
Finding the Angle Between Vectors
Using the geometric formula for the dot product, we can solve for the angle between two vectors:
$$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$$
Then:
$$\theta = \cos^{-1}\left(\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}\right)$$
This formula always gives an angle between $0°$ and $180°$ (or $0$ and $\pi$ radians), which is exactly what we want for the angle between two vectors.
Find the angle between $\vec{u} = \langle 4, 3 \rangle$ and $\vec{v} = \langle 1, 2 \rangle$.
Step 1: Calculate the dot product:
$$\vec{u} \cdot \vec{v} = (4)(1) + (3)(2) = 4 + 6 = 10$$
Step 2: Find the magnitudes:
$$|\vec{u}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$
$$|\vec{v}| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$$
Step 3: Apply the formula:
$$\cos \theta = \frac{10}{5 \cdot \sqrt{5}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$
Step 4: Find the angle:
$$\theta = \cos^{-1}\left(\frac{2\sqrt{5}}{5}\right) \approx 26.57°$$
Answer: The angle between the vectors is approximately $26.57°$ (or about $0.464$ radians).
Parallel and Perpendicular Vectors
The dot product gives us quick tests for two important relationships between vectors.
Perpendicular (Orthogonal) Vectors: Two vectors are perpendicular if and only if their dot product is zero:
$$\vec{u} \perp \vec{v} \iff \vec{u} \cdot \vec{v} = 0$$
Why does this work? If $\theta = 90°$, then $\cos 90° = 0$, so $|\vec{u}||\vec{v}|\cos 90° = 0$.
Parallel Vectors: Two vectors are parallel if one is a scalar multiple of the other. In other words, $\vec{u}$ and $\vec{v}$ are parallel if there exists some scalar $c$ such that:
$$\vec{u} = c\vec{v}$$
Parallel vectors point in the same direction (if $c > 0$) or opposite directions (if $c < 0$). We can also check using the dot product: $\vec{u}$ and $\vec{v}$ are parallel if and only if:
$$|\vec{u} \cdot \vec{v}| = |\vec{u}||\vec{v}|$$
This happens when $\cos \theta = 1$ or $\cos \theta = -1$, meaning $\theta = 0°$ or $\theta = 180°$.
Determine whether the following pairs of vectors are perpendicular, parallel, or neither:
a) $\vec{u} = \langle 3, 6 \rangle$ and $\vec{v} = \langle 4, -2 \rangle$
b) $\vec{u} = \langle 2, 5 \rangle$ and $\vec{v} = \langle 6, 15 \rangle$
c) $\vec{u} = \langle 1, 4 \rangle$ and $\vec{v} = \langle 2, 3 \rangle$
Part a) Check the dot product: $$\vec{u} \cdot \vec{v} = (3)(4) + (6)(-2) = 12 - 12 = 0$$
Since the dot product is zero, these vectors are perpendicular.
Part b) Check if one is a scalar multiple of the other: $$\frac{6}{2} = 3 \quad \text{and} \quad \frac{15}{5} = 3$$
Since $\vec{v} = 3\vec{u}$, these vectors are parallel.
Part c) Check the dot product: $$\vec{u} \cdot \vec{v} = (1)(2) + (4)(3) = 2 + 12 = 14 \neq 0$$
Not perpendicular. Check for parallel: $$\frac{2}{1} = 2 \quad \text{but} \quad \frac{3}{4} = 0.75$$
These ratios are different, so the vectors are neither parallel nor perpendicular.
Notation and Terminology
| Symbol/Term | How to Read It | What It Means |
|---|---|---|
| $\vec{v}$ | “vector v” | A quantity with magnitude and direction |
| $\langle a, b \rangle$ | “the vector a, b” | Component form of a vector |
| $|\vec{v}|$ or $|\vec{v}|$ | “the magnitude of v” | The length of the vector |
| $\hat{v}$ | “v hat” | A unit vector (magnitude = 1) in the direction of $\vec{v}$ |
| $\hat{i}$ | “i hat” | The unit vector $\langle 1, 0 \rangle$ |
| $\hat{j}$ | “j hat” | The unit vector $\langle 0, 1 \rangle$ |
| $\vec{u} \cdot \vec{v}$ | “u dot v” | The dot product (scalar product) of two vectors |
| $\theta$ | “theta” | The angle between two vectors |
| $\perp$ | “is perpendicular to” | Vectors meet at a $90°$ angle |
Key Properties and Rules
Magnitude Properties
- $|\vec{v}| \geq 0$ for all vectors (magnitude is never negative)
- $|\vec{v}| = 0$ if and only if $\vec{v} = \langle 0, 0 \rangle$ (the zero vector)
- $|c\vec{v}| = |c| \cdot |\vec{v}|$ (scaling a vector scales its magnitude)
Vector Addition Properties
- Commutative: $\vec{u} + \vec{v} = \vec{v} + \vec{u}$
- Associative: $(\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})$
- Zero vector identity: $\vec{v} + \vec{0} = \vec{v}$
- Additive inverse: $\vec{v} + (-\vec{v}) = \vec{0}$
Scalar Multiplication Properties
- Distributive over vectors: $c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$
- Distributive over scalars: $(c + d)\vec{v} = c\vec{v} + d\vec{v}$
- Associative: $c(d\vec{v}) = (cd)\vec{v}$
- Identity: $1 \cdot \vec{v} = \vec{v}$
Dot Product Properties
- Commutative: $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$
- Distributive: $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$
- Scalar factor: $(c\vec{u}) \cdot \vec{v} = c(\vec{u} \cdot \vec{v})$
- Self dot product: $\vec{v} \cdot \vec{v} = |\vec{v}|^2$
Relationship Formulas
- Magnitude from components: $|\vec{v}| = \sqrt{a^2 + b^2}$
- Direction angle: $\tan \theta = \frac{b}{a}$
- Dot product (algebraic): $\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2$
- Dot product (geometric): $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos \theta$
- Angle between vectors: $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$
- Unit vector: $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$
Real-World Applications
Navigation and Aviation
Pilots and sailors constantly work with vectors. When an airplane flies through the air, its actual path depends on two vectors: the direction and speed the plane is pointing (the airspeed vector) and the wind (another vector). The actual ground velocity is the vector sum of these two. A pilot heading north at 500 mph in a 100 mph crosswind from the west will not end up going due north. Understanding vectors is essential for computing the actual course and determining what heading to fly to reach a destination.
Physics and Forces
Forces are vectors. When you push a heavy box, the force you apply has both a magnitude (how hard you push) and a direction (which way). If two people push on a box from different directions, the box moves in the direction of the combined force, which is the vector sum of the individual forces. Engineers use this constantly when designing structures that must withstand multiple forces from different directions.
Computer Graphics
Video games and animated movies rely heavily on vectors. Every object in a 3D scene has a position vector. Lighting effects depend on the angle between surface normal vectors and light direction vectors (calculated using dot products). Character movement is described by velocity vectors. Even something as simple as determining whether a character is facing toward or away from a light source uses the sign of a dot product.
Sports Analysis
In sports like golf, tennis, or baseball, the trajectory of the ball depends on the velocity vector at the moment of impact. A golf drive has both a horizontal component (how far forward) and a vertical component (how high). Understanding these vector components helps coaches and players optimize their technique. Launch angle and exit velocity are vector concepts.
GPS and Mapping
Your smartphone’s GPS calculates your position and velocity as vectors. When giving directions, it computes the vector from your current location to waypoints along your route. The “distance remaining” is the magnitude of that vector, and the direction to go is the vector’s direction.
Robotics
Robots navigating a warehouse or a Mars rover exploring another planet use vectors to plan their movements. Each movement command is essentially a vector: move this far in this direction. The robot’s position is updated by adding these movement vectors. Collision avoidance involves checking whether velocity vectors will cause the robot’s path to intersect with obstacles.
Self-Test Problems
Problem 1: Find the magnitude of the vector $\vec{v} = \langle -5, 12 \rangle$.
Show Answer
$$|\vec{v}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$
Answer: The magnitude is 13.
Problem 2: Find the vector from point $A = (-1, 4)$ to point $B = (3, -2)$, then find its magnitude.
Show Answer
Step 1: Find the vector: $$\overrightarrow{AB} = \langle 3 - (-1), -2 - 4 \rangle = \langle 4, -6 \rangle$$
Step 2: Find the magnitude: $$|\overrightarrow{AB}| = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$
Answer: The vector is $\langle 4, -6 \rangle$ and its magnitude is $2\sqrt{13} \approx 7.21$.
Problem 3: Let $\vec{u} = \langle 2, -3 \rangle$ and $\vec{v} = \langle -4, 1 \rangle$. Calculate $\vec{u} + \vec{v}$, $\vec{u} - \vec{v}$, and $2\vec{u} + 3\vec{v}$.
Show Answer
Addition: $$\vec{u} + \vec{v} = \langle 2 + (-4), -3 + 1 \rangle = \langle -2, -2 \rangle$$
Subtraction: $$\vec{u} - \vec{v} = \langle 2 - (-4), -3 - 1 \rangle = \langle 6, -4 \rangle$$
Linear combination: $$2\vec{u} + 3\vec{v} = 2\langle 2, -3 \rangle + 3\langle -4, 1 \rangle$$ $$= \langle 4, -6 \rangle + \langle -12, 3 \rangle = \langle -8, -3 \rangle$$
Answers: $\vec{u} + \vec{v} = \langle -2, -2 \rangle$, $\vec{u} - \vec{v} = \langle 6, -4 \rangle$, $2\vec{u} + 3\vec{v} = \langle -8, -3 \rangle$
Problem 4: Find the unit vector in the direction of $\vec{v} = \langle 1, -2 \rangle$.
Show Answer
Step 1: Find the magnitude: $$|\vec{v}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$
Step 2: Divide by the magnitude: $$\hat{v} = \frac{\langle 1, -2 \rangle}{\sqrt{5}} = \left\langle \frac{1}{\sqrt{5}}, \frac{-2}{\sqrt{5}} \right\rangle = \left\langle \frac{\sqrt{5}}{5}, \frac{-2\sqrt{5}}{5} \right\rangle$$
Answer: The unit vector is $\left\langle \frac{\sqrt{5}}{5}, -\frac{2\sqrt{5}}{5} \right\rangle$ (or equivalently, $\left\langle \frac{1}{\sqrt{5}}, \frac{-2}{\sqrt{5}} \right\rangle$).
Problem 5: Find the dot product of $\vec{u} = \langle 5, -2 \rangle$ and $\vec{v} = \langle 3, 7 \rangle$. Then determine if the angle between them is acute, right, or obtuse.
Show Answer
Step 1: Calculate the dot product: $$\vec{u} \cdot \vec{v} = (5)(3) + (-2)(7) = 15 - 14 = 1$$
Step 2: Interpret the result:
- If $\vec{u} \cdot \vec{v} > 0$, the angle is acute (less than $90°$).
- If $\vec{u} \cdot \vec{v} = 0$, the angle is right (exactly $90°$).
- If $\vec{u} \cdot \vec{v} < 0$, the angle is obtuse (greater than $90°$).
Since $\vec{u} \cdot \vec{v} = 1 > 0$, the angle is acute.
Answer: The dot product is 1, and the angle between the vectors is acute.
Problem 6: Find the angle between $\vec{u} = \langle 2, 2 \rangle$ and $\vec{v} = \langle 0, 5 \rangle$.
Show Answer
Step 1: Calculate the dot product: $$\vec{u} \cdot \vec{v} = (2)(0) + (2)(5) = 0 + 10 = 10$$
Step 2: Find the magnitudes: $$|\vec{u}| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$ $$|\vec{v}| = \sqrt{0^2 + 5^2} = \sqrt{25} = 5$$
Step 3: Apply the formula: $$\cos \theta = \frac{10}{2\sqrt{2} \cdot 5} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
Step 4: Find the angle: $$\theta = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = 45°$$
Answer: The angle between the vectors is $45°$ (or $\frac{\pi}{4}$ radians).
Problem 7: Determine whether $\vec{u} = \langle 6, -9 \rangle$ and $\vec{v} = \langle -4, 6 \rangle$ are parallel, perpendicular, or neither.
Show Answer
Method 1: Check if one is a scalar multiple of the other.
$$\frac{-4}{6} = -\frac{2}{3} \quad \text{and} \quad \frac{6}{-9} = -\frac{2}{3}$$
Both ratios are equal, so $\vec{v} = -\frac{2}{3}\vec{u}$.
Method 2: Check the dot product: $$\vec{u} \cdot \vec{v} = (6)(-4) + (-9)(6) = -24 - 54 = -78 \neq 0$$
Since the dot product is not zero, they are not perpendicular. But since one is a scalar multiple of the other, they are parallel (pointing in opposite directions since the scalar is negative).
Answer: The vectors are parallel.
Summary
- A vector is a quantity with both magnitude (length) and direction. Scalars, by contrast, have only magnitude.
- Vectors can be written in component form $\langle a, b \rangle$, with an arrow $\vec{v}$, or in $\hat{i}$-$\hat{j}$ notation $a\hat{i} + b\hat{j}$.
- The magnitude of $\vec{v} = \langle a, b \rangle$ is $|\vec{v}| = \sqrt{a^2 + b^2}$.
- The direction angle is found using $\tan \theta = \frac{b}{a}$, being careful about which quadrant the vector lies in.
- A unit vector has magnitude 1. To find the unit vector in the direction of $\vec{v}$, compute $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$.
- The standard unit vectors are $\hat{i} = \langle 1, 0 \rangle$ and $\hat{j} = \langle 0, 1 \rangle$.
- Vector addition and subtraction work component by component: $\vec{u} + \vec{v} = \langle u_1 + v_1, u_2 + v_2 \rangle$.
- Scalar multiplication scales each component: $c\vec{v} = \langle cv_1, cv_2 \rangle$.
- The dot product $\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2$ gives a scalar, not a vector.
- The dot product also equals $|\vec{u}||\vec{v}|\cos \theta$, connecting algebra to geometry.
- To find the angle between vectors: $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$.
- Vectors are perpendicular if and only if their dot product is zero.
- Vectors are parallel if and only if one is a scalar multiple of the other.
- Vectors appear throughout physics, engineering, computer graphics, navigation, and many other fields where direction matters as much as size.