This section covers:

**Introduction to Vectors****Vector Operations****Applications of Vectors****Dot Product and Angle Between Two Vectors****3D Vectors – Vectors in Space (including Cross Product)****Parametric Form of the Equation of a Line in Space**

# Introduction to Vectors

A **vector** (also called a **direction vector**) is just is something that has both **magnitude** (length, or size) and **direction**. So it’s different than a regular number, since it really has two components to it. We see vectors represented by **arrows**, so we can remember that we need to get a length of a vector (the magnitude), as well as the direction (which way it’s pointing).

We use vectors in mathematics, engineering, and physics, since many times we need to know both the size of something and which way it’s going. For example, with an airplane, we can use a vector to measure the speed of the plane (the “size”) and the direction it’s flying.

**Geometric Vectors **are **directed line segments** in the ** xy**-plane, and, as an example, the vector from a point

**(initial point) to a point**

*A***(terminal point) can be represented by \(\overrightarrow{{AB}}\) .**

*B*So, for example, if ** A** is (2, 7) and

**is (–3, 8), the vector is**

*B***second point minus first point**, or \(\displaystyle \left\langle {{{x}_{2}}} \right.-{{x}_{1}},\left. {{{y}_{2}}-{{y}_{1}}} \right\rangle \), or \(\left\langle {-3-2} \right.,\left. {8-7} \right\rangle =\left\langle {-5,\left. 1 \right\rangle } \right.\). The “

**” part of the vector (–5) is called the**

*x***x-component**, and the ”

**” part (1) is called the**

*y***.**

*y*-componentNote also that vectors can also be written in the form *a i* +

*b*, so this vector can also be written as

**j****–5**, or

*i*+ 1*j***–5**.

*i*+*j*The **magnitude** of the vector, written \(\left\| {AB} \right\|\) is the **distance** between the two points (like the hypotenuse of a right triangle), or \(\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}\), or with the new vector \(\left\langle {x,} \right.\left. y \right\rangle \), it’s just \(\sqrt{{{{x}^{2}}+{{y}^{2}}}}\). So for our points ** A **and

**above, \(\left\| {AB} \right\|=\sqrt{{{{{\left( {-5} \right)}}^{2}}+{{1}^{2}}}}=\sqrt{{26}}\).**

*B*Now looking at this vector visually, do you see how we can use the **slope** of the line of the vector (from the initial point to the terminal point) to get the **direction** of the vector? Pretty cool! So we can just use \({{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)\) (**second part of vector over first part of vector**) to get the angle measurement of the vector’s direction. Remembering from the **Polar Coordinates, Equations and Graphs** section though, we have to be careful which quadrant the vector terminates in (“pretending” that the vector’s initial point is at the **origin**) to know how many degrees we should add to that **tangent** value when we use a **calculator**:

Here is all this visually. Note that we had to **add 180**° to the angle measurement we got from the calculator (–11.3°) since the vector would terminate in the 2^{nd} quadrant if we were to start at (0, 0). So we get 168.7°, which is the angle measurement from the **positive x axis** going **counterclockwise**.

168.7° from the positive ** x** axis can also be described as

**11.3° North of West**(

**11.3° N of W**, or

**W11.3°W**), since the closest axis to the angle is the negative

**axis (west) and we are going a little north of that:**

*x*(We saw a similar concept of this when we were working with **bearings** here in the **Law of Sines and Cosines, and Areas of Triangles** section).

Note that a vector that has a magnitude of 0 (and thus no direction) is called a **zero vector**. So hypothetically, the vector \(\overrightarrow{{AA}}\) would be a zero vector.

A **unit vector** is a vector with magnitude **1**; in some applications, it’s easier to work with unit vectors. To find the unit vector that is associated with a vector (has same direction, but magnitude of 1), use the following formula: \(u=\frac{v}{{\left\| v \right\|}}\) (just divide each component of a vector by its magnitude to get its unit vector). We’ll see some problems below.

# Vector Operations

## Adding and Subtracting Vectors

There are a couple of ways to **add** and **subtract** vectors. When we add vectors, geometrically, we just put the beginning point (initial point) of the second vector at the end point (terminal point) of the first vector, and see where we end up (new vector starts at beginning of one and ends at end of the other). If the vectors aren’t this way to begin with, we can move the second vector (as long as it has the same magnitude and direction, so it’s like a slide) to be this way.

You can think of adding vectors as connecting the **diagonal** of the **parallelogram** (a four-sided figure with two pairs of parallel sides) that contains the two vectors.

Do you see how when we add vectors geometrically, to get the **sum**, we can just add the ** x** components of the vector, and the

**components of the vectors?**

*y*When we **subtract** two vectors, we just take the vector that’s being subtracting, **reverse the direction** and **add** it to the first vector. This is because the **negative** of a vector is that vector with the same magnitude, but has an **opposite direction** (thus adding a vector and its negative results in a zero vector).

Note that to make a vector **negative**, you can just negate each of its components (** x** component and

**component) (see graph below).**

*y*## Multiplying Vectors by a Number (Scalar)

To **multiply a vector** by a number, or scalar, you simply stretch (or shrink if the absolute value of that number is less than 1), or you can simply multiply the ** x** component and

**component by that number. Notice also that the**

*y***magnitude**is multiplied by that scalar.

Do you see how two vectors that are **parallel **are just a multiple of each other?

**Multiplying by a negative number** changes the direction of that vector.

Here’s what **subtracting vectors** and also **multiplying vectors** by a scalar looks like:

Let’s put all this together to perform the following vector operations, given the vectors shown:

You may also see problems like this, where you have to tell whether the statement is true or false. Note that you want to look at where you end up in relation to where you started to see the resulting vector. if you end up **exactly where you started from**, the resulting vector is **0**.

Here are a couple more examples of **vector problems**. Notice in the second set of problems when we are given a **magnitude** and **direction** of a vector, and have to find that vector, we use the following equation, like we did when we here in the **Polar Coordinates, Equations and Graphs** section: , where \(\left\| {\,v} \right\|\) or the magnitude of a vector is like the “** r**” (radius) we saw for polar numbers: \(v=\left\| {\,v} \right\|\left( {\cos \alpha i+\sin \alpha j} \right)=\left( {\left\| {\,v} \right\|\cos \alpha } \right)i+\left( {\left\| {\,v} \right\|\sin \alpha } \right)j\). (

**Trigonometry**always seems to come back and haunt us!) We’ll We’ll leave our answers in

**a**form.

*i*+ b*j*# Applications of Vectors

Vectors are extremely important in many applications of science and engineering. Since vectors include both a length and a direction, many vector applications have to do with **vehicle motion **and** direction**.

We saw above that, given a **magnitude** and **direction**, we can find the vector with \(v=\left\| {\,v} \right\|\left( {\cos \alpha i+\sin \alpha j} \right)=\left( {\left\| {\,v} \right\|\cos \alpha } \right)i+\left( {\left\| {\,v} \right\|\sin \alpha } \right)j\), where \(\left\| {\,v} \right\|\) is the **speed**. This way we can **add and subtract vectors**, and get a resulting speed and direction for the new vector.

Remember that a **bearing **(we saw here in the **Law of Sines and Cosines, and Areas of Triangles** section), is typically expressed a measure of the **clockwise angle** that starts **due north **or** on the positive y axis** (initial side) and terminates a certain number of degrees (terminal side) from that due north starting place. (This is also written, as in the case of a bearing of 40° as “40° east of north”, or “N40°E”).

(A lot of times, the bearing includes more directions, such as 70° west of north, also written as N70°W. In this case, the angle will start due north (straight up, or on the positive ** y** axis) and go counterclockwise 70° (because it’s going west, or to the left, instead of east). Similarly, a bearing of 50° south of east, or E50°S, would be an angle that starts due east (on the positive

**axis) and go clockwise 50° clockwise (towards the south, or down). Also, if you see a bearing of southwest, for example, the angle would be 45° south of west, or 225° clockwise from north, and so on.)**

*x*Each time a moving object changes course, you have to draw another line to the north to map its new bearing.

When there’s a **tail wind**, remember that you have to **add** this vector to the vector that the object is trying to go on (its programmed or “steered” course), to get the **actual** vector of the object. So remember:

**Problem:**

A plane is flying on a bearing of 25° south of west at 500 miles per hour (**speed**). Express the **velocity** of the plane (as a vector).

**Solution:**

**Problem:**

A sailboat is sailing on a bearing of 30° north of west at 6.5 miles per hour (in still water). A tail wind blowing 20 miles per hour in the direction 40° south of west alters the course of the boat.

Express the **actual velocity** of the sailboat as a vector. Then determine the **actual speed** and **direction** of the boat.

**Solution:**

Note that if we were given a vector for the **actual **course of the boat and had to come up with the vector for which the boat should be “steered”, we would have to **subtract **the **wind** from the **actual** course.

Now let’s the problem we already did using **Law of Cosines** from the **Law of Sines and Cosines, and Areas of Triangles** section:

**Problem:**

A cruise ship travels at a **bearing** of 40°at 60 mph for 3 hours, and changes course to a bearing of 120°. It then travels 40 mph for 2 hours. Find the distance the ship is from its **original position** and also its **bearing **from the original position.** **

**Solution:**

In this problem, **Distance = rate x time**, since we are given rates and times and need to calculate distances.

Since no specific directions (like West of South) are given for these **bearings**, we will obtain the angles by measuring the **clockwise angle** that starts **due north **or** on the positive y axis **(

**East**of

**North**).

And remember that with a change of **bearing**, we have to draw another line to the north to map its new bearing.

Now that we have the angles, we can use **vector addition** to solve this problem; doing the problem with vectors is actually easier than using **Law of Cosines**:

# Dot Product and Angle Between Two Vectors

The** dot product **of two vectors \(v=ai+bj\) and \(w=ci+dj\) (sort of like multiplying two vectors) is defined as \(v\bullet \,w=ac+bd\); in other words, you multiply the two “** x**” parts of the vectors, and multiply the two “

**” parts, and then add them together. The result is a**

*y***scalar**(single number).

Here is an example: if \(v=-2i+3j\) and \(w=2+j\), the dot product \(v\bullet \,w=\left( {-2} \right)\left( 2 \right)+\left( 3 \right)\left( 1 \right)=-1\).

We use dot products to find the **angle measurements between two vectors**; the cosine of the angle between two vectors is the dot product of the vectors, divided by the product of each of their magnitudes:

(And we don’t need to worry about getting the correct quadrant when putting this in the calculator!)

So we might be able to this formula instead of, say, the **Law of Cosines**, for applications.

Note that if the **dot product of two vectors is 0**, the vectors form **right angles**, or are **orthogonal**, since the **cos of 90°** is **0** (and thus the whole expression will be 0).

And remember that we noted above that if two vectors are **parallel**, then one is a “multiple” of another, or \(v=aw\). So for example, the vector \(v=-2i+3j\) would be parallel to the vector \(v=-4i+6j\). If vectors are parallel, the angle between them is either 0 (if they are the same vector) or *π*.

Here are some example problems:

# 3D Vectors – Vectors in Space

We’ve been dealing with vectors (and everything else!) in the two dimensional plane, but “real life” is actually three dimensional, so we need to know how to work in **3D**, or **space**, too.

A 3D coordinate system is typically drawn like this, with the *z***-axis** going “up”. Note that the **positive** *x***-axis** comes forward at you, and the **positive** *y***-axis** to the right of that, if you’re looking head on. Maybe you can remember this by the expression “Exit. Why?” (*x* – *z* – *y* when looking head on).

**Geometric Vectors **in** 3D **are still **directed line segments**, but in the ** xyz**-plane. We still can find the vector between two coordinate points by “subtracting” the first vector from the second.

So, for example, if ** A** is (–4, 2, 7) and

**is (–3, 8, 0), the vector \(\overrightarrow{{AB}}\) is**

*B***second point minus first point**, or \(\displaystyle \left\langle {{{x}_{2}}} \right.-{{x}_{1}},\,\left. {{{y}_{2}}-{{y}_{1}},\,{{z}_{2}}-{{z}_{1}}} \right\rangle \) or \(\left\langle {-3-\left( {-4} \right)} \right.,\left. {8-2,0-7} \right\rangle =\left\langle {1,\left. {6,-7} \right\rangle } \right.\). Note also that vectors can also be written in the form

*a*+

**i***b*+

**j***c*, so this vector can also be written as

**k****.**

*i*+ 6*j*–7*k*The **magnitude** of the 3D vector, written \(\left\| {AB} \right\|\) is still the **distance** between the two points (like taking hypotenuse of a right triangle twice actually), or \(\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}+{{{\left( {{{z}_{2}}-{{z}_{1}}} \right)}}^{2}}}}\), or with the new vector \(\left\langle {x,} \right.\left. {y,z} \right\rangle \), it’s just \(\sqrt{{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\). So for our points ** A **and

**above, \(\left\| {AB} \right\|=\sqrt{{{{1}^{2}}+{{6}^{2}}+{{{\left( {-7} \right)}}^{2}}+}}=\sqrt{{86}}\).**

*B*## Vector Operations in Three Dimensions

Adding, subtracting **3D** vectors, and multiplying **3**D vectors by a scalar are done the same way as 2D vectors; you just have to work with **three components**.

Like for 2D vectors, the** dot product **of two vectors \(v=ai+bj+ck\) and \(w=di+ej+fk\) (sort of like multiplying two vectors) is defined as \(v\bullet \,w=ad+be+cf\); in other words, you multiply the two “** x**” parts of the vectors, multiply the two “

**” parts, multiply the two “**

*y***” parts, and then add them together. The result is a**

*z***scalar**(single number).

Again, like for 2D, we use dot products to find the **angle measurements between two vectors**; the cosine of the angle between two vectors is the dot product of the vectors, divided by the product of each of their magnitudes:

Here are some problems; included is how to get the **equation of a sphere**:

Writing a 3D vector in terms of its **magnitude** and **direction** is a little more complicated. Since we can’t really describe a 3D vector in terms of only a magnitude and one direction, we have to get what we call the **direction angles**:

Here are what these angles look like:

So it turns out for the vector **\(v=ai+bj+ck\)**, the direction angles \(\alpha ,\,\,\beta ,\,\,\text{and}\,\,\gamma \) are:

These cosine values are called the **direction cosines** for the vector **v**.

To **find the 3D vector** in terms of its **magnitude** and **direction cosines**, we use:

Now let’s do a problem:

## Cross Products of 3D Vectors

Also, only for vectors 3D vectors, we have what we call a **cross product of vectors** (also called **vector product**, since the result is still a vector) of two vectors \(v=ai+bj+ck\) and \(w=di+ej+fk\). The vector that is the cross product of two vectors is actually **orthogonal **(**perpendicular**) to both of the original vectors. This is also called the **normal vector**.

This looks crazy! But we can get this cross product using **determinants** of **matrices**. We learned about determinants of matrices here in the **The Matrix and Solving Systems with Matrices** section.

Remember that for a 2 by 2 matrix, we get the determinant this way:

And here is an example of how we got the determinant of a 3 by 3 matrix:

Here is an example of how we use a determinant to find the **cross product **of two vectors \(v=i+2j-4k\) and \(w=-i+5j+3k\):

So the vector \(\displaystyle 26i+j+7k\) is **orthogonal** (**perpendicular**, **normal**) to the vectors ** v** and

**above.**

*w*A few things to remember here. First, we must watch the order of the vectors when we are finding the cross products of vectors; *v* × *w* is not necessarily the same thing as *w* × *v*.

Also, we can use the **right hand rule** to find the **direction** of the cross product of two vectors by holding up your right hand and make your index finger, middle finger, and thumb all perpendicular to each other (easier said than done!). Then point your index finger in the direction of the first vector (such as *v*) and your middle finger in the direction of the second vector (such as *w*). Your thumb will point in the direction of *v* × *w*.

This is something you probably won’t need too much in your math classes, but it can become very handy in Physics. (And remember the directions of 3D vectors as shown in the coordinate system below).

We can use the cross product to find the **area of a 3D parallelogram. **If that parallelogram** **has two adjacent sides with vectors ** v** and

**, we can take the**

*w***magnitude**of the vectors’ cross product to find its area: \(\left\| {v\,\,\times \,\,w} \right\|\). We can also use this if given four vertices of a parallelogram; we would just have to find two adjacent sides of the parallelogram in vector form first.

Here are some cross product problems:

## The Equation of a Plane

You might also be asked to find the **equation of the plane **that passes through a given point and is **perpendicular to a certain vector**, or even the equation of a plane **containing three points**.

Remember that the equation of a line can be in the standard form \(ax+by=c\), so the equation of a **plane** can be in the form \(ax+by+cz=d\). (These are called **Cartesian** equations.)

To see what this plane might look like, we can see where it intersects each of the three axes by setting the other variables to 0, for example with the graph \(2x+6y+3z=12\) (set ** y** and

**equal to 0 and solve for**

*z***, and so on):**

*x*Remember that the **dot product** will be 0 if two orthogonal vectors. So it turns out that a **vector equation of the plane** is \(\displaystyle \left\langle {a,b,c} \right\rangle \bullet \left( {\left\langle {x,y,z} \right\rangle -\left\langle {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right\rangle } \right)=0\), or \(\displaystyle \left\langle {a,b,c} \right\rangle \bullet \left\langle {x-{{x}_{0}},y-{{y}_{0}},z-{{z}_{0}}} \right\rangle =0\), where \(\left\langle {a,b,c} \right\rangle \) is orthogonal to the plane (the **normal** vector) and \(\left( {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right)\) is a **point** on the plane.

Another way to get the **equation of the plane** for a vector **perpendicular** to a certain plane at point \(\left( {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right)\) is to use \(ax+by+cz=d\), where \(d=a{{x}_{0}}+b{{y}_{0}}+c{{z}_{0}}\) (plug in the point to get *d*).

OK, so this looks really complicated, so let’s do a problem to show it’s not too bad:

To find the equation of the **plane containing three points**, we first have to find **two vectors** defined by the points, find the cross product of the two vectors, and then use the Cartesian equation above to find ** d**:

# Parametric Form of the Equation of a Line in Space

We can get a vector form of an equation of a **line in 3D space** by using **parametric equations**. (We’ll review **Parametric Equations** here).

In two dimensions, we worked with a slope of the line and a point on the line (or the ** y**-intercept).

In 3D space, we can use a **3D vector** \(\left\langle {a,b,c} \right\rangle \) as the **slope of a line**, and define that line by an initial point \(\left\langle {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right\rangle \) (vector that goes through that point and the origin) and the vector \(\left\langle {a,b,c} \right\rangle \). Note that the vector \(\left\langle {a,b,c} \right\rangle \) will be **parallel** to the line we’re describing, just like a slope going through the origin is parallel to a 2D line. We have three different ways to write this 3D line using parametric equations:

Notice to get the **last** form, we **solve for t** in the second set of equations. Also note that to go from the

**last equation**to the

**first equation**, we have to set each to

**, and solve back for**

*t***,**

*x***, and**

*y***.**

*z*This looks a little complicated, so let’s do some problems to show it’s not too bad:

Here are more problems that look more difficult, but are actually easier, since we’re finding a plane **perpendicular** to the vector or line:

Here are a few more problems:

**Learn these rules, and practice, practice, practice!**

On to **Parametric Equations** – you are ready!

I don’t think that your right hand rule illustration is correct. The x and y axes do not match up between the hand and the coordinate system.

Thank you SO MUCH for finding this! Do you want to do more editing ? 🙂 Lisa