# Chapter 08: Polar Coordinates, Complex Numbers & Vectors

## §8.1: Polar Coordinates

### The Idea

So far, you’ve only really experienced one coordinate system—the Cartesian system, also known as rectangular coordinates. But there are more…many more!

With (2D) polar coordinates, rather than measuring linear distances along two axes, we instead measure the angle and direction of a point—the angle being measured from some axis (a ray), and the distance being measured from the endpoint of the ray.

### Converting from Rectangular to Polar

Let the ray be the positive x-axis, and let the ray’s endpoint be the origin. Call the angle $\theta$, and call the distance from the origin to the point r. The rectangular coordinates are $\left(x,y\right)$ and the polar coordinates are $\left(r,\theta \right)$.

There are some nice right triangles in there! The Pythagorean Theorem tells us that ${r}^{2}={x}^{2}+{y}^{2}⇒r=\sqrt{{x}^{2}+{y}^{2}}$. Right triangle trigonometry tells us that $tan\theta =\frac{y}{x}⇒\theta ={tan}^{-1}\left(\frac{y}{x}\right)$.

You may have to adjust the value that arctan spits out in order to get your angle pointing in the correct direction, though. In fact, because there are an infinite number of angles for which $tan\theta =\frac{y}{x}$, there are really an infinite number of polar coordinate pairs that describe the point! For example, the coordinates $\left(2,\frac{\pi }{4}\right)$ and $\left(2,\frac{9\pi }{4}\right)$ are the same point.

…and just to be weird, we can also allow the value of r to be negative. In that case, the point isn’t along the angle specified…rather, it’s in a direction a half-circle around! Thus, the coordinates $\left(2,\frac{\pi }{4}\right)$ and $\left(-2,\frac{5\pi }{4}\right)$ are the same point.

### Converting from Polar to Rectangular

Back to the diagram above now…since $cos\theta =\frac{x}{r}$, we can write $x=rcos\theta$. Similarly, $sin\theta =\frac{y}{r}⇒y=rsin\theta$.

### Transforming Equations

…but why stop at coordinates? Why not transform entire equations?

When writing rectangular equation, we like to write them in $y=$ form (if possible). When writing polar equations, we want to write them in $r=$ form (again, if possible).

In both cases, just use the conversions above!

$y=3$ becomes $rsin\theta =3⇒r=\frac{3}{sin\theta }⇒r=3csc\theta$.

$y={x}^{2}$ becomes $rsin\theta ={\left(rcos\theta \right)}^{2}⇒rsin\theta ={r}^{2}{cos}^{2}\theta ⇒\frac{sin\theta }{{cos}^{2}\theta }=r$.

$r=1$ becomes $\sqrt{{x}^{2}+{y}^{2}}=1⇒{x}^{2}+{y}^{2}=1$ (a circle).

$r=2cos\theta$ has to be tweaked a bit: ${r}^{2}=2rcos\theta ⇒{x}^{2}+{y}^{2}=2x⇒{x}^{2}-2x+{y}^{2}=0⇒\left({x}^{2}-2x+1\right)+{y}^{2}=1⇒{\left(x-1\right)}^{2}+{y}^{2}=1$ (also a circle).

### Examples

[1.] Convert to polar coordinates: $\left(\sqrt{3},-1\right)$.

$r=\sqrt{{\left(\sqrt{3}\right)}^{2}+{\left(-1\right)}^{2}}=\sqrt{3+1}=\sqrt{4}=2$$tan\theta =\frac{-1}{\sqrt{3}}$ …and that’s an angle I have memorized, so I don’t need a calculator! Specifically, $tan\theta =-\frac{1}{\sqrt{3}}⇒\theta =\frac{5\pi }{6}$. Thus, the coordinates are $\left(2,\frac{5\pi }{6}\right)$.

[2.] Convert to rectangular coordinates: $\left(-2,75°\right)$.

Since $75°$ isn’t an angle about which I have anything memorized, I’ll have to use my calculator.

$x=\left(-2\right)cos75°=-0.5176$ and $y=\left(-2\right)sin75°=-1.9319$, so the coordinates are $\left(-0.5176,-1.9319\right)$.

[3.] Rewrite as a polar equation: $x+y=0$.

Use the conversion equations first: $x+y=0⇒rcos\theta +rsin\theta =0$. We’d like to solve that for r…perhaps you see a looming issue! $rcos\theta +rsin\theta =0⇒r\left(cos\theta +sin\theta \right)=0$, and you (should) know not to divide away the sine and cosine from that.

There are two possibilities—either $r=0$ (which isn’t very interesting), or $cos\theta +sin\theta =0$ (which might be interesting). $cos\theta +sin\theta =0⇒sin\theta =-cos\theta ⇒\frac{sin\theta }{cos\theta }=-1⇒tan\theta =-1⇒\theta =-\frac{\pi }{4}$. The equation $\theta =-\frac{\pi }{4}$ does pass through the point where $r=0$, we won’t lose anything by just looking at the second factor! Thus, $x+y=0$ is equivalent to $\theta =-\frac{\pi }{4}$.

Yes, I know—I said that we wanted to solve for r. Wanted. Sometimes you just can’t.

[4.] Convert to a rectangular equation: ${r}^{2}=2{cos}^{2}\theta +3{sin}^{2}\theta$.

Let’s rewrite that just a bit: ${r}^{2}=2{\left(cos\theta \right)}^{2}+3{\left(sin\theta \right)}^{2}$. There’s just one problem: you need an r next to that sine and cosine in order to make use of the conversion equations. OK—let’s just multiply both sides by ${r}^{2}$, resulting in ${r}^{4}=2{\left(rcos\theta \right)}^{2}+3{\left(rsin\theta \right)}^{2}$ (I skipped a step in there—I hope you don’t mind).

Convert! ${r}^{4}={\left({r}^{2}\right)}^{2}={\left({x}^{2}+{y}^{2}\right)}^{2}={x}^{4}+2{x}^{2}{y}^{2}+{y}^{4}$, so our equation becomes ${x}^{4}+2{x}^{2}{y}^{2}+{y}^{4}=2{x}^{2}+3{y}^{2}⇒{x}^{4}-2{x}^{2}+2{x}^{2}{y}^{2}+{y}^{4}-3{y}^{2}=0$. Good luck solving that for y! Let me know how it works out for you.

## §8.2: Polar Equations and Graphs

Just as you learned how to make graphs way back in Algebra 1, so now you should begin with polar graphs. Pick some inputs, pop them into the equation and record the outputs; then graph the resulting coordinates.

Remember that in polar, we want angles to be the inputs and r values to be the outputs.

### Examples

[5.] Graph $r=4cos\theta$.

Let’s make a table for the angles that we know.

 $\theta$ $4cos\theta$ 0 $4cos0=4$ $\frac{\pi }{6}$ $4cos\frac{\pi }{6}=4\left(\frac{\sqrt{3}}{2}\right)=2\sqrt{3}$ $\frac{\pi }{4}$ $4cos\frac{\pi }{4}=4\left(\frac{1}{\sqrt{2}}\right)=\frac{{2}^{2}}{{2}^{\frac{1}{2}}}={2}^{\frac{3}{2}}$ $\frac{\pi }{3}$ $4cos\frac{\pi }{3}=4\left(\frac{1}{2}\right)=2$ $\frac{\pi }{2}$ $4cos\frac{\pi }{2}=4\left(0\right)=0$ $\frac{2\pi }{3}$ $4cos\frac{2\pi }{3}=4\left(-\frac{1}{2}\right)=-2$ $\frac{3\pi }{4}$ $4cos\frac{3\pi }{4}=4\left(-\frac{1}{\sqrt{2}}\right)=-\frac{{2}^{2}}{{2}^{\frac{1}{2}}}=-{2}^{\frac{3}{2}}$ $\frac{5\pi }{6}$ $4cos\frac{5\pi }{6}=4\left(-\frac{\sqrt{3}}{2}\right)=-2\sqrt{3}$ $\pi$ $4cos\pi =4\left(-1\right)=-4$ $\frac{7\pi }{6}$ $4cos\frac{7\pi }{6}=4\left(-\frac{\sqrt{3}}{2}\right)=-2\sqrt{3}$ $\frac{5\pi }{4}$ $4cos\frac{5\pi }{4}=4\left(-\frac{1}{\sqrt{2}}\right)=-\frac{{2}^{2}}{{2}^{\frac{1}{2}}}=-{2}^{\frac{3}{2}}$ $\frac{4\pi }{3}$ $4cos\frac{4\pi }{3}=4\left(-\frac{1}{2}\right)=-2$ $\frac{3\pi }{2}$ $4cos\frac{3\pi }{2}=4\left(0\right)=0$ $\frac{5\pi }{3}$ $4cos\frac{5\pi }{3}=4\left(\frac{1}{2}\right)=2$ $\frac{7\pi }{4}$ $4cos\frac{7\pi }{4}=4\left(\frac{1}{\sqrt{2}}\right)=\frac{{2}^{2}}{{2}^{\frac{1}{2}}}={2}^{\frac{3}{2}}$ $\frac{11\pi }{6}$ $4cos\frac{11\pi }{6}=4\left(\frac{\sqrt{3}}{2}\right)=2\sqrt{3}$

Without a calculator, you’ll need to know things like $\sqrt{2}\approx 1.414$ and $\sqrt{3}\approx 1.732$ in order to plot this…

[6.] Graph $r=1+cos\theta$.

This time, I won’t write out the table…

[7.] Graph $r=\frac{{\theta }^{2}}{2}$.

## §8.3: Complex Numbers

#### The Definition

A complex number something of the form $a+b\mathbf{i}$, where $a\in ℝ$$b\in ℝ$, and $\mathbf{i}\equiv \sqrt{-1}$ (the triple symbol means “is defined to be.”).

Some examples of complex numbers include $2-3\mathbf{i}$$7$ (which is really $7+0\mathbf{i}$ ), and $-5\mathbf{i}$. The complex numbers complete the number path that you’ve been following throughout your school years: $ℕ\subset ℤ\subset ℚ\subset ℝ\subset ℂ$.

(okay, okay! The natural numbers are a subset of the integers, the integers are a subset of the rationals; the rationals are a subset of the reals; and the reals are a subset of the complex numbers)

There is an unfortunate tendency among many students (and a few teachers) to forget that every real number is also a complex number, and thus think (or at least say) that numbers like $\pi$ are non-complex (i.e., complex and non-real are the same thing). The distinction that one needs is that some numbers are real, and some numbers are non-real.

There is another unfortunate tendency (among students and teachers) to refer to non-real numbers as being imaginary. There is only one imaginary number: $\mathbf{i}$ (which is the imaginary unit). Numbers like $-5\mathbf{i}$ are pure imaginary; numbers like 7 are real…both are complex.

### The Complex Plane

You graph real numbers on a number line (one number only needs one dimension). Thus, complex numbers—which require two real numbers to be defined—must require two dimensions in order to be graphed! Hence, the complex plane.

Plot the value of a horizontally and plot the value of b vertically. Easy!

### Different Forms

#### Rectangular Form

The form which I used to define complex numbers is called rectangular form.

#### Polar Form (Trigonometric)

…but you know that rectangular coordinates can be changed to polar coordinates! Since the value of a is graphed on the x-axis, a takes the place of x in our conversion formulas. Similarly, b takes the place of y. Thus, $a+b\mathbf{i}=\left(rcos\theta \right)+\mathbf{i}\left(rsin\theta \right)=r\left(cos\theta +\mathbf{i}\cdot sin\theta \right)$. This last bit is sometimes abbreviated to $r\cdot cis\left(\theta \right)$.

#### Polar Form (Euler)

It turns out that there is one more form…one that can be used to connect five of the most important mathematical constants! In Euler’s form, the number is written as $r\cdot {e}^{\mathbf{i}\theta }$. You do not need to know why this is the same thing; I’m certainly not going to try and explain it.

### Important Features

#### The Modulus

The value of r—the distance to the origin in the complex plane—is called the modulus. The symbol for that is $|z|$.

Yes, I know you think of those bars as being “absolute value.” Get over it. Those bars mean “size of,” and how we measure size depends greatly on the thing being measured.

From our previous work, it should not be hard to see that $|a+b\mathbf{i}|=\sqrt{{a}^{2}+{b}^{2}}$.

#### The Real Part

The value of a is called the Real Part of the complex number. This is sometimes abbreviated $Re\left(z\right)$. Thus, $Re\left(-2+7\mathbf{i}\right)=-2$.

#### The Imaginary Part

The value of b is called the Imaginary Part of the complex number. This is sometimes abbreviated $Im\left(z\right)$. Thus, $Im\left(-2+7\mathbf{i}\right)=7$.

#### The Argument

The value of $\theta$ is called the Argument of the complex number. This is sometimes abbreviated $Arg\left(z\right)$. Thus, $Arg\left(1-\mathbf{i}\right)=-\frac{\pi }{4}$.

### Conversion Between Forms

If you remember how to convert rectangular and polar coordinates, then you know how to convert amongst the various forms of complex numbers! Thus, I don’t need to work any examples.

What’s that? You want me to do a few anyway? Really?

Oh, all right. Check the examples section of this section.

### Operations

Every set of numbers has some operations…that’s a fundamental part of mathematics! Since the complex numbers are an extension of the reals, the same operations are defined.

To add or subtract complex numbers, add (or subtract) the corresponding real and imaginary parts. Thus, $\left(2+5\mathbf{i}\right)+\left(3-\mathbf{i}\right)=5+4\mathbf{i}$.

Notice that I did that in rectangular form—you don’t want to do that in any other form.

#### Multiplication

To multiply complex numbers in rectangular form, use the good old FOIL method. $\left(2+5\mathbf{i}\right)\left(3-\mathbf{i}\right)=6-2\mathbf{i}+15\mathbf{i}-5{\mathbf{i}}^{2}=6+13\mathbf{i}+5=11+13\mathbf{i}$

To multiply in either polar form, multiply the moduli (plural of modulus) and add the angles. In the trig form, $\left[2\left(cos\frac{\pi }{3}+\mathbf{i}sin\frac{\pi }{3}\right)\right]\left[3\left(cos\frac{\pi }{5}+\mathbf{i}sin\frac{\pi }{5}\right)\right]=6\left(cos\left(\frac{\pi }{3}+\frac{\pi }{5}\right)+\mathbf{i}sin\left(\frac{\pi }{3}+\frac{\pi }{5}\right)\right)=6\left(cos\frac{8\pi }{15}+\mathbf{i}sin\frac{8\pi }{15}\right)$; in Euler’s form, $2{e}^{\mathbf{i}\frac{\pi }{3}}\cdot 3{e}^{\mathbf{i}\frac{\pi }{5}}=6{e}^{\mathbf{i}\left(\frac{\pi }{3}+\frac{\pi }{5}\right)}=6{e}^{i\frac{8\pi }{15}}$.

#### Division

In either polar form, since multiplication results in adding angles, division must result in…that’s right! Subtraction! Divide the moduli and subtract the angles.

In rectangular form, use the complex conjugate of the denominator to eliminate the non-real parts in the denominator, thus resulting in a new rectangular form complex number.

$\frac{1+\mathbf{i}}{2-3\mathbf{i}}=\frac{1+\mathbf{i}}{2-3\mathbf{i}}\cdot \frac{2+3\mathbf{i}}{2+3\mathbf{i}}=\frac{-2+5\mathbf{i}}{4+9}=-\frac{2}{13}+\frac{5}{13}\mathbf{i}$

#### Exponentiation

Raising a complex number to an exponent is an awful task if you are using rectangular form. Better to make use of DeMoivre’s Theorem with (either) polar form!

DeMoivre’s Theorem says that ${\left[r\left(cos\theta +\mathbf{i}sin\theta \right)\right]}^{n}={r}^{n}\left(cos\left(n\theta \right)+\mathbf{i}sin\left(n\theta \right)\right)$, or ${\left(r{e}^{\mathbf{i}\theta }\right)}^{n}=r{e}^{\mathbf{i}\left(n\theta \right)}$.

#### Roots

If raising a complex number to a power means multiplying the angle, then taking the root of a complex number must mean…that’s right! Divide the angle.

There is just one hitch.

When raising a number to a power, there is only one result. However, when taking the root of a number, there are multiple results. In particular, there will be n answers for the nth root of a complex number.

Now—how to find all of those other values…remember when we solved problems like $cos\left(2x\right)=\frac{1}{2}$? We had to go around the circle twice, then take all of those answers and divide by two. The same sort of thing happens here! I’ll do an example in the example section below.

### Examples

[8.] Multiply: $\left[3\left(cos\frac{2\pi }{3}+\mathbf{i}\cdot sin\frac{2\pi }{3}\right)\right]\cdot \left[2\left(cos\frac{\pi }{5}+\mathbf{i}\cdot sin\frac{\pi }{5}\right)\right]$.

$\left[3\left(cos\frac{2\pi }{3}+\mathbf{i}\cdot sin\frac{2\pi }{3}\right)\right]\cdot \left[2\left(cos\frac{\pi }{5}+\mathbf{i}\cdot sin\frac{\pi }{5}\right)\right]=6\left(cos\left(\frac{2\pi }{3}+\frac{\pi }{5}\right)+\mathbf{i}\cdot sin\left(\frac{2\pi }{3}+\frac{\pi }{5}\right)\right)$ $=6\left(cos\frac{13\pi }{15}+\mathbf{i}\cdot sin\frac{13\pi }{15}\right)$.

[9.] Divide: $\frac{3\left(cos120°+\mathbf{i}\cdot sin120°\right)}{5\left(cos45°+\mathbf{i}\cdot sin45°\right)}$.

$\frac{3\left(cos120°+\mathbf{i}\cdot sin120°\right)}{5\left(cos45°+\mathbf{i}\cdot sin45°\right)}=\left(\frac{3}{5}\right)\left(cos\left(120°-45°\right)+\mathbf{i}\cdot sin\left(120°-45°\right)\right)=\left(\frac{3}{5}\right)\left(cos75°+\mathbf{i}\cdot sin75°\right)$

[10.] Convert to trig form: $-5+5\mathbf{i}$.

There are two possibilities here—either this number comes out with an argument that is a nice angle (i.e., an angle about which you have some things memorized), or it isn’t. The only way that you can do this by hand is in the former case. In the latter case, you must use a calculator.

In either case, I’ll start by finding (and factoring out) the modulus: $|-5+5\mathbf{i}|=\sqrt{{\left(-5\right)}^{2}+{5}^{2}}=\sqrt{50}=5\sqrt{2}$$-5+5\mathbf{i}=5\sqrt{2}\left(\frac{-5}{5\sqrt{2}}+\frac{5}{5\sqrt{2}}\mathbf{i}\right)=5\sqrt{2}\left(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\mathbf{i}\right)$.

If the angle is a nice one, you should now be looking at some familiar coordinates…and I am! Those are the coordinates for $\frac{3\pi }{4}$. Thus, the number is $5\sqrt{2}\left(cos\frac{3\pi }{4}+\mathbf{i}\cdot sin\frac{3\pi }{4}\right)$.

[11.] Convert to trig form: $-7-5\mathbf{i}$.

$|-7-5\mathbf{i}|=\sqrt{{\left(-7\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{49+25}=\sqrt{74}$$-7-5\mathbf{i}=\sqrt{74}\left(-\frac{7}{\sqrt{74}}-\frac{5}{\sqrt{74}}\mathbf{i}\right)$. That doesn’t look familiar, so I’ll have to go to the calculator to find the angle. Be careful! This number lies in the third quadrant, and arctan can’t give an answer back there. ${tan}^{-1}\left(\frac{-5}{-7}\right)=0.6202$, but that is in the first quadrant…so I’ll need to add a half circle to get the correct angle: ${tan}^{-1}\left(\frac{-5}{-7}\right)+\pi =3.7602$.

The number is $\sqrt{74}\left(cos\left(3.7602\right)+\mathbf{i}\cdot sin\left(3.7602\right)\right)$.

[12.] Evaluate: ${\left(1-\mathbf{i}\sqrt{3}\right)}^{7}$.

First of all, notice that I put the imaginary unit in front of the root. The textbook doesn’t do that! I put it in front so that it is crystal clear as to whether the imaginary unit is inside or outside of the root. OK—back to the answer.

First of all, you really want to convert this back to trig form. $|1-\mathbf{i}\sqrt{3}|=\sqrt{{1}^{2}+{\left(-\sqrt{3}\right)}^{2}}=\sqrt{4}=2$${\left(1-\mathbf{i}\sqrt{3}\right)}^{7}={\left(2\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\mathbf{i}\right)\right)}^{7}$. I recognize those coordinates! ${\left(1-\mathbf{i}\sqrt{3}\right)}^{7}={\left(2\left(cos\frac{7\pi }{4}+\mathbf{i}\cdot sin\frac{7\pi }{4}\right)\right)}^{7}$, which now allows me to take the power more easily: ${\left(2\left(cos\frac{7\pi }{4}+\mathbf{i}\cdot sin\frac{7\pi }{4}\right)\right)}^{7}={2}^{7}\left(cos\left(7\cdot \frac{7\pi }{4}\right)+\mathbf{i}\cdot sin\left(7\cdot \frac{7\pi }{4}\right)\right)=128\left(cos\frac{49\pi }{4}+\mathbf{i}\cdot sin\frac{49\pi }{4}\right)$. If you’re not careful (or just lazy) you might stop there—but you can go on, since that angle is coterminal with another angle which we all know and love. $128\left(cos\frac{49\pi }{4}+\mathbf{i}\cdot sin\frac{49\pi }{4}\right)=128\left(cos\frac{\pi }{4}+\mathbf{i}\cdot sin\frac{\pi }{4}\right)=128\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\mathbf{i}\right)=\frac{128}{\sqrt{2}}+\frac{128}{\sqrt{2}}\mathbf{i}$

[13.] Find the 4th roots of 3.

Again, start by converting to trig form. $3=3+0\mathbf{i}=3\left(1+0\mathbf{i}\right)=3\left(cos0+\mathbf{i}\cdot sin0\right)$ … but to get all of the solutions, we’ll need to run that angle around the unit circle four times. Thus, the numbers which I’ll start with (before taking the root) are $3\left(cos0+\mathbf{i}\cdot sin0\right)$$3\left(cos2\pi +\mathbf{i}\cdot sin2\pi \right)$$3\left(cos4\pi +\mathbf{i}\cdot sin4\pi \right)$ and $3\left(cos6\pi +\mathbf{i}\cdot sin6\pi \right)$. Now, take the fourth root of each of those: $\sqrt[4]{3}\left(cos\frac{0}{4}+\mathbf{i}\cdot sin\frac{0}{4}\right)$$\sqrt[4]{3}\left(cos\frac{2\pi }{4}+\mathbf{i}\cdot sin\frac{2\pi }{4}\right)$$\sqrt[4]{3}\left(cos\frac{4\pi }{4}+\mathbf{i}\cdot sin\frac{4\pi }{4}\right)$ and $\sqrt[4]{3}\left(cos\frac{6\pi }{4}+\mathbf{i}\cdot sin\frac{6\pi }{4}\right)$. Reduce as much as possible: $\sqrt[4]{3}\left(cos\frac{0}{4}+\mathbf{i}\cdot sin\frac{0}{4}\right)=\sqrt[4]{3}\left(cos0+\mathbf{i}\cdot sin0\right)=\sqrt[4]{3}\left(1+0\right)=\sqrt[4]{3}$$\sqrt[4]{3}\left(cos\frac{2\pi }{4}+\mathbf{i}\cdot sin\frac{2\pi }{4}\right)=\sqrt[4]{3}\left(cos\frac{\pi }{2}+\mathbf{i}\cdot sin\frac{\pi }{2}\right)=\sqrt[4]{3}\left(0+\mathbf{i}\cdot 1\right)=\mathbf{i}\sqrt[4]{3}$$\sqrt[4]{3}\left(cos\frac{4\pi }{4}+\mathbf{i}\cdot sin\frac{4\pi }{4}\right)=\sqrt[4]{3}\left(cos\pi +\mathbf{i}\cdot sin\pi \right)=\sqrt[4]{3}\left(-1+\mathbf{i}\cdot 0\right)=-\sqrt[4]{3}$ and $\sqrt[4]{3}\left(cos\frac{6\pi }{4}+\mathbf{i}\cdot sin\frac{6\pi }{4}\right)=\sqrt[4]{3}\left(cos\frac{3\pi }{2}+\mathbf{i}\cdot sin\frac{3\pi }{2}\right)=\sqrt[4]{3}\left(0+\mathbf{i}\cdot \left(-1\right)\right)=-\mathbf{i}\sqrt[4]{3}$.

Thus, the fourth roots of 3 are $±\sqrt[4]{3}$ and $±\mathbf{i}\sqrt[4]{3}$.

## §8.4: Vectors

### The Definition

A scalar is a pure quantity. Mass is a scalar. Integers are scalars.

A vector is a quantity and a direction. Weight (force) is a vector. The complex numbers are vectors…

### Geometric Vectors

#### Representation

One way to think about vectors is as directed line segments. The endpoints of the segment are called the initial and terminal points.

Two vectors are equal if they have the same magnitude (quantity) and direction—thus, the actual initial and terminal points are mostly irrelevant! You can slide a vector around all you like, as long as you don’t change the length or orientation. Thus, the following image shows a bunch of equal vectors.

When writing about vectors, we use a little arrow above the name of the vector. The name can be a single letter, or it can be the names of its initial and terminal points: $\stackrel{\to }{a}$ or $\stackrel{\to }{PQ}$.

#### Operations

Vector addition is performed with the “head-to-tail” method. To add $\stackrel{\to }{a}+\stackrel{\to }{b}$, slide the initial point of $\stackrel{\to }{b}$ (the “tail”) to coincide with the terminal point (the “head”) of $\stackrel{\to }{a}$. The sum of the vectors is the vector determined by connecting the very first tail to the very last head. In the following diagram, vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are in black, and their sum is in red.

##### Scalar Product

When you multiply a vector by a scalar, this is called a scalar product. Geometrically, you do it by copying the vector as many times as needed. In the diagram below, I’ve illustrated $3\cdot \stackrel{\to }{a}$ (once with three copies of $\stackrel{\to }{a}$, and once with a single segment for the result).

Don’t forget that multiplying by a negative just changes the direction by a half-circle!

##### Subtraction

…and don’t forget that $\stackrel{\to }{a}-\stackrel{\to }{b}=\stackrel{\to }{a}+\left(-1\right)\cdot \stackrel{\to }{b}$.

### Algebraic Vectors

#### Representation

Of course, you don’t want to do all of this with little pictures…in fact, it would probably easier if you could reduce all of this to numbers!

Well, you can.

Remember that you can slide those vector segments around to your heart’s delight…well, why not slide the tail down to the origin? That way, we could just use the coordinates of the head to represent the vector. We write this as $〈x,y〉$. This is sometimes referred to as component form.

#### Operations

##### Scalar Product

To do scalar multiplication, just distribute the scalar to each component.

##### Subtraction

…and to subtract, just subtract components.

Isn’t that simple?

### Important Features

#### Magnitude

In the geometric representation, the length of the directed line segment is the magnitude of the vector. If you use coordinates, then that is the distance between the initial and terminal points. If the initial point is at the origin, then that distance is simple!

$|〈x,y〉|=\sqrt{{x}^{2}+{y}^{2}}$

(this is sometimes called the norm of the vector)

(…and some people like to use double bars: $‖〈x,y〉‖=\sqrt{{x}^{2}+{y}^{2}}$ )

#### Direction

In the geometric representation, the direction is the angle that the directed line segment makes with the positive x-axis. This is the same angle that we used for polar coordinates (and other things)! Thus, the angle is $\theta ={tan}^{-1}\left(\frac{y}{x}\right)$ (don’t forget that you may have to adjust the value that arctan actually gives you…).

### Unit Vectors

A unit vector is a vector with length one. In two dimensions, there are two very important unit vectors—the horizontal one ( $\stackrel{\to }{i}$ ), and the vertical one ( $\stackrel{\to }{j}$ ).

Every vector $〈x,y〉$ can be written as a sum of these two special unit vectors: $〈x,y〉=x\cdot \stackrel{\to }{i}+y\cdot \stackrel{\to }{j}$.

### Examples

[14.] Add: $〈6,2〉+〈5,-4〉$.

$〈6,2〉+〈5,-4〉=〈6+5,2-4〉=〈11,-2〉$.

[15.] Multiply: $-5\cdot 〈5,-3〉$.

$-5\cdot 〈5,-3〉=〈\left(-5\right)5,\left(-5\right)\left(-3\right)〉=〈-25,15〉$.

[16.] Rewrite using the unit vectors $\stackrel{\to }{i}$ and $\stackrel{\to }{j}$$〈-2,-1〉$.

$〈-2,-1〉=-2\stackrel{\to }{i}-\stackrel{\to }{j}$.

[17.] Find the magnitude and direction of $3\stackrel{\to }{i}+6\stackrel{\to }{j}$.

The magnitude is $|3\stackrel{\to }{i}+6\stackrel{\to }{j}|=\sqrt{{3}^{2}+{6}^{2}}=\sqrt{9+36}=\sqrt{45}\approx 6.708$. The direction will be in the first quadrant, so the result from arctan will not need adjusting. The angle is ${tan}^{-1}\left(\frac{6}{3}\right)\approx 1.107$ (that’s in radians, of course).

[18.] Find the unit vector having the same direction as $-6\stackrel{\to }{i}+\stackrel{\to }{j}$.

First, find the length: $|-6\stackrel{\to }{i}+\stackrel{\to }{j}|=\sqrt{{\left(6\right)}^{2}+{1}^{2}}=\sqrt{37}$. Now just do a scalar product! The answer is $\frac{1}{\sqrt{37}}\left(-6\stackrel{\to }{i}+\stackrel{\to }{j}\right)=-\frac{6}{\sqrt{37}}\stackrel{\to }{i}+\frac{1}{\sqrt{37}}\stackrel{\to }{j}$. I’ll let you check to see that this is, in fact, a unit vector.

## §8.5: The Dot Product

### The Definition

$〈a,b〉\cdot 〈c,d〉=ac+bd$

### Properties

When two vectors are orthogonal (i.e., perpendicular), the dot product will be zero.

### Application

The angle between two vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is $\theta ={cos}^{-1}\left(\frac{\stackrel{\to }{a}\cdot \stackrel{\to }{b}}{|\stackrel{\to }{a}|\cdot |\stackrel{\to }{b}|}\right)$.

Of course, you could just find the angle of each vector individually, and then find the difference in those angles yourself…

### Examples

[19.] Evaluate: $〈5,-8〉\cdot 〈1,2〉$.

$〈5,-8〉\cdot 〈1,2〉=\left(5\right)\left(1\right)+\left(-8\right)\left(2\right)=5-16=-11$.

[20.] Find the angle between $3\stackrel{\to }{i}-4\stackrel{\to }{j}$ and $5\stackrel{\to }{i}+2\stackrel{\to }{j}$.

We know that $cos\theta =\frac{〈3,-4〉\cdot 〈5,2〉}{|〈3,4〉|\cdot |〈5,2〉|}=\frac{7}{\left(5\right)\left(\sqrt{29}\right)}$, or $cos\theta \approx 0.26$, so $\theta \approx {cos}^{-1}\left(0.26\right)\approx 1.3078$.

[21.] An airplane is attempting to fly $20°$ South of East at 520 kph. The prevailing wind is blowing at 65kph at $40°$ South of East. What is the actual course and speed of the airplane?

Let’s call the attempted course vector $\stackrel{\to }{a}$, and the wind vector $\stackrel{\to }{w}$. The actual course and speed will be the sum of these vectors! You will have to adjust that angle into standard form. $\stackrel{\to }{a}=〈520cos340°,520sin340°〉\approx 〈488.64,-177.85〉$ and $\stackrel{\to }{w}=〈65cos320°,65sin320°〉\approx 〈49.79,-41.78〉$. The result is $\stackrel{\to }{a}+\stackrel{\to }{w}=〈488.64+49.79,-177.85-41.78〉=〈538.43,-219.63〉$ …but that doesn’t really answer the question. We really ought to give the answer with a speed and direction! The speed is the magnitude: $|〈538.43,-219.63〉|\approx 581.505$. The direction is on the fourth quadrant. Arctan gives angles in that zone, but they’re negative…we should probably translate them to something more readable. ${tan}^{-1}\left(\frac{-219.63}{538.43}\right)=-22.191°$ …this is better described as $22.191°$ South of East.

## §8.6: Vectors in Space

### Forms

Why stop at two dimensions? Why not try three?

It’s a little hard to draw in three dimensions, so we don’t use geometric diagrams to talk about vectors in 3-space. We do use component form and unit vector form.

#### Unit Vectors

An example: $2\stackrel{\to }{i}+3\stackrel{\to }{j}-\stackrel{\to }{k}$.

#### Components

The same example: $〈2,3,-1〉$.

### Operations

This works just as it does in 2-space!

#### Scalar Product

…and this does, too!

…and this!

#### Dot Product

The dot product extends quite nicely: $〈a,b,c〉\cdot 〈d,e,f〉=ad+be+cf$.

### Important Features

#### Magnitude

Even though we don’t draw them, the whole idea behind the component form is that we’ve slid the vector up to the origin, and thus only need to record the location of the head (which now requires three coordinates). Thus, finding the magnitude is still just using the distance formula: $|〈a,b,c〉|=\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}$.

#### Directions

The concept of direction doesn’t translate quite as well. I prefer a form different from that which the book uses…but I suppose I’ll just explain this version rather than going off on an irrelevant tangent.

(Ha ha, very funny. You know that you secretly enjoy my little diversions!)

Remember from geometry that two lines (usually) determine a unique plane (you do remember, don’t you?). Well then, a vector and the positive x-axis are two rays which determine a plane. Let the angle $\alpha$ be the angle between the vector and the positive x-axis in this plane. It turns out that for the vector $〈a,b,c〉$$cos\alpha =\frac{a}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}$.

Repeat that process with the positive y-axis, and again with the positive z-axis—that produces two more angles, $\beta$ and $\gamma$, so that $cos\beta =\frac{b}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}$ and $cos\gamma =\frac{c}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}$.

Note that we could have done the same thing in 2-space…letting $\delta$ be the angle between the vector and the positive x-axis, and letting $\epsilon$ be the angle between the vector and the positive y-axis. With the Pythagorean Identity, a little substitution and knowledge of complementary angles, we could have produced the (true) statement that ${cos}^{2}\delta +{cos}^{2}\epsilon =1$. It is for this same reason that we always know (in 3-space) that ${cos}^{2}\alpha +{cos}^{2}\beta +{cos}^{2}\gamma =1$.

### Examples

[22.] Add: $\left(2\stackrel{\to }{i}+5\stackrel{\to }{j}-4\stackrel{\to }{k}\right)+\left(-2\stackrel{\to }{i}-3\stackrel{\to }{j}-5\stackrel{\to }{k}\right)$.

$\left(2\stackrel{\to }{i}+5\stackrel{\to }{j}-4\stackrel{\to }{k}\right)+\left(-2\stackrel{\to }{i}-3\stackrel{\to }{j}-5\stackrel{\to }{k}\right)=0\stackrel{\to }{i}+2\stackrel{\to }{j}-9\stackrel{\to }{k}=2\stackrel{\to }{j}-9\stackrel{\to }{k}$.

[23.] Multiply: $7\cdot 〈2,3,5〉$.

$7\cdot 〈2,3,5〉=〈14,21,35〉$.

[24.] Convert to component form: $-2\stackrel{\to }{i}+\stackrel{\to }{j}-7\stackrel{\to }{k}$.

$-2\stackrel{\to }{i}+\stackrel{\to }{j}-7\stackrel{\to }{k}=〈-2,1,-7〉$.

[25.] Find the magnitude and direction angles for $4\stackrel{\to }{i}+7\stackrel{\to }{k}$.

The magnitude is $|4\stackrel{\to }{i}+7\stackrel{\to }{k}|=\sqrt{{0}^{2}+{4}^{2}+{7}^{2}}=\sqrt{65}$. The direction angles are $\alpha ={cos}^{-1}\left(\frac{0}{\sqrt{65}}\right)=0$$\beta ={cos}^{-1}\left(\frac{4}{\sqrt{65}}\right)=1.052$ and $\gamma ={cos}^{-1}\left(\frac{7}{\sqrt{65}}\right)=0.519$.

[26.] Evaluate $〈2,5,-4〉\cdot 〈-2,-3,-5〉$, and find the angle between the two vectors.

The dot product is $〈2,5,-4〉\cdot 〈-2,-3,-5〉=\left(-4\right)+\left(-15\right)+\left(20\right)=1$. The angle between the vectors causes $cos\theta =\frac{1}{|〈2,5,-4〉|\cdot |〈-2,-3,-5〉|}=\frac{1}{\left(3\sqrt{5}\right)\left(\sqrt{38}\right)}$, or $cos\theta =0.0242$ …so $\theta ={cos}^{-1}\left(0.0242\right)=1.5466$.

## §8.7: The Cross Product

### The Definition

$〈a,b,c〉×〈d,e,f〉=|\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ a& b& c\\ d& e& f\end{array}|$

### The Geometric Interpretations

#### Orthogonal Vectors

The cross product of two vectors produces a vector which is perpendicular to the plane containing the original vectors.

It was just mentioned a bit ago that two rays determine a unique plane…of course, there is an exception to this: if the vectors are parallel, then an infinite number of planes contain those vectors (remember that we can slide vectors around—if they are parallel, we can slide one directly on top of the other one).

When two vectors are parallel, then the cross product will be zero.

#### Area of a Parallelogram

The magnitude of the cross product of two vectors returns the area of the parallelogram which has side lengths equal to the magnitudes of the original vectors.

Thus, for parallelogram $ABCD$, the area of $ABCD=|\stackrel{\to }{AB}×\stackrel{\to }{AC}|$.

### Examples

[27.] Evaluate: $〈3,4,-7〉×〈-2,1,3〉$.

$〈3,4,-7〉×〈-2,1,3〉=|\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ 3& 4& -7\\ -2& 1& 3\end{array}|=$ $\left(4\right)\left(3\right)\stackrel{\to }{i}+\left(-7\right)\left(-2\right)\stackrel{\to }{j}+\left(3\right)\left(1\right)\stackrel{\to }{k}-\left(-7\right)\left(1\right)\stackrel{\to }{i}-\left(3\right)\left(3\right)\stackrel{\to }{j}-\left(4\right)\left(-2\right)\stackrel{\to }{k}=$ $\left(12+7\right)\stackrel{\to }{i}+\left(14-9\right)\stackrel{\to }{j}+\left(3+8\right)\stackrel{\to }{k}=19\stackrel{\to }{i}+5\stackrel{\to }{j}+11\stackrel{\to }{k}$.

[28.] Find a vector that is orthogonal to both $\stackrel{\to }{i}-\stackrel{\to }{j}+2\stackrel{\to }{k}$ and $2\stackrel{\to }{i}+3\stackrel{\to }{j}-4\stackrel{\to }{k}$.

In other words, find the cross product! I won’t be quite as verbose this time. $|\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ 1& -1& 2\\ 2& 3& -4\end{array}|=4\stackrel{\to }{i}+4\stackrel{\to }{j}+3\stackrel{\to }{k}-6\stackrel{\to }{i}+4\stackrel{\to }{j}+2\stackrel{\to }{k}=-2\stackrel{\to }{i}+8\stackrel{\to }{j}+5\stackrel{\to }{k}$.

[29.] Find the area of the parallelogram ABCD with vertices $A:\left(-1,2\right)$$B:\left(3,4\right)$$C:\left(6,1\right)$ and $D:\left(2,-1\right)$.

This is the magnitude of the cross product for two sides. I’ll construct vectors $\stackrel{\to }{AB}$ and $\stackrel{\to }{AD}$, then find their cross product’s magnitude. $\stackrel{\to }{AB}=〈3-\left(-1\right),4-2〉=〈4,2〉$ and $\stackrel{\to }{AD}=〈6-\left(-1\right),1-2〉=〈7,-1〉$. Note that in 3-space, these are $\stackrel{\to }{AB}=〈4,2,0〉$ and $\stackrel{\to }{AD}=〈7,-1,0〉$.

The cross product (without any work shown this time) is $|\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ 4& 2& 0\\ 7& -1& 0\end{array}|=-18\stackrel{\to }{k}$; the magnitude of that vector is 18. Hence, the area of that parallelogram is 18 square units.

Page last updated 2012-04-01