# Introduction to Vectors

## Introduction to Vectors

A vector (also called a direction vector) is a quantity that has both magnitude (length, or size) and direction. It’s different than a regular number, since it 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 $A$ (initial point) to a point $B$ (terminal point) can be represented by $\overrightarrow{{AB}}$. For example, if $A$ is $(2,7)$ and $B$ is $(-3,8)$, the vector  is 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 “$x$” part of the vector (–5) is called the $x$-component, and the ”$y$” part (1) is called the $y$-component. This $\left\langle {x,} \right.\left. y \right\rangle$ form is called component form.

We usually call vectors with single letters, like $\overrightarrow{\text{u}}$, $\overrightarrow{\text{v}}$ or $\overrightarrow{\text{w}}$, or just u, v, w. Note also that vectors can also be written in the form $\text{ai}+\text{bj}$ (called the linear combination of the unit vectors $\text{i}$ and $\text{j}$), so this vector can also be written as $-5\text{i}+1\text{j}$, or $-5\text{i}+\text{j}$.

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 $\sqrt{{{{x}^{2}}+{{y}^{2}}}}$. For our points $A$ and $B$ above, $\left\| {AB} \right\|=\sqrt{{{{{\left( {-5} \right)}}^{2}}+{{1}^{2}}}}=\sqrt{{26}}$.

Here is 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! We can use $\displaystyle {{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)$ (second part of vector over first part of vector) to get the angle measurement of the vector’s direction. 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 2nd quadrant if we were to start at $(0,0)$ (see rules that we used from the Polar Coordinates, Equations and Graphs section). We get 168.7°, which is the angle measurement from the positive $\boldsymbol {x}$-axis going counterclockwise.

Note that 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°N), since the closest axis to the angle is the negative $x$-axis (west) and we are going a little north of that. (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).

A few more basics about vectors…

A vector that has a magnitude of 0 (and thus no direction) is called a zero vector. Thus, 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: $\displaystyle \text{u}=\frac{\text{v}}{{\left\| \text{v} \right\|}}$ (just divide each component of a vector by its magnitude to get its unit vector). We’ll see some problems below.

Also note that if two vectors are parallel, they have the exactly the same direction, or opposite directions; we’ll see this below.

## Vector Operations

There are a couple of ways to add and subtract vectors. To add vectors, geometrically, put the beginning point (initial point, or “tail”) of the second vector at the end point (terminal point, or “head”) of the first vector, and see where we end up (new vector starts at the original initial point and ends at the terminal point of the second vector). If the vectors aren’t this way to begin with, move the second vector (as long as it has the same magnitude and direction, so it’s like a slide) to be this way. This is called the “head-to-tail” method.

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 $y$-components of the vectors?

To subtract two vectors, reverse the direction of the vector that’s being subtracted (the second vector), 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 $y$-component) (see graph below).

### Multiplying Vectors by a Number (Scalar)

To multiply a vector by a number, or scalar, stretch (or shrink if the absolute value of that number is less than 1) the vector that many times. You can also multiply the $x$-component and $y$-component by the scalar. Notice also that the magnitude is multiplied by that scalar. Multiplying by a negative number also changes the direction of that vector.

Can you see how two vectors that are parallel are always a multiple of each other (with a multiple of 1 if the vectors are the same size)?

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. And don’t forget that if you end up exactly where you started from, the resulting vector is 0.

Here are a couple more examples of vector problems. To find a vector given a magnitude and direction, use the following equation (like in the Polar Coordinates, Equations and Graphs section), where $\left\| {\text{v}} \right\|$ or the magnitude of a vector is like the “$r$” (radius) for polar numbers:

\begin{align}\text{v}&=\left\| {\,\text{v}} \right\|\left( {\cos \alpha \text{i}+\sin \alpha \text{j}} \right)\\&=\left( {\left\| {\,\text{v}} \right\|\cos \alpha } \right)\text{i}+\left( {\left\| {\,\text{v}} \right\|\sin \alpha } \right)\text{j}\end{align}

We can leave our answers in $\text{a}\text{i}+\text{b}\text{j}$ form.

## 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 $\left\langle {\left\| \text{v} \right\|\cos \theta ,\,\,\left\| \text{v} \right\|\sin \theta } \right\rangle$, where $\left\| \text{v} \right\|$ 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 (like 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 $\boldsymbol{y}$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”).

Note that the bearing may include more directions, such as 70° west of north, also written as N70°W. In this case, the angle starts due north (straight up, or on the positive $y$-axis) and goes 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, starts due east (on the positive $x$-axis) and goes clockwise 50° clockwise (towards the south, or down). Also, if you see a bearing of southwest, for example, the angle is 45° south of west, or 225° clockwise from north, and so on.

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, 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. Remember:

ACTUAL COURSE = PROGRAMMED COURSE + COURSE of WIND
PROGRAMMED COURSE = ACTUAL COURSE – COURSE of WIND

Here are some problems:

Let’s try a problem we have done using Law of Cosines (Trig) in the Law of Sines and Cosines, and Areas of Triangles section: it is probably easier doing this problem with Trig.

Also notice that we need to use a definition of navigation bearing with respect to vectors. It is defined as the positive angle (0 to 360 degrees) measured clockwise with respect to the north (positive $y$-axis).

## Dot Product and Angle Between Two Vectors

The dot product of two vectors $\text{u}=\text{ai}+\text{bj}$ and $\text{v}=\text{ci}+\text{dj}$ is defined as $\text{u}\bullet \text{v}=\text{ac}+\text{bd}$; in other words, you multiply the two “$x$” parts of the vectors, and multiply the two “$y$” parts, and then add them together. The result is a scalar (single number). Here is an example: If $\displaystyle \text{u}=-2\text{i}+3\text{j}$ and $\text{v}=2\text{i}+\text{j}$, the dot product $\text{u}\bullet \text{v}=\left( {-2} \right)\left( 2 \right)+\left( 3 \right)\left( 1 \right)=-1$.

Dot Products are useful 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:

$\displaystyle \begin{array}{c}\cos \theta =\displaystyle \frac{{\text{u}\,\bullet \text{v}}}{{\left\| {\text{u}} \right\|\left\| {\text{v}} \right\|}}\,\,\,\,\,\,\,\,\\\theta ={{\cos }^{{-1}}}\left( {\displaystyle \frac{{\text{u}\bullet \text{v}}}{{\left\| {\text{u}} \right\|\left\| {\text{v}} \right\|}}} \right)\end{array}$

(And we don’t need to worry about getting the correct quadrant when putting this in the calculator!) We might be able to use 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).

Remember also that if two vectors are parallel, then one is a “multiple” of another, or $\text{u}=a\text{v}$. For example, the vector $\text{u}=-2\text{i}+3\text{j}$ would be parallel to the vector $\text{v}=-4\text{i}+6\text{j}$. If vectors are parallel, the angle between them is either 0 (if they are the same vector) or $\pi$.

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 positive $\boldsymbol{z}$axis going “up”. Note that the positive $\boldsymbol{x}$axis comes forward at you, and the positive $\boldsymbol{y}$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.

For example, if $A$ is $(-4,2,7)$ and $B$ is $(-3,8,0)$, the vector $\overrightarrow{{AB}}$ is 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 $\text{ai}+\text{bj}+\text{ck}$, so this vector can also be written as $\text{i}+6\text{j}-7\text{k}$. To the right is what that vector looks like visual, courtesy of CalcPlot3D:

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

### Vector Operations in Three Dimensions

Adding, subtracting 3D vectors, and multiplying 3D 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 $\text{u}=\text{ai}+\text{bj}+\text{ck}$ and $\text{v}=\text{di}+\text{ej}+\text{fk}$ is defined as $\text{u}\bullet \text{v}=\text{ad}+\text{be}+\text{cf}$; in other words, you multiply the two “$x$” parts of the vectors, multiply the two “$y$” parts, multiply the two “$z$” parts, and then add them together. The result is a 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:

$\displaystyle \begin{array}{c}\cos \theta =\displaystyle \frac{{\text{u}\,\bullet \text{v}}}{{\left\| {\text{u}} \right\|\left\| {\text{v}} \right\|}}\,\,\,\,\,\,\,\,\\\theta ={{\cos }^{{-1}}}\left( {\displaystyle \frac{{\text{u}\bullet \text{v}}}{{\left\| {\text{u}} \right\|\left\| {\text{v}} \right\|}}} \right)\end{array}$

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:

$\displaystyle \begin{array}{l}\alpha =\,\text{angle between v and i }\,\text{(positive }x-\text{axis)}\\\beta =\,\text{angle between v and j }\,\text{(positive }y-\text{axis)}\\\gamma =\,\text{angle between v and k }\,\text{(positive }z-\text{axis)}\end{array}$.    Here are what these angles look like:

It turns out for the vector $\text{v}=\text{ai}+\text{bj}+\text{ck}$, the respective cosines of direction angles $\alpha ,\,\beta ,\,\text{and}\,\gamma$ are:

These cosine values are called the direction cosines for the vector v. From this, the actual angles are the ${{\cos }^{{-1}}}$ of these values.

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 (instead of a dot product), 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 $\text{u}={{\text{a}}_{\text{1}}}\text{i}+{{\text{b}}_{\text{1}}}\text{j}+{{\text{c}}_{\text{1}}}\text{k}$ and $\text{v}={{\text{a}}_{\text{2}}}\text{i}+{{\text{b}}_{\text{2}}}\text{j}+{{\text{c}}_{\text{2}}}\text{k}$. 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.

Some reasons we might need cross products of vectors is in Physics or Engineering, for example, is to find orthogonal vectors (to find equations of planes), to find areas of parallelograms, or to calculate moments of force around a point or line (over my head!). Here is the formula:

Note: If vectors are defined as $\text{u}={{\text{a}}_{\text{1}}}\text{i}+{{\text{a}}_{2}}\text{j}+{{\text{a}}_{3}}\text{k}$ and $\text{v}={{\text{b}}_{1}}\text{i}+{{\text{b}}_{\text{2}}}\text{j}+{{\text{b}}_{3}}\text{k}$, you’ll see this equation as $\begin{array}{l}\text{u}\,\,\times \text{v}=\left( {{{\text{a}}_{2}}{{b}_{3}}-{{\text{a}}_{3}}{{b}_{2}}} \right)\text{i}-\left( {{{\text{a}}_{3}}{{\text{b}}_{1}}-{{\text{a}}_{1}}{{\text{b}}_{3}}} \right)\text{j}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( {{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{2}}}-{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{1}}}} \right)\text{k}\end{array}$.

An easier way to get the cross product is to use determinants of matrices. We learned about determinants of matrices here in the The Matrix and Solving Systems with Matrices section:

Here is an example of how we use a determinant to find the cross product of two vectors $\text{u}=\text{i}+2\text{j}-4\text{k}$ and $\text{v}=-\text{i}+5\text{j}+3\text{k}$:

The vector $\displaystyle 26\text{i}+\text{j}+7\text{k}$ is orthogonal (perpendicular, normal) to the vectors u and v above.

A few things to remember here. First, we must watch the order of the vectors when we are finding the cross products of vectors; $\text{u}\times \text{v}$ is not necessarily the same thing as $\text{v}\times \text{u}$.

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 u) and your middle finger in the direction of the second vector (such as v). Your thumb will point in the direction of $\text{u}\times \text{v}$. 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).

As well as finding orthogonal vectors, we can use the cross product to find the area of a 3D parallelogram. If that parallelogram has two adjacent sides with vectors u and v, we can take the magnitude of the vectors’ cross product to find its area: $\left\| {u\times v} \right\|$. 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.) Equations of planes are very useful in engineering as many times objects are represented as meshes of triangles, and each triangle defines a plane.

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 $z$ equal to 0 and solve for $x$, and so on):

It turns out that a vector equation of the plane is: $\displaystyle \left\langle {a,b,c} \right\rangle \bullet \left\langle {x-{{x}_{0}},y-{{y}_{0}},z-{{z}_{0}}} \right\rangle =0,\,\,\text{or}\,\,a\left( {x-{{x}_{0}}} \right)+b\left( {y-{{y}_{0}}} \right)+c\left( {z-{{z}_{0}}\,} \right)=0$, where $\left\langle {a,b,c} \right\rangle$ (also written as $\displaystyle a\text{i}+b\text{j}+c\text{k}$) is orthogonal to the plane (the normal vector – we can use the cross product to get this, as shown above!) and $\left( {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right)$ is a point on the plane.

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 $d$ in the equation $ax+by+cz=d$:

We saw that if two planes are perpendicular, the dot product of their normal vectors is 0. If two planes are parallel, their normal vectors are the same, or multiples of each other (with a different “$d$”). If two planes are identical, then the whole plane equation (including the “$d$”) is the same or a multiple of the other.

Here’s a type of problem you may see:

## 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.

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. We can then define that line in space by an initial point $\left\langle {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right\rangle$, (the position vector that goes through that point and the origin), and the direction vector $\left\langle {a,b,c} \right\rangle$. Note that the vector $\left\langle {a,b,c} \right\rangle$ is parallel to the line we’re describing, just like a slope going through the origin is parallel to a 2D line. We can think of $t$ as a scalar. We have three different ways to write this 3D line using parametric equations:

$\displaystyle \left\langle {x,y,z} \right\rangle =\left\langle {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right\rangle +t\left\langle {a,b,c} \right\rangle$            $\displaystyle \begin{array}{c}x={{x}_{0}}+at\\y={{y}_{0}}+bt\\z={{z}_{0}}+ct\end{array}$            $\displaystyle \frac{{x-{{x}_{0}}}}{a}=\frac{{y-{{y}_{0}}}}{b}=\frac{{z-{{z}_{0}}}}{c}\,\,\,\,(=t)$

Notice to get the last form, solve for $\boldsymbol {t}$ in the second set of equations. Also note that to go from the last equation to the first equation, set each to $t$, and solve back for $x$, $y$, and $z$.

Here are some problems:

Here are some more complicated problems; these can get a little tricky!

Learn these rules, and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) and click on the Search icon to the right of “How should I answer?” on the top. Type in “Vector” to see different vector problems that can be solved. this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Parametric Equations – you are ready!

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