Alright everyone, welcome back. In our first lecture, we learned to locate specific points in 3D space. Today, we're going to introduce a powerful new concept to describe not just where something is, but also where it's going and how fast. This is the idea of a vector.
A vector is a quantity that has both magnitude (length or size) and direction. Think of the velocity of a spacecraft or the force of gravity—these can't be described by a single number. This is what makes vectors so useful in physics, engineering, and of course, calculus.
So, what is a vector? Geometrically, you can think of it as an arrow, or a directed line segment. It has an initial point (where it starts) and a terminal point (where it ends).
Now, here’s the most important conceptual leap you need to make today. If you look at the graph above, you'll see several different arrows. Notice that vector $\mathbf{v}$ and vector $\mathbf{w}$ are equivalent. Why? Because they represent the same magnitude and direction, even though they are located in different parts of the plane. A vector is defined only by its length and direction, not by its specific starting location.
Since the starting point doesn't matter, we can make our lives much easier by agreeing to always start our vectors at the origin. When a vector's initial point is at the origin, we say it is in standard position. Now, the vector is completely defined by its terminal point. This is where the connection between points and vectors becomes clear. While the point $(v_1, v_2)$ is a fixed location, the vector $\langle v_1, v_2 \rangle$ represents the displacement, or the journey, from the origin to that point. We'll be going back and forth between $\mathbb{R}^2$ and $\mathbb{R}^3$ a lot in this course, so you should get comfortable thinking about vectors in both spaces.
Now that we have a solid idea of what a vector is, let's see how we can combine them using algebra. The rules for adding vectors and multiplying them by a number are very intuitive.
Let $\mathbf{a} = \langle 2, 0, -1 \rangle$ and $\mathbf{b} = \langle 1, 5, 3 \rangle$. Let's find $\mathbf{a} + \mathbf{b}$ and $-3\mathbf{a}$.
Vector Addition:
$$ \mathbf{a} + \mathbf{b} = \langle 2+1, 0+5, -1+3 \rangle = \langle 3, 5, 2 \rangle $$
Scalar Multiplication:
$$ -3\mathbf{a} = \langle -3(2), -3(0), -3(-1) \rangle = \langle -6, 0, 3 \rangle $$
To add two vectors, you simply add their corresponding components. Geometrically, this is the same as placing the vectors tail-to-tip. The resulting sum, called the resultant vector, is the vector from the first tail to the last tip. You can also see it as the diagonal of the parallelogram formed by the two vectors.
To multiply a vector by a scalar (a real number) $c$, you just multiply each component by that number. Multiplying by a positive scalar $c > 1$ stretches the vector. Multiplying by a scalar $0 < c < 1$ shrinks it. And as you might guess, multiplying by a negative scalar like -1 reverses the vector's direction completely.
Let's start with an intuitive example. Suppose you are standing at point $A(1, 2)$ and someone gives you the instructions "go 3 units east (positive x-direction) and 4 units north (positive y-direction)". Where do you end up? You would simply add these displacements to your starting coordinates to find your ending point, $B(1+3, 2+4) = B(4, 6)$.
Now, let's reverse the question. If you start at $A(1, 2)$ and end at $B(4, 6)$, what were the instructions? You would subtract the starting point from the ending point: $\langle 4-1, 6-2 \rangle = \langle 3, 4 \rangle$. This is exactly how we find the component form of a vector between any two points.
The vector $\vec{AB}$ from an initial point $A(x_1, y_1)$ to a terminal point $B(x_2, y_2)$ is:
$$ \vec{AB} = \langle x_2 - x_1, y_2 - y_1 \rangle $$Let's find the vector from point $A(1, 2)$ to point $B(5, 5)$.
We use the formula by subtracting the components of the initial point (A) from the terminal point (B):
$$ \vec{AB} = \langle 5 - 1, 5 - 2 \rangle = \langle 4, 3 \rangle $$
This means the journey from A to B is equivalent to a journey of 4 units in the x-direction and 3 units in the y-direction from the origin.
Problem: Find the component form of the vector with initial point $P(-3, 4)$ and terminal point $Q(2, -1)$.
We've talked a lot about a vector's two key properties: direction and magnitude. We know the components give us the direction. To find the magnitude, or length, we just use the distance formula.
Find the magnitude of the vector from $A(1, 0, 2)$ to $B(4, 4, 3)$.
First, we need to find the component form of the vector in standard position:
$$ \vec{AB} = \langle 4-1, 4-0, 3-2 \rangle = \langle 3, 4, 1 \rangle $$
Now, we find the magnitude of this component vector using the distance formula from the origin:
$$ |\vec{AB}| = \sqrt{3^2 + 4^2 + 1^2} = \sqrt{9 + 16 + 1} = \sqrt{26} $$
Problem: Find the magnitude of the vector from $P(-1, 5)$ to $Q(3, 2)$.
To analyze vectors in a systematic way, it's helpful to use a standard reference frame, which is our familiar $xy$ or $xyz$ coordinate system. We can define a set of fundamental "direction vectors," which are unit vectors that point along each of the positive coordinate axes. We call these special vectors $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$. Using these, we can represent any vector as a sum of these basic directions.
Notice that $\mathbf{i}$ in $\mathbb{R}^2$ is not the same as $\mathbf{i}$ in $\mathbb{R}^3$. One is a two dimensional object, and the other is a three dimensional object. However, any vector can be written as a sum of scaled versions of these basis vectors. This is called a linear combination.
In $\mathbb{R}^2$: Let's break down the vector $\mathbf{v} = \langle -3, 4 \rangle$.
First, we can split it into a sum of two vectors, one for each component:
$$ \langle -3, 4 \rangle = \langle -3, 0 \rangle + \langle 0, 4 \rangle $$
Next, we can factor out the scalar from each component vector:
$$ = -3 \langle 1, 0 \rangle + 4 \langle 0, 1 \rangle $$
Finally, we replace the component vectors with the standard basis vectors $\mathbf{i}$ and $\mathbf{j}$:
$$ = -3\mathbf{i} + 4\mathbf{j} $$
In $\mathbb{R}^3$: The same logic applies. Let's decompose $\mathbf{w} = \langle 2, -1, 5 \rangle$.
$$ \langle 2, -1, 5 \rangle = \langle 2, 0, 0 \rangle + \langle 0, -1, 0 \rangle + \langle 0, 0, 5 \rangle $$
$$ = 2 \langle 1, 0, 0 \rangle + (-1) \langle 0, 1, 0 \rangle + 5 \langle 0, 0, 1 \rangle $$
$$ = 2\mathbf{i} - \mathbf{j} + 5\mathbf{k} $$
Sometimes, we are only interested in a vector's direction, not its length. For this, we use a unit vector, which is any vector with a magnitude of exactly 1. We will use them frequently in later chapters to define directions for derivatives and integrals.
To find a unit vector $\mathbf{u}$ in the same direction as a given vector $\mathbf{v}$, we simply divide the vector $\mathbf{v}$ by its own magnitude. This process is called normalizing the vector.
Let's find a unit vector in the direction of the vector $\mathbf{v} = \langle 3, 4 \rangle$.
First, we need to find the magnitude of $\mathbf{v}$:
$$ |\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$
Now that we have the magnitude, we divide our original vector by it:
$$ \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{\langle 3, 4 \rangle}{5} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle $$
The vector $\mathbf{u} = \langle 3/5, 4/5 \rangle$ points in the same direction as $\mathbf{v}$ but has a length of 1.
Problem: Find a unit vector in the direction of $\mathbf{v} = \langle 5, -5 \rangle$.
Find the vector that has the same direction as $\mathbf{v} = \langle 6, 2, -3 \rangle$ but has a length of 4.
Step 1: Find the magnitude of the original vector. This tells us its current length.
$$ |\mathbf{v}| = \sqrt{6^2 + 2^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 $$
Step 2: Find the unit vector. We find the "pure direction" by dividing the vector by its length.
$$ \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{\langle 6, 2, -3 \rangle}{7} = \left\langle \frac{6}{7}, \frac{2}{7}, -\frac{3}{7} \right\rangle $$
Step 3: Scale the unit vector to the desired length. Now we take this unit vector (which has length 1) and multiply it by our target length of 4.
$$ \mathbf{w} = 4\mathbf{u} = 4 \left\langle \frac{6}{7}, \frac{2}{7}, -\frac{3}{7} \right\rangle = \left\langle \frac{24}{7}, \frac{8}{7}, -\frac{12}{7} \right\rangle $$
The vector $\mathbf{w}$ is our final answer. It points in the same direction as $\mathbf{v}$ but has a length of 4.
One of the most powerful applications of vectors is breaking them down into components that align with a coordinate system. This is especially useful in physics and engineering. By using the standard basis vectors, we can decompose any force vector into its horizontal and vertical components.
If a child pulls a sled through the snow on a level path with a force of 50 Newtons exerted at an angle of 38 degrees above the horizontal, find the horizontal and vertical components of the force.
Step 1: Visualize the problem. We can draw the force vector $\mathbf{F}$ starting at the origin. It has a magnitude of $|\mathbf{F}| = 50$ and makes an angle of $\theta = 38^\circ$ with the positive x-axis. The horizontal component ($F_x$) is the adjacent side of this right triangle, and the vertical component ($F_y$) is the opposite side.
Step 2: Use trigonometry to find the components. The horizontal component is the magnitude times the cosine of the angle: $$ F_x = |\mathbf{F}| \cos(\theta) = 50 \cos(38^\circ) $$ The vertical component is the magnitude times the sine of the angle: $$ F_y = |\mathbf{F}| \sin(\theta) = 50 \sin(38^\circ) $$
Step 3: Calculate the values. Using a calculator, we find: $$ F_x \approx 50(0.7880) \approx 39.4 \text{ N} $$ $$ F_y \approx 50(0.6157) \approx 30.8 \text{ N} $$
Conclusion: The horizontal component of the force is about 39.4 N, and the vertical component is about 30.8 N. We can write the force vector as $\mathbf{F} \approx \langle 39.4, 30.8 \rangle$ or $\mathbf{F} \approx 39.4\mathbf{i} + 30.8\mathbf{j}$.
Problem: A quarterback throws a football with an angle of elevation of 40° and a speed of 60 ft/s. Find the horizontal and vertical components of the velocity vector.
So today we've taken a big step into the world of multivariable calculus. Vectors give us a brand new language for describing quantities that have both magnitude and direction. We've learned how to represent them, how to do basic algebra with them, and how to find their length. Mastering these skills is essential, because everything we do from here on out—from describing motion to calculating forces and fields—will be built on this foundation. Next time, we will explore a new way to multiply vectors: the dot product. 🚀