Vectors

Some quantities are just a size: a temperature of ${20^\circ}$, a mass of ${3}$ kg. Others also have a direction: a wind of ${40}$ mi/h from the west, a force of ${50}$ newtons straight up. A vector is a quantity with both a size and a direction. This page shows how to describe vectors with numbers and how to add them.

What is a vector?

We draw a vector as an arrow. Its length shows the size, and it points in the direction. The arrow from a point ${P}$ to a point ${Q}$ is written ${\overrightarrow{PQ}}$. ${P}$ is the initial point (the tail), and ${Q}$ is the terminal point (the tip).

The vector ${\mathbf{v}=\overrightarrow{PQ}}$ starts at ${P}$ and ends at ${Q}$.

Vectors are written in bold, like ${\mathbf{u}}$ and ${\mathbf{v}}$ (by hand, with an arrow on top: ${\vec{v}}$). Two vectors are equal if they have the same length and the same direction, even if they start at different points. So you may slide a vector anywhere without changing it.

A plain number, like ${3}$ or ${-2}$, is called a scalar, to contrast it with a vector.

Components

Slide a vector so that its tail is at the origin. Its tip is then at some point ${(a,b)}$. We write the vector as

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

The numbers ${a}$ and ${b}$ are the components of ${\mathbf{v}}$: how far it goes across and how far up. The angle brackets ${\langle\ \rangle}$ show that it is a vector, not a point.

The vector ${\langle 3,4\rangle}$: ${3}$ units right and ${4}$ units up. Its length is ${5}$.

By the Pythagorean theorem, the length of ${\mathbf{v}}$, called its magnitude and written ${|\mathbf{v}|}$, is

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

If a vector goes from ${P(x_1,y_1)}$ to ${Q(x_2,y_2)}$, subtract the coordinates of the start from those of the end:

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

Example 1: Find the components and the magnitude of the vector from ${P(1,2)}$ to ${Q(4,-2)}$.

Solution:

$\begin{align*}\overrightarrow{PQ}&=\langle 4-1,\ -2-2\rangle\\&=\langle 3,-4\rangle\end{align*}$

$\begin{align*}|\overrightarrow{PQ}|&=\sqrt{3^2+(-4)^2}\\&=\sqrt{25}=5\end{align*}$

Adding vectors and multiplying by scalars

To add two vectors, place the tail of the second at the tip of the first. The sum goes from the tail of the first to the tip of the second. If ${\mathbf{u}}$ and ${\mathbf{v}}$ are two moves, then ${\mathbf{u}+\mathbf{v}}$ is the result of making both.

To add, put the tail of ${\mathbf{v}}$ at the tip of ${\mathbf{u}}$. The sum ${\mathbf{u}+\mathbf{v}}$ (green) goes from the start to the end. The other route (dashed) gives the same sum.

Multiplying a vector by a scalar ${c}$ stretches it by ${|c|}$. If ${c}$ is negative, it also turns the vector around.

The vector ${2\mathbf{v}}$ is twice as long as ${\mathbf{v}}$. The vector ${-\mathbf{v}}$ points the opposite way.

In components, both are done one component at a time:

$\begin{align*}&\langle a_1,b_1\rangle+\langle a_2,b_2\rangle\\&=\langle a_1+a_2,\ b_1+b_2\rangle\end{align*}$

$c\,\langle a,b\rangle=\langle ca,\ cb\rangle$

Subtraction is $\mathbf{u}-\mathbf{v}=\mathbf{u}+(-1)\mathbf{v}$: subtract the components.

Example 2: Let ${\mathbf{u}=\langle 2,-1\rangle}$ and ${\mathbf{v}=\langle -3,4\rangle}$. Find ${\mathbf{u}+\mathbf{v}}$, ${2\mathbf{u}-3\mathbf{v}}$, and ${|\mathbf{u}+\mathbf{v}|}$.

Solution:

$\begin{align*}&\mathbf{u}+\mathbf{v}\\&=\langle 2+(-3),\ -1+4\rangle\\&=\langle -1,3\rangle\end{align*}$

$\begin{align*}2\mathbf{u}-3\mathbf{v}&=\langle 4,-2\rangle-\langle -9,12\rangle\\&=\langle 13,-14\rangle\end{align*}$

$|\mathbf{u}+\mathbf{v}|=\sqrt{(-1)^2+3^2}=\sqrt{10}$

Unit vectors

A unit vector has length ${1}$. To get the unit vector in the same direction as ${\mathbf{v}}$, divide ${\mathbf{v}}$ by its length: ${\dfrac{\mathbf{v}}{|\mathbf{v}|}}$.

Two unit vectors get special names: ${\mathbf{i}=\langle 1,0\rangle}$ points right, and ${\mathbf{j}=\langle 0,1\rangle}$ points up. Every vector is a combination of them:

$\langle a,b\rangle=a\mathbf{i}+b\mathbf{j}$

Example 3: Find the unit vector in the direction of ${\mathbf{v}=\langle -3,4\rangle}$. Write it with ${\mathbf{i}}$ and ${\mathbf{j}}$.

Solution:

${|\mathbf{v}|=\sqrt{9+16}=5}$. Divide each component by ${5}$:

$\dfrac{\mathbf{v}}{|\mathbf{v}|}=\left\langle -\dfrac{3}{5},\dfrac{4}{5}\right\rangle=-\dfrac{3}{5}\mathbf{i}+\dfrac{4}{5}\mathbf{j}$

Magnitude and direction

The direction angle ${\theta}$ of a vector is the angle it makes with the positive ${x}$-axis, measured counterclockwise.

The direction angle ${\theta}$ is measured from the positive ${x}$-axis, like an angle in standard position.

A vector with length ${|\mathbf{v}|}$ and direction angle ${\theta}$ has components given by cosine and sine, exactly like a point on a circle of radius ${|\mathbf{v}|}$:

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

Going the other way, ${\tan\theta=\dfrac{b}{a}}$. Use the quadrant of ${(a,b)}$ to choose the right angle.

Example 4:

(a) Find the components of a vector of length ${10}$ with direction angle ${120^\circ}$.

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

Solution:

(a)

$\begin{align*}&\langle 10\cos 120^\circ,\ 10\sin 120^\circ\rangle\\&=\langle -5,\ 5\sqrt{3}\rangle\end{align*}$

(b) ${\tan\theta=\dfrac{4}{-3}}$. The point ${(-3,4)}$ is in quadrant II. The reference angle is ${\tan^{-1}\dfrac{4}{3}\approx 53.13^\circ}$, so

$\theta\approx 180^\circ-53.13^\circ=126.87^\circ$

Applications: velocity and force

Velocity is a vector: its magnitude is the speed, and its direction is the way you are going. When a plane flies through moving air, its true velocity over the ground is its velocity through the air plus the wind's velocity.

Example 5: A plane heads N ${30^\circ}$ E at an airspeed of ${300}$ mi/h. A wind blows toward the east at ${40}$ mi/h. Find the plane's true speed and direction.

The true velocity (green) is the sum of the plane’s velocity in the air (blue) and the wind (orange).

Solution:

Put east along the ${x}$-axis and north along the ${y}$-axis. N ${30^\circ}$ E is ${30^\circ}$ from north, so the direction angle is ${60^\circ}$:

$\begin{align*}\text{plane}&=\langle 300\cos 60^\circ,\ 300\sin 60^\circ\rangle\\&\approx\langle 150,\ 259.81\rangle\end{align*}$

The wind is ${\langle 40,0\rangle}$. Add:

$\text{ground}\approx\langle 190,\ 259.81\rangle$

Speed: ${\sqrt{190^2+259.81^2}\approx 321.9}$ mi/h.

Direction:

$\theta=\tan^{-1}\dfrac{259.81}{190}\approx 53.8^\circ$

That is ${53.8^\circ}$ from east, so ${36.2^\circ}$ from north (since ${90^\circ-53.8^\circ=36.2^\circ}$): a bearing of N ${36.2^\circ}$ E.

Forces add the same way. The total of several forces acting on an object is called the resultant. If the resultant is the zero vector ${\langle 0,0\rangle}$, the object is in equilibrium: the forces balance.

Example 6: Two ropes pull on a post. One pulls with a force of ${30}$ newtons toward the east. The other pulls with ${40}$ newtons toward the north. Find the resultant force.

Solution:

$\langle 30,0\rangle+\langle 0,40\rangle=\langle 30,40\rangle$

Its magnitude is ${\sqrt{900+1600}=50}$ newtons. Its direction angle is ${\tan^{-1}\dfrac{40}{30}\approx 53.1^\circ}$, north of east.

Summary

A vector has a magnitude and a direction. In components, ${\mathbf{v}=\langle a,b\rangle}$, and ${|\mathbf{v}|=\sqrt{a^2+b^2}}$.
The vector from ${(x_1,y_1)}$ to ${(x_2,y_2)}$ is ${\langle x_2-x_1,\ y_2-y_1\rangle}$.
Add vectors and multiply by scalars one component at a time.
A vector with length ${|\mathbf{v}|}$ and direction angle ${\theta}$ is $\langle |\mathbf{v}|\cos\theta,\ |\mathbf{v}|\sin\theta\rangle$.
Velocities and forces add as vectors.

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