Math Core

Some quantities are described completely by a single number: a temperature of $72^\circ$ , a mass of $5$ kilograms. Others need a direction too. A wind blowing $20$ miles per hour from the west is very different from one blowing $20$ miles per hour from the north. A vector packages a size and a direction together, and trigonometry is the tool that lets you move between the two descriptions.

What a vector is

Draw an arrow from a point $P$ to a point $Q$ . The arrow has a length and it points somewhere. That is all a vector is: a quantity with magnitude (length) and direction. The starting point $P$ is the initial point and the ending point $Q$ is the terminal point. You write the vector as $\overrightarrow{PQ}$ , or give it a bold or arrowed name like $\mathbf{v}$ or $\vec{v}$ .

Two vectors are equal when they have the same magnitude and the same direction, even if they start in different places. Sliding an arrow around the plane without turning or stretching it does not change the vector.

The vector from P(1, 1) to Q(4, 5) moves 3 units right and 4 units up, so v = ⟨3, 4⟩.Open in grapher →

Component form

Because the location of the arrow does not matter, you can describe a vector by how far it moves horizontally and vertically. Those two numbers are its components.

Definition

Component form

If a vector has initial point $P(x_1, y_1)$ and terminal point $Q(x_2, y_2)$ , its component form is

\overrightarrow{PQ} = \langle x_2 - x_1,\; y_2 - y_1 \rangle.

A vector $\mathbf{v} = \langle a, b \rangle$ drawn from the origin ends at the point $(a, b)$ . This is the vector's standard position.

The angle brackets $\langle \; \rangle$ remind you that $\langle 3, 4 \rangle$ is a vector (a movement), not the point $(3, 4)$ .

Magnitude and direction angle

The components form the legs of a right triangle, and the vector is its hypotenuse. So the Pythagorean theorem gives the length, and right-triangle trig gives the angle.

Magnitude and direction

For $\mathbf{v} = \langle a, b \rangle$ :

\|\mathbf{v}\| = \sqrt{a^2 + b^2} \qquad \tan\theta = \frac{b}{a}

where the direction angle $\theta$ is measured counterclockwise from the positive $x$ -axis. Going the other way, a vector with magnitude $\|\mathbf{v}\|$ and direction angle $\theta$ has components

\mathbf{v} = \langle \|\mathbf{v}\|\cos\theta,\; \|\mathbf{v}\|\sin\theta \rangle.

The second formula is just the unit circle scaled up: a point at angle $\theta$ on a circle of radius $r$ is $(r\cos\theta, r\sin\theta)$ .

Common mistake

Your calculator's $\tan^{-1}$ only returns angles between $-90^\circ$ and $90^\circ$ . For a vector in Quadrant II or III, add $180^\circ$ to the calculator's answer; for Quadrant IV, add $360^\circ$ if you want an angle from $0^\circ$ to $360^\circ$ . Always sketch the vector first so you know which quadrant it points into.

Worked example: From two points to magnitude and direction

Find the component form, magnitude and direction angle of the vector from $P(5, 2)$ to $Q(1, 6)$ .

Components. $\langle 1 - 5,\; 6 - 2 \rangle = \langle -4, 4 \rangle$ .

Magnitude. $\sqrt{(-4)^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \approx 5.66$ .

Direction. $\tan\theta = \dfrac{4}{-4} = -1$ . The calculator gives $-45^\circ$ , but the vector points left and up, into Quadrant II. So $\theta = -45^\circ + 180^\circ = 135^\circ$ .

Worked example: From magnitude and direction to components

A hiker walks $8$ km on a heading that makes a $210^\circ$ angle with the positive $x$ -axis (east). Write the displacement in component form.

\mathbf{v} = \langle 8\cos 210^\circ,\; 8\sin 210^\circ \rangle = \left\langle 8\left(-\tfrac{\sqrt{3}}{2}\right),\; 8\left(-\tfrac{1}{2}\right) \right\rangle = \langle -4\sqrt{3}, -4 \rangle.

The hiker ends about $6.93$ km west and $4$ km south of the start.

Adding vectors and multiplying by scalars

To add two vectors geometrically, place them tip to tail: start the second where the first ends. The sum, called the resultant, runs from the start of the first to the end of the second. In components, you just add matching parts.

Tip to tail: u = ⟨3, 1⟩ followed by v = ⟨1, 3⟩ gives the resultant u + v = ⟨4, 4⟩ (dashed).Open in grapher →

Multiplying a vector by a real number $k$ (a scalar) stretches it by a factor of $|k|$ . If $k$ is negative, the vector also flips to point the opposite way.

\langle a, b \rangle + \langle c, d \rangle = \langle a + c,\; b + d \rangle \qquad k\langle a, b \rangle = \langle ka,\; kb \rangle

Subtraction works the same way: $\mathbf{u} - \mathbf{v} = \mathbf{u} + (-1)\mathbf{v}$ .

Unit vectors and i, j notation

A unit vector has magnitude $1$ . To get the unit vector pointing the same way as $\mathbf{v}$ , divide $\mathbf{v}$ by its own magnitude: $\dfrac{\mathbf{v}}{\|\mathbf{v}\|}$ .

Two unit vectors are special: $\mathbf{i} = \langle 1, 0 \rangle$ points right and $\mathbf{j} = \langle 0, 1 \rangle$ points up. Any vector can be written using them: $\langle a, b \rangle = a\mathbf{i} + b\mathbf{j}$ . So $\langle 5, -2 \rangle$ and $5\mathbf{i} - 2\mathbf{j}$ are the same vector.

Worked example: Finding a resultant force

Two ropes pull on a crate. One pulls with $60$ newtons at $0^\circ$ and the other with $40$ newtons at $90^\circ$ . Find the magnitude and direction of the resultant force.

Write each force in components: $\mathbf{F}_1 = \langle 60, 0 \rangle$ and $\mathbf{F}_2 = \langle 0, 40 \rangle$ . Add them: $\mathbf{F} = \langle 60, 40 \rangle$ .

\|\mathbf{F}\| = \sqrt{60^2 + 40^2} = \sqrt{5200} \approx 72.1 \text{ N} \qquad \theta = \tan^{-1}\frac{40}{60} \approx 33.7^\circ

The crate feels about $72.1$ N of force at about $33.7^\circ$ above the first rope's direction.

Tip

To check a magnitude-and-angle conversion, go back the other way. From $72.1$ N at $33.7^\circ$ : $72.1\cos 33.7^\circ \approx 60$ and $72.1\sin 33.7^\circ \approx 40$ . The components match.

Practice

Practice 1

Find the component form of the vector with initial point $P(2, -1)$ and terminal point $Q(7, 3)$ . Enter it as $(a, b)$ .

Enter a point like (2, -3)

Practice 2

Find the magnitude of $\mathbf{v} = \langle -6, 8 \rangle$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Let $\mathbf{u} = \langle 3, -2 \rangle$ and $\mathbf{v} = \langle -1, 5 \rangle$ . Find $2\mathbf{u} - \mathbf{v}$ . Enter it as $(a, b)$ .

Enter a point like (2, -3)

Practice 4

Find the direction angle, in degrees from $0^\circ$ to $360^\circ$ , of $\mathbf{v} = \langle -3, 3 \rangle$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

A vector has magnitude $10$ and direction angle $150^\circ$ . Find its component form. Enter it as $(a, b)$ ; exact values or decimals to the nearest hundredth are fine.

Enter a point like (2, -3)

Practice 6

Find the unit vector in the same direction as $\mathbf{v} = \langle 5, -12 \rangle$ . Enter it as $(a, b)$ .

Enter a point like (2, -3)

Practice 7

Find the direction angle of $4\mathbf{i} - 7\mathbf{j}$ , in degrees from $0^\circ$ to $360^\circ$ , rounded to the nearest tenth.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Two forces act on an object: $20$ pounds at $0^\circ$ and $30$ pounds at $60^\circ$ . Find the magnitude of the resultant force, rounded to the nearest tenth of a pound.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.