1. What a vector represents
A quantity carrying two pieces of information at once: a magnitude (how big) and a direction (which way). Visually, an arrow drawn in the plane. The arrow's length encodes magnitude; its orientation encodes direction.
Compare these two statements:
- "The car's speed is 60 mph." — a scalar. One number, no direction. Just a magnitude on a measuring stick.
- "The car's velocity is 60 mph due east." — a vector. The same magnitude, plus an arrow telling you where the car is going.
Both statements are useful, but they're not interchangeable. If two cars each travel at 60 mph for an hour, they end up 120 miles apart only if they go in opposite directions; if they go the same direction, they stay side-by-side. Direction matters, and the vector is the object that carries it.
Notation: vectors are usually written in bold ($\mathbf{v}$) in print, or with an arrow ($\vec{v}$) when you're handwriting. Their magnitude is written with bars: $|\mathbf{v}|$ or sometimes just $v$ in italic.
A vector encodes only magnitude and direction — not where the arrow is drawn. Two arrows of the same length and same orientation, drawn anywhere on the page, represent the same vector. This is why we can slide arrows around freely when we add them.
2. Two representations of the same arrow
Every 2-D vector can be described in two equivalent ways, and the entire chapter is about moving fluently between them.
| Form | Symbol | What it tells you | Natural for |
|---|---|---|---|
| Magnitude–direction | $(|\mathbf{v}|, \theta)$ | How long the arrow is and which way it points | "A plane flies 500 knots at heading 045°" |
| Component | $\langle v_x, v_y \rangle$ | How far the arrow reaches horizontally and vertically | Adding vectors; passing them to a computer; physics calculations |
Notation note: components are often written in angle brackets $\langle v_x, v_y \rangle$ to distinguish them from a point $(x, y)$. But you'll see them written as $(v_x, v_y)$ too — context decides.
The two forms describe the same arrow, and trigonometry is the dictionary between them. The bridge is a right triangle whose hypotenuse is the vector itself.
3. Decomposing into components
Suppose you're given a vector in magnitude–direction form: $|\mathbf{v}| = 10$, $\theta = 35°$ measured counterclockwise from the positive $x$-axis. You want the components. Drop a perpendicular from the tip of the arrow to the $x$-axis. You've made a right triangle with hypotenuse $|\mathbf{v}|$ and angle $\theta$ at the tail. The two legs are precisely the components.
By the definitions of sine and cosine:
$$ v_x = |\mathbf{v}| \cos \theta \qquad v_y = |\mathbf{v}| \sin \theta $$That's the whole story. Magnitude times cosine gives the horizontal component; magnitude times sine gives the vertical component. The reason it works is the unit circle — sine and cosine were defined exactly so this would work.
Plugging in our numbers:
$$ v_x = 10 \cos 35° \approx 8.19 \qquad v_y = 10 \sin 35° \approx 5.74 $$So the vector is $\langle 8.19,\; 5.74 \rangle$ in component form.
The formulas $v_x = |\mathbf{v}|\cos\theta$ and $v_y = |\mathbf{v}|\sin\theta$ work for any $\theta$, not just acute angles. If $\theta = 135°$, then $\cos 135° < 0$ and the $x$-component comes out negative — exactly right, because that vector points into the second quadrant. You don't have to think about signs; trig handles it.
4. Reassembling: magnitude and angle from components
Going the other direction is just as clean. Given $\langle v_x, v_y \rangle$, you want $|\mathbf{v}|$ and $\theta$.
The magnitude is a straight application of the Pythagorean theorem on that same right triangle:
$$ |\mathbf{v}| = \sqrt{v_x^{\,2} + v_y^{\,2}} $$The angle is where things get subtle. The naive formula is
$$ \theta = \arctan\!\left(\frac{v_y}{v_x}\right) $$and for vectors in the first quadrant ($v_x > 0$, $v_y > 0$) this works fine. But $\arctan$ only returns angles in $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ — its range is one half of the plane. If your vector lives in the second or third quadrant, $\arctan(v_y / v_x)$ silently gives you the wrong answer by $180°$, because $\arctan$ can't tell apart $\langle -3, -4 \rangle$ from $\langle 3, 4 \rangle$ — both give the same ratio $v_y/v_x$.
Enter atan2
Every programming language and scientific calculator has a function called atan2(y, x) that takes both components and returns the angle in the correct quadrant — over the full range $(-\pi, \pi]$. It works by looking at the signs of $v_x$ and $v_y$ separately:
| Quadrant | $v_x$ | $v_y$ | Angle returned |
|---|---|---|---|
| I | + | + | between $0$ and $\pi/2$ |
| II | − | + | between $\pi/2$ and $\pi$ |
| III | − | − | between $-\pi$ and $-\pi/2$ |
| IV | + | − | between $-\pi/2$ and $0$ |
So the safe formula — the one to remember — is:
$$ \theta = \operatorname{atan2}(v_y,\; v_x) $$Whenever you're recovering an angle from components, reach for atan2, not arctan. The single time arctan is safe is when you've already confirmed by other means that the vector is in the right half-plane.
$\langle -3, -4 \rangle$ points into the third quadrant, around $-127°$ from the positive $x$-axis. But $\arctan(-4/-3) = \arctan(4/3) \approx 53°$ — pointing into the first quadrant. The $\arctan$ function literally cannot tell those two vectors apart. atan2(-4, -3) returns about $-2.21$ rad ($\approx -127°$), which is what you actually want.
5. Vector addition
Adding two vectors $\mathbf{u}$ and $\mathbf{v}$ has a beautiful geometric picture: head-to-tail. Draw $\mathbf{u}$ anywhere. Pick up $\mathbf{v}$ (remember, position doesn't matter) and slide it so its tail is at $\mathbf{u}$'s head. The sum $\mathbf{u} + \mathbf{v}$ is the new arrow from $\mathbf{u}$'s original tail to $\mathbf{v}$'s new head.
That's the picture. The algebra is even simpler — vectors add componentwise:
$$ \mathbf{u} + \mathbf{v} = \langle u_x + v_x,\; u_y + v_y \rangle $$If $\mathbf{u} = \langle 3, 2 \rangle$ and $\mathbf{v} = \langle 1, 4 \rangle$, then $\mathbf{u} + \mathbf{v} = \langle 4, 6 \rangle$. Done. No trigonometry needed at the addition step — that's the payoff of working in components.
The horizontal trip you make following $\mathbf{u}$ then $\mathbf{v}$ is $u_x + v_x$. The vertical trip is $u_y + v_y$. The combined trip — the arrow from where you started to where you ended — therefore has those exact components. The two descriptions are literally describing the same journey.
To subtract, just flip the second vector and add: $\mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}) = \langle u_x - v_x,\; u_y - v_y \rangle$.
6. Scalar multiplication
Multiplying a vector by a number (a scalar) scales its length. If $c$ is a positive real number,
$$ c\,\mathbf{v} = \langle c\,v_x,\; c\,v_y \rangle $$has the same direction as $\mathbf{v}$ but $c$ times the magnitude. So $2\mathbf{v}$ points the same way and is twice as long; $\tfrac{1}{2}\mathbf{v}$ points the same way and is half as long.
If $c$ is negative, the vector reverses direction (and scales by $|c|$). The vector $-\mathbf{v}$ has the same magnitude as $\mathbf{v}$ but points the opposite way. This is what makes subtraction work.
And the magnitude scales obediently:
$$ |c\,\mathbf{v}| = |c|\,|\mathbf{v}| $$Note the absolute value on $c$ — magnitude is never negative, even when the scalar is.
7. Unit vectors and the standard basis
A vector of magnitude exactly $1$. Pure direction, no scale. Often written with a hat: $\hat{\mathbf{v}}$.
To turn any nonzero vector into a unit vector pointing the same way, divide by its own magnitude:
$$ \hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} = \left\langle \frac{v_x}{|\mathbf{v}|},\; \frac{v_y}{|\mathbf{v}|} \right\rangle $$This operation is called normalizing. The result has the same direction as $\mathbf{v}$ but magnitude $1$. Useful whenever you want to talk about direction without dragging magnitude along — for example, "which way is the gradient pointing?" or "what direction is the camera facing?"
The standard basis: $\mathbf{i}$ and $\mathbf{j}$
Two particular unit vectors get reserved names:
$$ \mathbf{i} = \langle 1, 0 \rangle \qquad \mathbf{j} = \langle 0, 1 \rangle $$$\mathbf{i}$ points one unit along the positive $x$-axis; $\mathbf{j}$ points one unit along the positive $y$-axis. Together they form the standard basis for the 2-D plane.
Any vector can be written as a sum of multiples of these two:
$$ \langle v_x, v_y \rangle = v_x\,\mathbf{i} + v_y\,\mathbf{j} $$So $\langle 3, 4 \rangle$ is the same as $3\mathbf{i} + 4\mathbf{j}$. The two notations carry identical information; physics texts and engineering texts tend to favor the $\mathbf{i}, \mathbf{j}$ form, mathematics texts the angle-bracket form. Pick whichever your context expects.
The fact that every 2-D vector is a unique combination $v_x\mathbf{i} + v_y\mathbf{j}$ is your first taste of a deep idea: $\mathbf{i}$ and $\mathbf{j}$ are a basis, and the components $(v_x, v_y)$ are coordinates with respect to that basis. Pick a different pair of independent vectors and you get different coordinates for the same arrow — the geometry is real, the numbers depend on your choice of basis. This idea is the entire spine of linear algebra.
8. Applications: navigation, forces, motion
Navigation: airspeed plus wind
A plane's airspeed vector is how it moves relative to the air around it. The wind vector is how the air moves relative to the ground. The plane's ground velocity — what you'd measure with GPS — is the sum:
$$ \mathbf{v}_{\text{ground}} = \mathbf{v}_{\text{air}} + \mathbf{v}_{\text{wind}} $$A 500-knot plane heading due east into a 50-knot headwind from the east has ground speed only $450$ knots. The same plane with a $50$-knot tailwind from the west has ground speed $550$ knots. A crosswind tilts the actual ground track away from the plane's heading — which is why pilots have to crab into the wind to fly a straight ground course.
Statics: forces in equilibrium
An object sitting still has many forces acting on it, but the vector sum of all of them is zero:
$$ \sum \mathbf{F}_i = \mathbf{0} $$This is a vector equation, which means it's really two scalar equations: $\sum F_{ix} = 0$ and $\sum F_{iy} = 0$. To solve it you decompose every force into components, then sum the $x$'s and the $y$'s separately. Trigonometry shows up because each force is usually given in magnitude–direction form ("the rope pulls with $50$ N at $30°$ above horizontal").
Displacement vs velocity
Displacement is a vector from where you started to where you ended — it cares about the straight-line difference, not the path. Velocity is the time derivative of displacement, also a vector. Speed is the magnitude of velocity; distance traveled along a curvy path is generally not the magnitude of displacement. Mixing them up is the source of many physics blunders, especially when an object reverses direction.
9. Common pitfalls
Asking "is this vector positive or negative?" is a category error. Vectors have direction, not sign. A component can be negative (meaning that component points along the negative axis), but the vector itself isn't $\pm$ — it points somewhere, and "somewhere" is richer information than a single bit.
arctan instead of atan2
The single most common quadrant bug. $\arctan(v_y/v_x)$ throws away the individual signs of $v_x$ and $v_y$ — it only sees the ratio. Vectors in quadrants II and III come back as quadrant IV and I respectively, off by $180°$. Always use atan2(v_y, v_x) unless you've explicitly handled the quadrant yourself.
"Wind from the north" means the wind is moving southward — the arrow points away from north. "Heading 045°" (in aviation) means the plane is moving toward the northeast — the arrow points toward $045°$. Meteorological conventions report wind by where it comes from; aviation reports heading by where the nose is pointed toward. Mixing these conventions is a classic word-problem disaster. Sketch the arrow before you compute anything.
Math measures angles counterclockwise from the positive $x$-axis. Navigation often measures clockwise from north. Compass bearings are different from heading angles, which are different from polar angles. Before you plug into $|\mathbf{v}|\cos\theta$ and $|\mathbf{v}|\sin\theta$, convert your angle into the math convention. Doing trig on a navigation bearing without translating first gives nonsense.
If two forces of $3$ N and $4$ N pull on an object, the resultant magnitude is $7$ N only if they're parallel. If they're perpendicular it's $5$ N. If they're antiparallel it's $1$ N. You can never reduce "add the magnitudes" to a one-liner — you have to add the vectors, then take the magnitude of the sum.
10. Worked examples
Try each one before opening the solution. Sketching the arrow first is half the work.
Example 1 · A 50 mph wind blowing from N30°E — find its components
Setup. "From N30°E" means the wind is coming from a direction $30°$ east of due north, so it is moving toward the opposite direction: $30°$ west of due south, i.e. S30°W. We want to write this as a vector in standard math convention (counterclockwise from positive $x$-axis, with $+x$ = east, $+y$ = north).
Step 1 · Convert to a math angle. S30°W points into the third quadrant. Measured counterclockwise from $+x$ (east), it's $180° + 60° = 240°$.
Step 2 · Decompose. With $|\mathbf{v}| = 50$ and $\theta = 240°$:
$$ v_x = 50 \cos 240° = 50 \cdot \left(-\tfrac{1}{2}\right) = -25 $$ $$ v_y = 50 \sin 240° = 50 \cdot \left(-\tfrac{\sqrt{3}}{2}\right) \approx -43.30 $$Result. The wind vector is $\langle -25,\; -43.30 \rangle$ mph: $25$ mph westward and about $43$ mph southward. Sign check: westward and southward — yes, that's S30°W. ✓
Example 2 · Add two displacement vectors
You walk $\mathbf{a} = \langle 4,\; 3 \rangle$ blocks (4 east, 3 north), then $\mathbf{b} = \langle -1,\; 5 \rangle$ blocks (1 west, 5 north). What is your total displacement from the start?
Step 1 · Add componentwise.
$$ \mathbf{a} + \mathbf{b} = \langle 4 + (-1),\; 3 + 5 \rangle = \langle 3,\; 8 \rangle $$Step 2 · Magnitude.
$$ |\mathbf{a} + \mathbf{b}| = \sqrt{3^2 + 8^2} = \sqrt{73} \approx 8.54 \text{ blocks} $$Step 3 · Direction. Both components positive, so quadrant I; atan2 and arctan agree here.
You ended up $\sqrt{73} \approx 8.54$ blocks from start, at about $69°$ north of east.
Example 3 · Resultant of two forces at 60° to each other
Two ropes pull on a ring. Rope A pulls with $30$ N due east. Rope B pulls with $40$ N at $60°$ north of east. What is the net force?
Step 1 · Components of A. Along the $+x$-axis, so
$$ \mathbf{A} = \langle 30,\; 0 \rangle $$Step 2 · Components of B. Magnitude $40$, angle $60°$ from $+x$:
$$ B_x = 40 \cos 60° = 20 \qquad B_y = 40 \sin 60° \approx 34.64 $$ $$ \mathbf{B} = \langle 20,\; 34.64 \rangle $$Step 3 · Add.
$$ \mathbf{R} = \mathbf{A} + \mathbf{B} = \langle 50,\; 34.64 \rangle $$Step 4 · Magnitude and angle of R.
$$ |\mathbf{R}| = \sqrt{50^2 + 34.64^2} \approx 60.83 \text{ N} $$ $$ \theta_{\mathbf{R}} = \operatorname{atan2}(34.64,\; 50) \approx 34.7° $$The two ropes together pull with about $60.8$ N at $34.7°$ north of east. Notice the resultant magnitude $(60.8)$ is less than $30 + 40 = 70$ — the ropes aren't perfectly aligned, so some of their pull cancels.
Example 4 · Airplane heading vs ground track in a crosswind
A plane points due north with airspeed $200$ knots. A steady wind blows from the west at $30$ knots (meaning the air is moving eastward at $30$ knots). What's the plane's ground velocity?
Step 1 · Airspeed vector. Due north, magnitude $200$:
$$ \mathbf{v}_{\text{air}} = \langle 0,\; 200 \rangle $$Step 2 · Wind vector. Moving eastward at $30$:
$$ \mathbf{v}_{\text{wind}} = \langle 30,\; 0 \rangle $$Step 3 · Ground velocity.
$$ \mathbf{v}_{\text{ground}} = \langle 0 + 30,\; 200 + 0 \rangle = \langle 30,\; 200 \rangle $$Step 4 · Ground speed and track.
$$ |\mathbf{v}_{\text{ground}}| = \sqrt{30^2 + 200^2} \approx 202.24 \text{ knots} $$ $$ \theta = \operatorname{atan2}(200,\; 30) \approx 81.5° \text{ from } +x \text{ (east)} $$That's about $8.5°$ east of due north. The plane is being pushed sideways by the crosswind — its nose points north, but it's actually moving slightly to the northeast. To keep a true northward ground track, the pilot would need to crab: aim the nose slightly west of north so the wind blows the plane straight onto the desired course.
Example 5 · Recovering a vector with a quadrant trap
A force has components $\langle -6,\; -8 \rangle$ N. Find its magnitude and direction.
Step 1 · Magnitude.
$$ |\mathbf{F}| = \sqrt{(-6)^2 + (-8)^2} = \sqrt{100} = 10 \text{ N} $$Step 2 · Angle — the careful way. Both components negative, so the vector points into quadrant III. Using atan2:
(Equivalently $+233.13°$ if you prefer angles in $[0°, 360°)$.)
Why not just arctan? Watch what happens:
That's a quadrant-I angle — the answer for $\langle 6, 8 \rangle$, not for $\langle -6, -8 \rangle$. The two vectors point in opposite directions but have the same $v_y/v_x$ ratio, and $\arctan$ alone can't tell them apart. To rescue $\arctan$ you'd manually add $180°$ when both components are negative. atan2 does that bookkeeping for you.