Math Core

Some quantities need only a size: a temperature, a mass, a distance. Others need a size and a direction: a wind of 30 mph out of the west, a 50-newton pull up a ramp, a plane flying northeast. Vectors are the tool for these quantities, and the trigonometry you already know is what connects a vector's size and direction to its components.

What a vector is

Definition

Vector

A vector is a quantity with both magnitude (length) and direction. You draw it as an arrow from an initial point to a terminal point. Two arrows with the same length and direction represent the same vector, no matter where they start.

Vectors are written in bold, like $\mathbf{v}$ , or with an arrow, like $\vec{v}$ . A plain number, such as $3$ or $-2.5$ , is called a scalar to contrast it with a vector.

Component form

Because only length and direction matter, you can slide any vector so that it starts at the origin. Its terminal point then tells you everything. If a vector runs from $P(x_1, y_1)$ to $Q(x_2, y_2)$ , its component form is

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

The first component is the horizontal change and the second is the vertical change. Angle brackets distinguish a vector $\langle 6, 8 \rangle$ from the point $(6, 8)$ . (When you type a vector as an answer on this site, use parentheses: (6, 8).)

The magnitude of $\mathbf{v} = \langle a, b \rangle$ comes straight from the Pythagorean theorem:

\|\mathbf{v}\| = \sqrt{a^2 + b^2}.

The vector from P to Q moves 6 right and 8 up, so v = ⟨6, 8⟩ and its length is 10.

Worked example: Component form and magnitude

Find the component form and magnitude of the vector from $P(-2, 1)$ to $Q(4, 9)$ .

\mathbf{v} = \langle 4 - (-2),\ 9 - 1 \rangle = \langle 6, 8 \rangle, \qquad \|\mathbf{v}\| = \sqrt{36 + 64} = 10.

Common mistake

Subtract in the order terminal minus initial. Reversing the order gives $\langle -6, -8 \rangle$ , a vector with the same length pointing the opposite way.

Vector arithmetic

Operations work component by component. For $\mathbf{u} = \langle u_1, u_2 \rangle$ , $\mathbf{v} = \langle v_1, v_2 \rangle$ and a scalar $k$ :

\mathbf{u} + \mathbf{v} = \langle u_1 + v_1,\ u_2 + v_2 \rangle, \qquad k\mathbf{v} = \langle k v_1,\ k v_2 \rangle.

Geometrically, you add vectors head to tail: draw $\mathbf{u}$ , start $\mathbf{v}$ where $\mathbf{u}$ ends, and the sum runs from the start of $\mathbf{u}$ to the end of $\mathbf{v}$ . The sum is called the resultant. Multiplying by $k$ stretches the vector by a factor of $\lvert k \rvert$ , and reverses it when $k < 0$ .

Head to tail: u + v runs from the tail of u to the head of v.

A unit vector has magnitude 1. To get the unit vector in the direction of $\mathbf{v}$ , divide by its length:

\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}.

The unit vectors along the axes have special names: $\mathbf{i} = \langle 1, 0 \rangle$ and $\mathbf{j} = \langle 0, 1 \rangle$ . Any vector can be written with them, since $\langle a, b \rangle = a\mathbf{i} + b\mathbf{j}$ . For example, $\langle 3, -7 \rangle = 3\mathbf{i} - 7\mathbf{j}$ .

Worked example: Combining vectors

Let $\mathbf{u} = \langle 3, -1 \rangle$ and $\mathbf{v} = \langle -2, 5 \rangle$ . Find $2\mathbf{u} - 3\mathbf{v}$ and the unit vector in the direction of $\mathbf{u}$ .

2\mathbf{u} - 3\mathbf{v} = \langle 6, -2 \rangle - \langle -6, 15 \rangle = \langle 12, -17 \rangle.

Since $\|\mathbf{u}\| = \sqrt{9 + 1} = \sqrt{10}$ , the unit vector is

\frac{\mathbf{u}}{\|\mathbf{u}\|} = \left\langle \frac{3}{\sqrt{10}},\ \frac{-1}{\sqrt{10}} \right\rangle \approx \langle 0.949,\ -0.316 \rangle.

Magnitude and direction

The direction angle $\theta$ of a vector is the angle it makes with the positive $x$ -axis, measured counterclockwise, just like an angle in standard position. This is where trigonometry comes in. A vector of length $\|\mathbf{v}\|$ at angle $\theta$ ends at the point $\|\mathbf{v}\|$ times the unit-circle point for $\theta$ .

Converting between the two descriptions

From magnitude and direction to components:

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

From components $\langle a, b \rangle$ to direction: $\tan\theta = \dfrac{b}{a}$ , with $\theta$ placed in the quadrant where the point $(a, b)$ lies.

Worked example: Both directions of the conversion

(a) A vector has magnitude 20 and direction angle $150^\circ$ . Find its components.

\mathbf{v} = \langle 20\cos 150^\circ,\ 20\sin 150^\circ \rangle = \langle -10\sqrt{3},\ 10 \rangle \approx \langle -17.32,\ 10 \rangle.

(b) Find the direction angle of $\mathbf{w} = \langle -3, -4 \rangle$ .

The calculator gives $\tan^{-1}\!\left(\dfrac{-4}{-3}\right) = \tan^{-1}\!\left(\dfrac{4}{3}\right) \approx 53.13^\circ$ , which is in Quadrant I. But $\langle -3, -4 \rangle$ points into Quadrant III, so add $180^\circ$ : $\theta \approx 233.13^\circ$ .

Common mistake

The inverse tangent only returns angles between $-90^\circ$ and $90^\circ$ . If the first component $a$ is negative, add $180^\circ$ to the calculator's answer. If the result is negative, add $360^\circ$ to get an angle from $0^\circ$ to $360^\circ$ . Always sketch the vector to see which quadrant it is in.

Applications: adding velocities and forces

When two forces act on an object, or when a plane flies through moving air, the total effect is the vector sum. The method is always the same: convert each vector to components, add, then convert back to magnitude and direction.

Worked example: A plane in a crosswind

A plane's velocity relative to the air is 300 mph at a direction angle of $60^\circ$ . A wind blows at 40 mph toward the west (direction angle $180^\circ$ ). Find the plane's actual speed and direction.

The wind pushes the plane's path slightly to the west.

Components of each velocity:

\begin{aligned} \mathbf{p} &= \langle 300\cos 60^\circ,\ 300\sin 60^\circ \rangle \approx \langle 150,\ 259.81 \rangle, \\ \mathbf{w} &= \langle 40\cos 180^\circ,\ 40\sin 180^\circ \rangle = \langle -40,\ 0 \rangle. \end{aligned}

Resultant: $\mathbf{p} + \mathbf{w} \approx \langle 110,\ 259.81 \rangle$ .

Speed: $\sqrt{110^2 + 259.81^2} \approx 282.1$ mph. Direction: $\tan^{-1}\!\left(\dfrac{259.81}{110}\right) \approx 67.1^\circ$ (Quadrant I, so no adjustment).

Tip

Keep at least four decimal places in intermediate components and round only the final answer. Rounding the components early can shift the final angle by a tenth of a degree or more.

Practice

Practice 1

Find the component form of the vector from $A(3, -2)$ to $B(-1, 5)$ . Enter it as $(a, b)$ .

Enter a point like (2, -3)

Practice 2

Find the magnitude of $\mathbf{v} = \langle -5, 12 \rangle$ .

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

Practice 3

Let $\mathbf{u} = \langle 2, -4 \rangle$ and $\mathbf{v} = \langle -1, 6 \rangle$ . Find $3\mathbf{u} + \mathbf{v}$ . Enter it as $(a, b)$ .

Enter a point like (2, -3)

Practice 4

Find the unit vector in the direction of $\langle 8, -6 \rangle$ . Enter it as $(a, b)$ .

Enter a point like (2, -3)

Practice 5

A vector has magnitude 12 and direction angle $225^\circ$ . Find its components, rounded to the nearest hundredth. Enter them as $(a, b)$ .

Enter a point like (2, -3)

Practice 6

Find the direction angle of $\langle -3, 5 \rangle$ , in degrees from $0^\circ$ to $360^\circ$ , to the nearest tenth.

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

Practice 7

Two forces act on a crate: 60 lb at a direction angle of $0^\circ$ and 45 lb at a direction angle of $50^\circ$ . Find the magnitude of the resultant force, to the nearest tenth of a pound.

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

Practice 8

A boat heads due north (direction angle $90^\circ$ ) at 12 mph while a current pushes it due east (direction angle $0^\circ$ ) at 5 mph. Find the direction angle of the boat's actual path, to the nearest tenth of a degree.

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