Introduction to Vectors

Some quantities are described completely by a single number: a mass of $12$ kg, a temperature of $21\,^\circ\text{C}$ . Others also need a direction: “drive $40$ km” isn’t enough to find your way, but “drive $40$ km north” is. Quantities with both a size and a direction are called vectors, and they’re the language of forces, motion, navigation, GPS, and computer graphics.

Key ideas

Scalars and vectors

A scalar has magnitude (size) only. A vector has both magnitude and direction.

Scalar (size only)	Matching vector (size and direction)
distance: $5$ km	displacement: $5$ km east
speed: $90$ km/h	velocity: $90$ km/h on a bearing of $230^\circ$
mass: $50$ kg	weight (a force): $490$ N straight down
time, temperature, area, volume, energy	force, acceleration

A GPS unit uses both: it reports your speed as a single number, but to predict where you’ll be in five minutes it needs your velocity, which includes the direction you’re heading.

Drawing a vector

A vector is drawn as a directed line segment: an arrow. The length of the arrow (to some scale) shows the magnitude, and the way it points shows the direction. The starting point is the tail and the arrowhead end is the head (or tip).

Notation

These are the conventions used on every vector page of this site:

Notation	Meaning
$\vec{v}$ , $\vec{u}$ , $\vec{F}$	a vector named by a letter with an arrow over it
$\overrightarrow{AB}$	the vector from point $A$ (tail) to point $B$ (head)
$\lvert\vec{v}\rvert$ , $\lvert\overrightarrow{AB}\rvert$	the magnitude of the vector, a scalar that is never negative

Some books print vectors in bold ( $\mathbf{v}$ ) instead of using an arrow. Order matters: $\overrightarrow{AB}$ starts at $A$ , while $\overrightarrow{BA}$ starts at $B$ .

Equal and opposite vectors

Two vectors are equal if they have the same magnitude and the same direction. Where they’re drawn doesn’t matter: you can slide a vector anywhere without changing it.

The opposite of $\vec{u}$ , written $-\vec{u}$ , has the same magnitude but points the opposite way. In particular,

\overrightarrow{BA} = -\overrightarrow{AB}

Both blue arrows are the same vector

\vec{u}

. The orange arrow is

-\vec{u}

Two vectors with the same or opposite directions are called parallel (or collinear). You’ll use this idea a lot in scalar multiplication.

Describing a direction

There are three common ways to describe the direction of a vector in a plane:

True bearing: the angle measured clockwise from north, from $0^\circ$ up to (not including) $360^\circ$ . It’s usually written with three digits, like $050^\circ$ or $320^\circ$ .
Quadrant bearing: start facing north or south, then turn some angle (from $0^\circ$ to $90^\circ$ ) toward east or west. “N $40^\circ$ W” means: face north, then turn $40^\circ$ toward the west.
Angle from the positive $x$ -axis: on a coordinate grid, the angle $\theta$ measured counterclockwise from the positive $x$ -axis, as in trigonometry. This is the form you’ll use with Cartesian vectors.

One direction, three descriptions: a true bearing of

320^\circ

, N

40^\circ

W, or

\theta = 130^\circ

from the positive

x

-axis.

To convert, a sketch is the safest tool. Two facts help:

North is $\theta = 90^\circ$ and bearings go the other way round, so $\theta = 90^\circ - \text{bearing}$ (add $360^\circ$ if the result is negative).
The opposite direction is $180^\circ$ away: add or subtract $180^\circ$ from a bearing, or swap N with S and E with W in a quadrant bearing.

Some textbooks write a vector’s direction in square brackets after its magnitude, like 40 km/h [N $40^\circ$ W]. This site writes it in words (” $40$ km/h at N $40^\circ$ W”) so that square brackets can be saved for Cartesian components.

Worked examples

Example 1: Scalar or vector?

Decide whether each quantity is a scalar or a vector.

(a) A car’s speedometer reads $80$ km/h.
(b) A plane flies at $650$ km/h on a bearing of $275^\circ$ .
(c) A bag of flour has a mass of $10$ kg.
(d) You push a box with a force of $30$ N to the east.
(e) A swimmer swims $400$ m in a pool.

Solution.

(a) Scalar: a speed, with no direction.

(b) Vector: a velocity, with magnitude $650$ km/h and a direction.

(d) Vector: a force with a direction.

(e) Scalar: $400$ m is the distance swum. After $8$ lengths of a $50$ m pool, the swimmer’s displacement (change in position) is zero.

Example 2: Converting between direction forms

Write each direction in the other two forms.

(a) a true bearing of $320^\circ$
(b) S $25^\circ$ E
(c) $\theta = 210^\circ$ from the positive $x$ -axis

Solution. Sketch each one on a compass.

(a) $320^\circ$ clockwise from north is $360^\circ - 320^\circ = 40^\circ$ short of north, on the west side: N $40^\circ$ W. From the positive $x$ -axis, $\theta = 90^\circ - 320^\circ = -230^\circ$ , and adding $360^\circ$ gives $\theta = 130^\circ$ . (This is the direction in the figure above.)

(b) Face south (a bearing of $180^\circ$ ) and turn $25^\circ$ toward the east. Turning from south to east is turning counterclockwise on a compass, so the bearing goes down: $180^\circ - 25^\circ = 155^\circ$ . Then $\theta = 90^\circ - 155^\circ = -65^\circ$ , or $\theta = 295^\circ$ .

(c) $\theta = 210^\circ$ points down and to the left, $30^\circ$ below the negative $x$ -axis. The true bearing is $90^\circ - 210^\circ = -120^\circ$ , plus $360^\circ$ , which is $240^\circ$ . That’s $60^\circ$ past south toward the west: S $60^\circ$ W.

	True bearing	Quadrant bearing	Angle from positive $x$ -axis
(a)	$320^\circ$	N $40^\circ$ W	$130^\circ$
(b)	$155^\circ$	S $25^\circ$ E	$295^\circ$
(c)	$240^\circ$	S $60^\circ$ W	$210^\circ$

Example 3: Equal and opposite vectors in a parallelogram

$ABCD$ is a parallelogram (vertices in order around the shape) whose diagonals meet at $E$ . Name

(a) a vector equal to $\overrightarrow{AB}$ ,
(b) a vector equal to $\overrightarrow{AE}$ ,
(c) two vectors opposite to $\overrightarrow{BC}$ .
(d) Is $\overrightarrow{AB} = \overrightarrow{CD}$ ?

Solution. Sketch the parallelogram first.

(a) Opposite sides of a parallelogram are parallel and equal in length. $\overrightarrow{AB}$ and $\overrightarrow{DC}$ also point the same way, so $\overrightarrow{AB} = \overrightarrow{DC}$ .

(b) The diagonals of a parallelogram bisect each other, so $E$ is the midpoint of $AC$ . Then $\overrightarrow{AE} = \overrightarrow{EC}$ : same length, same direction.

(c) $\overrightarrow{CB}$ is opposite to $\overrightarrow{BC}$ . Since $\overrightarrow{AD} = \overrightarrow{BC}$ , its reverse $\overrightarrow{DA}$ is also opposite to $\overrightarrow{BC}$ .

(d) No. They have the same magnitude, but $\overrightarrow{CD}$ points the opposite way to $\overrightarrow{AB}$ . In fact $\overrightarrow{CD} = -\overrightarrow{AB}$ .

Example 4: Distance vs displacement

A hiker walks $3$ km north and then $4$ km east. Find the total distance walked and the hiker’s displacement from the start.

Solution. The distance is a scalar: $3 + 4 = 7$ km.

The displacement is the vector from the start to the finish. The two legs form a right angle, so its magnitude comes from the Pythagorean theorem:

\sqrt{3^2 + 4^2} = \sqrt{25} = 5 \text{ km}

For the direction, let $\theta$ be the angle at the start between north and the displacement. The side opposite $\theta$ is the $4$ km east leg and the adjacent side is the $3$ km north leg:

\tan\theta = \frac{4}{3} \quad\Rightarrow\quad \theta \approx 53.1^\circ

The displacement is $5$ km at N $53.1^\circ$ E (a true bearing of about $053.1^\circ$ ), to one decimal place. Notice that $5 \ne 7$ : the length of the path isn’t the same as how far you end up from where you started. Adding vectors like this is the subject of vector addition and subtraction.

Common mistakes

Treating speed and velocity (or distance and displacement) as the same thing. Speed and distance are scalars. Velocity and displacement are vectors and must include a direction. ” $60$ km/h” is a speed; ” $60$ km/h due west” is a velocity.

Measuring a true bearing from the wrong place or the wrong way. True bearings always start at north and go clockwise. Angles from the positive $x$ -axis start at east and go counterclockwise. A quick sketch prevents mixing them up.

Reading N $40^\circ$ W as W $40^\circ$ N. N $40^\circ$ W starts at north and turns $40^\circ$ toward the west, so it’s closer to north. W $40^\circ$ N would start at west and be closer to west (it equals N $50^\circ$ W). Standard quadrant bearings always start from N or S.

Thinking equal vectors must be in the same place. Vectors are equal when their magnitudes and directions match. Position doesn’t matter, which is why you can slide vectors around when adding them.

Mixing up $\overrightarrow{AB}$ and $\overrightarrow{BA}$ . They have the same length but opposite directions: $\overrightarrow{BA} = -\overrightarrow{AB}$ . The first letter is always the tail.

Giving a negative magnitude. $\lvert\vec{v}\rvert$ is a length, so it’s never negative. A minus sign in $-\vec{v}$ changes the direction, not the size: $\lvert -\vec{v}\rvert = \lvert\vec{v}\rvert$ .

Practice

1. (Warm-up) Scalar or vector?

(a) An acceleration of $9.8$ m/s² straight down
(b) A juice box holding $250$ mL
(c) A wind of $30$ km/h from the northwest
(d) A room temperature of $22\,^\circ\text{C}$

Solution

(a) Vector: it has a direction (down).

(b) Scalar: volume has no direction.

(d) Scalar.

2. (Warm-up) Write each true bearing as a quadrant bearing.

(a) $070^\circ$
(b) $200^\circ$
(c) $315^\circ$

Solution

(a) $70^\circ$ clockwise from north, toward the east: N $70^\circ$ E.

(b) $200^\circ$ is $20^\circ$ past south ( $180^\circ$ ), toward the west: S $20^\circ$ W.

3. (Core) Write each quadrant bearing as a true bearing.

(a) S $35^\circ$ E
(b) N $15^\circ$ W
(c) S $80^\circ$ W

Solution

(a) From south ( $180^\circ$ ), turning toward east reduces the bearing: $180^\circ - 35^\circ = 145^\circ$ .

(b) From north, turning $15^\circ$ toward west: $360^\circ - 15^\circ = 345^\circ$ .

4. (Core) Find the angle $\theta$ , measured counterclockwise from the positive $x$ -axis (with north along the positive $y$ -axis), for each direction.

(a) a true bearing of $030^\circ$
(b) N $50^\circ$ W
(c) a true bearing of $250^\circ$

Solution

Use $\theta = 90^\circ - \text{bearing}$ , adding $360^\circ$ if needed.

(a) $\theta = 90^\circ - 30^\circ = 60^\circ$ .

(b) N $50^\circ$ W is a bearing of $360^\circ - 50^\circ = 310^\circ$ . Then $\theta = 90^\circ - 310^\circ = -220^\circ$ , so $\theta = 140^\circ$ . Check: $140^\circ$ is $50^\circ$ past the positive $y$ -axis (north) toward the negative $x$ -axis (west). ✓

(c) $\theta = 90^\circ - 250^\circ = -160^\circ$ , so $\theta = 200^\circ$ . Check: a bearing of $250^\circ$ is S $70^\circ$ W, which is $20^\circ$ below west, and $200^\circ$ is $20^\circ$ past the negative $x$ -axis. ✓

5. (Core) A ball’s velocity is $\vec{v}$ : $12$ m/s at N $25^\circ$ E. Describe $-\vec{v}$ using a quadrant bearing and a true bearing.

Solution

$-\vec{v}$ has the same magnitude, $12$ m/s, and the opposite direction. Swap N with S and E with W: S $25^\circ$ W.

As a true bearing, $\vec{v}$ is at $025^\circ$ , so $-\vec{v}$ is at $25^\circ + 180^\circ = 205^\circ$ . Check: S $25^\circ$ W is $180^\circ + 25^\circ = 205^\circ$ . ✓

6. (Core) $PQRS$ is a rectangle (vertices in order) whose diagonals meet at $T$ .

(a) Name a vector equal to $\overrightarrow{PQ}$ .
(b) Name a vector equal to $\overrightarrow{PT}$ .
(c) Name two vectors opposite to $\overrightarrow{QR}$ .
(d) Is $\overrightarrow{PR} = \overrightarrow{QS}$ ? Explain.

Solution

(a) $\overrightarrow{SR}$ : opposite sides are parallel and equal, and $S \to R$ points the same way as $P \to Q$ .

(b) $\overrightarrow{TR}$ : the diagonals bisect each other, so $T$ is the midpoint of $PR$ .

(d) No. The diagonals of a rectangle are equal in length, so $\lvert\overrightarrow{PR}\rvert = \lvert\overrightarrow{QS}\rvert$ , but they point in different directions, so the vectors aren’t equal.

7. (Core) A cyclist rides $6$ km west and then $8$ km south. Find the total distance ridden and the cyclist’s displacement, giving the direction as a quadrant bearing and a true bearing to one decimal place.

Solution

Distance: $6 + 8 = 14$ km.

The legs meet at a right angle, so the displacement has magnitude $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ km.

At the start, the angle $\alpha$ between south and the displacement has the $6$ km west leg opposite it and the $8$ km south leg adjacent:

\tan\alpha = \frac{6}{8} \quad\Rightarrow\quad \alpha \approx 36.9^\circ

The displacement is $10$ km at S $36.9^\circ$ W, which is a true bearing of $180^\circ + 36.9^\circ = 216.9^\circ$ .

8. (Challenge) Show that these three vectors are all equal:

$\vec{a}$ : $50$ km/h on a true bearing of $110^\circ$
$\vec{b}$ : $50$ km/h at S $70^\circ$ E
$\vec{c}$ : $50$ km/h at $\theta = -20^\circ$ from the positive $x$ -axis

Solution

All three have magnitude $50$ km/h, so compare the directions as true bearings.

$\vec{b}$ : from south ( $180^\circ$ ), turn $70^\circ$ toward east: $180^\circ - 70^\circ = 110^\circ$ .

$\vec{c}$ : bearing $= 90^\circ - \theta = 90^\circ - (-20^\circ) = 110^\circ$ .

All three point on a bearing of $110^\circ$ with the same magnitude, so $\vec{a} = \vec{b} = \vec{c}$ .

9. (Challenge) A lighthouse keeper reports a ship $12$ km from the lighthouse on a bearing of $300^\circ$ .

(a) What is the bearing of the lighthouse from the ship?
(b) Explain why “the ship is $12$ km from the lighthouse” alone doesn’t tell you where the ship is.

Solution

(a) The displacement from the ship to the lighthouse is the opposite of the displacement from the lighthouse to the ship. So turn around by $180^\circ$ (adding $180^\circ$ would pass $360^\circ$ , so subtract instead): $300^\circ - 180^\circ = 120^\circ$ . The lighthouse is on a bearing of $120^\circ$ (S $60^\circ$ E) from the ship.

(b) ” $12$ km away” gives only a magnitude. Every point on a circle of radius $12$ km centred at the lighthouse is $12$ km away. The direction (the bearing of $300^\circ$ ) picks out one point on that circle, which is why position needs a vector, not just a distance.