maths.free › Multivariable Calculus › 2. Vectors in Space › The Dot Product

The Dot Product

Calculate the dot product of two given vectors.

The Dot Product

This section only expects students to have familiarity with the notation and operations of vectors from . This section emphasizes the dot product as a tool in measuring the angle between vectors and in various forms of projections of vectors.

As fundamentally important as vectors are to most topics in multivariable calculus and great breadth of experience with vectors that students come into this course with, we suggest at least a week on the combined topics in , , and .

There are likely more activities in this section than any one instructor would do for this material, but we have erred on the side of including a larger set of activities in order to allow instructors to personalize how they would like to introduce topics or have students practice calculations. For example, some instructors might view the dot product as only a step on the way to using projections of vectors or some others might skip the orthogonal projection idea altogether. Students familiar with linear algebra may spend more effort on the geometric elements of measuring angles and projections or ensuring they understand how to draw the projection and orthogonal complements.

The idea of angle between vectors is emphasized in and and will be used in later sections about the directional derivative. There are several activities, like or , that are fairly generic activity focused on calculations involving dot products.

Introduction

In , we considered vector addition and scalar multiplication algebraically and interpreted the corresponding geometric meaning. In this section, we will introduce the first of two types of vector multiplication that we will study. This multiplication produces a scalar that tells us a remarkable amount about the geometric relationship between the vectors that are multiplied. We will start investigating these properties in the Preview Activity, but first, we provide the definition of this form of multiplication for two-dimensional vectors.

Consider the two-dimensional vectors $\vu = \langle u_1,u_2\rangle$ and $\vv = \langle v_1,v_2\rangle$. The dot product of $\vu$ and $\vv$ is denoted $\vu\cdot\vv$ and defined to be \[\begin{aligned}\end{aligned}\]. For instance, we have $\langle 2,-1\rangle\cdot \langle 1,4\rangle = 2\cdot 1 +(-1)\cdot 4 = -2$. At first glance, this may feel like we've lost information, as we have taken two objects with both magnitude and direction, multiplied them, and obtained a scalar. However, this scalar allows us to draw many useful conclusions, as you will now begin exploring.

In , you have started developing an understanding of how the dot product gives us information about the angle between two vectors in the plane. The rest of this section will formalize this idea about the dot product of two vectors and explore properties of this measurement.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

The Dot Product

The idea of multiplying corresponding components of vectors and summing them can be extended to any dimension, which gives us the definition of the dot product.

As we will see shortly, the dot product arises in physics to calculate the work done by a vector force in a given direction. It might be more natural to define the dot product in this context. However, it is more convenient from a mathematical perspective to define the dot product algebraically and then view work as an application of this definition.

To illustrate the idea of computing a dot product in more than three dimensions, we find that \[\begin{aligned}\langle 3, 0, 1, 5 \rangle\cdot\langle -2, 1, 4, -2 \rangle \amp= 3\cdot(-2) + 0\cdot1 + 1\cdot4 +5\cdot (-2) \\ \amp= -6 + 0 + 4 -10= -12\end{aligned}\].

Notice that the resulting quantity is a scalar. In the next section, we will explore another product of vectors for which the resulting quantity is a vector. For this reason, you will sometimes see the phrase scalar product used in place of dot product.

The dot product gives us a way to define a product of two vectors, and we will soon formalize the ideas from about what the dot product tells us about the angle between vectors. First, however, we state some useful properties the dot product that show how it behaves in ways that are similar to the product of real numbers. In fact, all of these properties are consequences of properties of addition and multiplication of real numbers.

Let $\vu$, $\vv$, and $\vw$ be vectors in $\R^n$. Then

$\vu \cdot \vv = \vv \cdot \vu$ (The dot product is commutative.)
$(\vu + \vv) \cdot \vw = (\vu \cdot \vw) + (\vv \cdot \vw)$ (The dot product can be distributed across vector addition.)
if $c$ is a scalar, then $(c\vu) \cdot \vw = c(\vu \cdot \vw)$ (The dot product is associative with scalar multiplication.)

In , you observed that for two-dimensional vectors, the dot product of a vector with itself is the square of the vector's magnitude. This is so useful, that we state it below.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

The Angle between Vectors

As we saw in , the sign of the dot product gives us information about the type of angle formed by two vectors. We also saw that the dot product alone does not give us sufficient information to measure the angle. However, with a bit more information about the vectors, we are able to measure the angle, as we will soon see. If we are given two vectors $\vu$ and $\vv$, there are two angles that these vectors create, as depicted in . We will call $\theta$, the smaller of these angles, the angle between these vectors. Notice that $\theta$ lies between 0 and $\pi$ because we would consider the angle from the other side if the angle was greater than $\pi$.

Proposition

Given vectors $\vu$ and $\vv$, let $\theta$ be the smallest angle between $\vu$ and $\vv$. We have the following relationship: \[\begin{aligned}\end{aligned}\] Solving for the angle $\theta$, we have: \[\begin{aligned}\end{aligned}\]

Proof

Recall that the Law of Cosines says that if we consider any triangle with side lengths $a,b,c$ and $\theta$ the measure of the angle formed by the sides of length $a$ and $b$, then $c^2=a^2+b^2-2ab\cos(\theta)$. To determine the angle $\theta$ here, we may apply the Law of Cosines to the triangle shown in .

Using the fact that the dot product of a vector with itself gives us the square of its length, together with other properties of the dot product, we find: \[\begin{aligned}\vecmag{\vu-\vv}^2 =\mathstrut \amp \vecmag{\vu}^2 + \vecmag{\vv}^2 - 2\vecmag{\vu}\vecmag{\vv}\cos(\theta) \\ (\vu-\vv)\cdot(\vu-\vv) =\mathstrut \amp \vu\cdot\vu + \vv\cdot\vv - 2\vecmag{\vu}\vecmag{\vv}\cos(\theta) \\ \vu\cdot(\vu-\vv) - \vv\cdot(\vu-\vv)=\mathstrut \amp \vu\cdot\vu + \vv\cdot\vv - 2\vecmag{\vu}\vecmag{\vv}\cos(\theta) \\ \vu\cdot\vu - 2\vu\cdot\vv + \vv\cdot\vv =\mathstrut \amp \vu\cdot\vu + \vv\cdot\vv - 2\vecmag{\vu}\vecmag{\vv}\cos(\theta) \\ -2\vu\cdot\vv =\mathstrut \amp -2\vecmag{\vu}\vecmag{\vv}\cos(\theta) \\ \vu\cdot\vv =\mathstrut \amp \vecmag{\vu}\vecmag{\vv}\cos(\theta)\end{aligned}\]

The powerful conclusion from is that the dot product is a simple algebraic calculation that provides us with important geometric information about the angle between the vectors that would be difficult to determine otherwise. While you normally need to consider if the angle that comes out of an inverse trigonometric function needs to be adjusted based on the context of a problem, we do not need to worry about that with because the angle measurement and the range of arccosine are both from $0$ to $\pi$.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Projections

Suppose we want to split the vector $\vu$ into parts that are parallel to $\vv$ and orthogonal to $\vv$, which we will call $\vw_1$ and $\vw_2$ respectively. In other words, we want to write $\vu=\vw_1+\vw_2$ where $\vw_1=k \vv$ and $\vw_2 \cdot \vv =0$. Geometrically, this will look like one of the possibilities shown in .

Notice that if the angle between $\vu$ and $\vv$ is obtuse, then the amount of $\vu$ that is parallel to $\vv$ will be in the direction opposite $\vv$. Before we look at how to calculate the vectors $\vw_1$ and $\vw_2$, we will do an activity about drawing the graphical representation of this splitting of a vector into parts that are parallel to and orthogonal to another vector.

Activity

In , you can see illustrations of splitting $\vu$ into two parts: $\vw_1$, which is parallel to $\vv$, and $\vw_2$, which is orthogonal to $\vv$. Use this figure for reference as you do the following.

We know from the previous subsection that there is a third configuration of vectors, which occurs when $\vu$ and $\vv$ are orthogonal. Suppose that $\vu$ and $\vv$ are nonzero orthogonal vectors. What would $\vw_1$ and $\vw_2$ be in this case?

We want to switch the roles of $\vu$ and $\vv$ for the examples in the previous parts. Specifically, for these configuration of vectors, we want to split $\vv$ into parts that are parallel to $\vu$, which we will call $\vz_1$, and orthogonal to $\vu$, which we will call $\vz_2$. On , draw $\vz_1$ and $\vz_2$ for each configuration.

Based on your drawing, is it the case that $\vw_1 = \vz_1$? What about $\vw_2$ and $\vz_2$?

Now that we have visually explored the relationship between $\vu$, $\vv$, $\vw_1$, and $\vw_2$, we will see how we can find formulas that allow us to compute $\vw_1$ and $\vw_2$ in terms of $\vu$ and $\vv$.

Activity

Suppose that $\vu$ and $\vv$ are vectors in $\R^n$. We wish to find $\vw_1$ and $\vw_2$ so that $\vw_1$ is parallel to $\vv$ and $\vw_2$ is orthogonal to $\vv$, as shown in .

Explain why $\vu=\vw_1 + \vw_2$.

Compute $\vu\cdot \vv$ as $(\vw_1+\vw_2)\cdot \vv$. Simplify your answer as much as possible, using the fact that $\vw_2$ is orthogonal to $\vv$.

Since $\vw_1$ is parallel to $\vv$, there is a scalar $k$ so that $\vw_1 = k\vv$. Substitute $k\vv$ for $\vw_1$ in your answer to the previous part and then solve for $k$.

Give a formula for $\vw_1$ purely in terms of $\vu$ and $\vv$.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Work, Force, and Displacement

In physics, work is a measure of the energy required to apply a force to an object through a displacement. In , we can see a diagram showing the force applied to a refrigerator (in orange) that is used to move the refrigerator from point $A$ to point $B$. The change in location, $\overrightarrow{AB}$, is called the displacement. Notice that not all of the force is in the direction of the displacement. The work done on the refrigerator only depends on the amount of force that is in the direction of $\overrightarrow{AB}$.

The work required to move the refrigerator is \[\begin{aligned}\end{aligned}\].

This means that the work is determined only by the magnitude of the force applied parallel to the displacement and the length of the displacement. In fact, work is closely related to the projection of the force vector onto the displacement vector. (More precisely, it is the product of the component of the force vector in the direction of the displacement vector times the magnitude of the displacement vector.)

The units commonly used for work are Joules or Newton-meters in the metric system and foot-pounds in customary units. (In electrical settings, units such as coulomb-volts or kilowatt-hours may be used.) Note that these units come from work being the dot product of a force vector, using units of Newtons or pounds, and a displacement vector, using units of meters or feet.

Activity

Determine the work done by a 25 pound force acting at a $30^{\circ}$ angle to the direction of the object's motion, if the object is pulled 10 feet.

Determine if more work or less work is done if the angle to the direction of the object's motion is $60^\circ$.

The Dot Product and Its Properties

We have already learned how to add and subtract vectors. In this chapter, we investigate two types of vector multiplication. The first type of vector multiplication is called the dot product, based on the notation we use for it, and it is defined as follows:

Note that if $\text{u}$ and $\text{v}$ are two-dimensional vectors, we calculate the dot product in a similar fashion. Thus, if $\text{u}=〈{u}_{1},{u}_{2}〉$ and $\text{v}=〈{v}_{1},{v}_{2}〉,$ then

\[\text{u}\cdot \text{v}={u}_{1}{v}_{1}+{u}_{2}{v}_{2}.\]

When two vectors are combined under addition or subtraction, the result is a vector. When two vectors are combined using the dot product, the result is a scalar. For this reason, the dot product is often called the scalar product. It may also be called the inner product.

Example

Try it.

Find the dot product of $\text{u}=〈3,5,2〉$ and $\text{v}=〈-1,3,0〉.$
Find the scalar product of $\text{p}=10\text{i}-4\text{j}+7\text{k}$ and $\text{q}=-2\text{i}+\text{j}+6\text{k}.$

Solution

Substitute the vector components into the formula for the dot product:
\[\begin{array}{ll}\text{u}\cdot \text{v} & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3} \\ & =3(-1)+5(3)+2(0)=-3+15+0=12.\end{array}\]
The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:
\[\begin{array}{ll}\text{p}\cdot \text{q} & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3} \\ & =10(-2)+(-4)(1)+(7)(6)=-20-4+42=18.\end{array}\]

Like vector addition and subtraction, the dot product has several algebraic properties. We prove three of these properties and leave the rest as exercises.

Condensed — the full section is in OpenStax Calculus Volume 3.

Using the Dot Product to Find the Angle between Two Vectors

When two nonzero vectors are placed in standard position, whether in two dimensions or three dimensions, they form an angle between them (). The dot product provides a way to find the measure of this angle. This property is a result of the fact that we can express the dot product in terms of the cosine of the angle formed by two vectors.

Condensed — the full section is in OpenStax Calculus Volume 3.

Projections

As we have seen, addition combines two vectors to create a resultant vector. But what if we are given a vector and we need to find its component parts? We use vector projections to perform the opposite process; they can break down a vector into its components. The magnitude of a vector projection is a scalar projection. For example, if a child is pulling the handle of a wagon at a 55° angle, we can use projections to determine how much of the force on the handle is actually moving the wagon forward (). We return to this example and learn how to solve it after we see how to calculate projections.

Definition

The vector projection of $\text{v}$ onto $\text{u}$ is the vector labeled proj_uv in . It has the same initial point as $\text{u}$ and $\text{v}$ and the same direction as $\text{u}$, and represents the component of $\text{v}$ that acts in the direction of $\text{u}$. If $\theta$ represents the angle between $\text{u}$ and $\text{v}$, then, by properties of triangles, we know the length of ${\text{proj}}_{\text{u}}\text{v}$ is $‖{\text{proj}}_{\text{u}}\text{v}‖=‖\text{v}‖\cdot \left|\text{cos}\right|\ \theta .$ Note that when the angle $\theta$ between $\text{u}$ and $\text{v}$ is an obtuse angle, the projection will be in the opposite direction of $\text{u}$. When expressing $\text{cos}\ \theta$ in terms of the dot product, this becomes

\[\begin{array}{ll}‖{\text{proj}}_{\text{u}}\text{v}‖ & =‖\text{v}‖\cdot \left|\text{cos}\right|\ \theta \\ & =‖\text{v}‖(\frac{|\text{u}\cdot \text{v}|}{‖\text{u}‖‖\text{v}‖}) \\ & =\frac{|\text{u}\cdot \text{v}|}{‖\text{u}‖}.\end{array}\]

We now multiply by a unit vector in the direction of $\text{u}$ to get ${\text{proj}}_{\text{u}}\text{v}\text{:}$

\[{\text{proj}}_{\text{u}}\text{v}=\frac{\text{u}\cdot \text{v}}{‖\text{u}‖}(\frac{1}{‖\text{u}‖}\text{u})=\frac{\text{u}\cdot \text{v}}{{‖\text{u}‖}^{2}}\text{u}.\]

The length of this vector is also known as the scalar projection of $\text{v}$ onto $\text{u}$ and is denoted by

\[{\text{comp}}_{\text{u}}\text{v}=\frac{\text{u}\cdot \text{v}}{‖\text{u}‖}.\]

Sometimes it is useful to decompose vectors—that is, to break a vector apart into a sum. This process is called the resolution of a vector into components. Projections allow us to identify two orthogonal vectors having a desired sum. For example, let $\text{v}=〈6,-4〉$ and let $\text{u}=〈3,1〉.$ We want to decompose the vector $\text{v}$ into orthogonal components such that one of the component vectors has the same direction as $\text{u}$.

Now consider the vector $\text{q}=\text{v}-\text{p}.$ We have

Clearly, by the way we defined $\text{q}$, we have $\text{v}=\text{q}+\text{p},$ and

Condensed — the full section is in OpenStax Calculus Volume 3.

Work

Now that we understand dot products, we can see how to apply them to real-life situations. The most common application of the dot product of two vectors is in the calculation of work.

From physics, we know that work is done when an object is moved by a force. When the force is constant and applied in the same direction the object moves, then we define the work done as the product of the force and the distance the object travels: $W=Fd.$ We saw several examples of this type in earlier chapters. Now imagine the direction of the force is different from the direction of motion, as with the example of a child pulling a wagon. To find the work done, we need to multiply the component of the force that acts in the direction of the motion by the magnitude of the displacement. The dot product allows us to do just that. If we represent an applied force by a vector F and the displacement of an object by a vector s, then the work done by the force is the dot product of F and s.

Let’s revisit the problem of the child’s wagon introduced earlier. Suppose a child is pulling a wagon with a force having a magnitude of 8 lb on the handle at an angle of 55°. If the child pulls the wagon 50 ft, find the work done by the force ().

We have

\[W=‖\text{F}‖‖\overset{\to}{PQ}‖\text{cos}\ \theta =8(50)(\text{cos}(55\text{^{\circ}}))\approx 229\ \text{ft}\cdot \text{lb}.\]

In U.S. standard units, we measure the magnitude of force $‖\text{F}‖$ in pounds. The magnitude of the displacement vector $‖\overset{\to}{PQ}‖$ tells us how far the object moved, and it is measured in feet. The customary unit of measure for work, then, is the foot-pound. One foot-pound is the amount of work required to move an object weighing 1 lb a distance of 1 ft straight up. In the metric system, the unit of measure for force is the newton (N), and the unit of measure of magnitude for work is a newton-meter (N·m), or a joule (J).

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

The dot product, or scalar product, of two vectors $\text{u}=〈{u}_{1},{u}_{2},{u}_{3}〉$ and $\text{v}=〈{v}_{1},{v}_{2},{v}_{3}〉$ is $\text{u}\cdot \text{v}={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3}.$
The dot product satisfies the following properties:
- $\text{u}\cdot \text{v}=\text{v}\cdot \text{u}$
- $\text{u}\cdot (\text{v}+\text{w})=\text{u}\cdot \text{v}+\text{u}\cdot \text{w}$
- $c(\text{u}\cdot \text{v})=(c\text{u})\cdot \text{v}=\text{u}\cdot (c\text{v})$
- $\text{v}\cdot \text{v}={‖\text{v}‖}^{2}$
The dot product of two vectors can be expressed, alternatively, as $\text{u}\cdot \text{v}=‖\text{u}‖‖\text{v}‖\text{cos}\ \theta .$ This form of the dot product is useful for finding the measure of the angle formed by two vectors.
Vectors $\text{u}$ and $\text{v}$ are orthogonal if $\text{u}\cdot \text{v}=0.$
The angles formed by a nonzero vector and the coordinate axes are called the direction angles for the vector. The cosines of these angles are known as the direction cosines.
The vector projection of $\text{v}$ onto $\text{u}$ is the vector ${\text{proj}}_{\text{u}}\text{v}=\frac{\text{u}\cdot \text{v}}{{‖\text{u}‖}^{2}}\text{u}.$ The magnitude of this vector is known as the scalar projection of $\text{v}$ onto $\text{u}$, given by ${\text{comp}}_{\text{u}}\text{v}=\frac{\text{u}\cdot \text{v}}{‖\text{u}‖}.$
Work is done when a force is applied to an object, causing displacement. When the force is represented by the vector F and the displacement is represented by the vector s, then the work done W is given by the formula $W=\text{F}\cdot s=‖\text{F}‖‖\text{s}‖\text{cos}\ \theta .$

Key Equations

Dot product of u and v	$\begin{array}{ll}\text{u}\cdot \text{v} & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3} \\ & =‖\text{u}‖‖\text{v}‖\text{cos}\ \theta \end{array}$
Cosine of the angle formed by $\text{u}$ and $\text{v}$	$\text{cos}\ \theta =\frac{\text{u}\cdot \text{v}}{‖\text{u}‖‖\text{v}‖}$
Vector projection of $\text{v}$ onto $\text{u}$	${\text{proj}}_{\text{u}}\text{v}=\frac{\text{u}\cdot \text{v}}{{‖\text{u}‖}^{2}}\text{u}$
Scalar projection of $\text{v}$ onto $\text{u}$	${\text{comp}}_{\text{u}}\text{v}=\frac{\text{u}\cdot \text{v}}{‖\text{u}‖}$
Work done by a force F to move an object through displacement vector $\overset{\to}{PQ}$	$W=\text{F}\cdot \overset{\to}{PQ}=‖\text{F}‖‖\overset{\to}{PQ}‖\text{cos}\ \theta$

The Dot Product

For the following exercises, the vectors $\text{u}$ and $\text{v}$ are given. Calculate the dot product $\text{u}\cdot \text{v}.$

For the following exercises, the vectors a, b, and c are given. Determine the vectors $(\text{a}\cdot \text{b})\text{c}$ and $(\text{a}\cdot \text{c})\text{b}.$ Express the vectors in component form.

For the following exercises, the two-dimensional vectors a and b are given.

Find the measure of the angle $\theta$ between a and b. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.
Is $\theta$ an acute angle?

For the following exercises, find the measure of the angle between the three-dimensional vectors a and b. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.

For the following exercises determine whether the given vectors are orthogonal.

For the following exercises, determine which (if any) pairs of the following vectors are orthogonal.

For the following problems, the vector $\text{u}$ is given.

Find the direction cosines for the vector $\text{u}$.
Find the direction angles for the vector $\text{u}$ expressed in degrees. (Round the answer to the nearest integer.)

Find the vector projection $\text{w}={\text{proj}}_{\text{u}}\text{v}$ of vector $\text{v}$ onto vector $\text{u}$. Express your answer in component form.
Find the scalar projection ${\text{comp}}_{\text{u}}\text{v}$ of vector $\text{v}$ onto vector u.

Condensed — the full section is in OpenStax Calculus Volume 3.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

When running a sprint, the racers may be aided or slowed by the wind. The wind assistance is a measure of the wind speed that is helping push the runners down the track. It is much easier to run a very fast race if the wind is blowing hard in the direction of the race. So that world records aren't dependent on the weather conditions, times are only recorded as record times if the wind aiding the runners is less than or equal to 2 meters per second. Wind speed for a race is recorded by a wind gauge that is set up close to the track. It is important to note, however, that weather is not always as cooperative as we might like. The wind does not always blow exactly in the direction of the track, so the gauge must account for the angle the wind makes with the track. Suppose a 4 mile per hour wind is blowing to aid runners by making a $38^{\circ}$ angle with the race track. Determine if any times set during such a race would qualify as records.

Vis svaret

Convert miles per hour to meters per second. Then $W = 1.78816 * cos(38) = 1.409 mph$ which is less than 2 mph.
Molecular geometry is the geometry determined by arrangements of atoms in molecules. Molecular geometry includes measurements like bond angle, bond length, and torsional angles. These attributes influence several properties of molecules, such as reactivity, color, and polarity.
As an example of the molecular geometry of a molecule, consider the methane $\text{CH}_4$ molecule, as illustrated in Figure. According to the Valence Shell Electron Repulsion (VSEPR) model, atoms that surround single different atoms do so in a way that positions them as far apart as possible. This means that the hydrogen atoms in the methane molecule arrange themselves at the vertices of a regular tetrahedron. The bond angle for methane is the angle determined by two consecutive hydrogen atoms and the central carbon atom. To determine the bond angle for methane, we can place the center carbon atom at the point $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right)$ and the hydrogen atoms at the points $(0,0,0)$, $(1,1,0)$, $(1,0,1)$, and $(0,1,1)$. Find the bond angle for methane to the nearest tenth of a degree.

Vis svaret

The vector from $(0, 0, 0) to (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ is $\va = \langle \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2} \rangle$.
The vector from $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) to (0, 1, 1)$ is $\vb = \langle \frac{-1}{2}, \frac{1}{2}, \frac{1}{2} \rangle$.
$\Vert\va\Vert = \sqrt{(\frac{-1}{2})^2 + (\frac{-1}{2})^2 + (\frac{-1}{2})^2} = \sqrt{\frac{3}{4}}, \Vert\vb\Vert = \sqrt{(\frac{-1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{3}{4}}$.
$\theta = cos^-1(\frac{(\frac{-1}{2})(\frac{-1}{2}) + (\frac{-1}{2})(\frac{1}{2}) + (\frac{-1}{2})(\frac{1}{2})}{(\sqrt{\frac{3}{4}}) * (\sqrt{\frac{3}{4}})}) = 109.47^(\circ)$.
1. Find the dot product of $\text{u}=〈3,5,2〉$ and $\text{v}=〈-1,3,0〉.$
2. Find the scalar product of $\text{p}=10\text{i}-4\text{j}+7\text{k}$ and $\text{q}=-2\text{i}+\text{j}+6\text{k}.$
Vis svaret
1. Substitute the vector components into the formula for the dot product:
  \[\begin{array}{ll}\text{u}\cdot \text{v} & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3} \\ & =3(-1)+5(3)+2(0)=-3+15+0=12.\end{array}\]
2. The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:
  \[\begin{array}{ll}\text{p}\cdot \text{q} & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3} \\ & =10(-2)+(-4)(1)+(7)(6)=-20-4+42=18.\end{array}\]
Find $\text{u}\cdot \text{v},$ where $\text{u}=〈2,9,-1〉$ and $\text{v}=〈-3,1,-4〉.$

Vis svaret

7
Let $\text{a}=〈1,2,-3〉,$ $\text{b}=〈0,2,4〉,$ and $\text{c}=〈5,-1,3〉.$ Find each of the following products.
1. $(\text{a}\cdot \text{b})\text{c}$
2. $\text{a}\cdot (2\text{c})$
3. ${‖\text{b}‖}^{2}$
Vis svaret
1. Note that this expression asks for the scalar multiple of c by $\text{a}\cdot \text{b}\text{:}$
  \[\begin{array}{ll}(\text{a}\cdot \text{b})\text{c} & =(〈1,2,-3〉\cdot 〈0,2,4〉)〈5,-1,3〉 \\ & =(1(0)+2(2)+(-3)(4))〈5,-1,3〉 \\ & =-8〈5,-1,3〉 \\ & =〈-40,8,-24〉.\end{array}\]
2. This expression is a dot product of vector a and scalar multiple 2c:
  \[\begin{array}{ll}\text{a}\cdot (2\text{c}) & =2(\text{a}\cdot \text{c}) \\ & =2(〈1,2,-3〉\cdot 〈5,-1,3〉) \\ & =2(1(5)+2(-1)+(-3)(3)) \\ & =2(-6)=-12.\end{array}\]
3. Simplifying this expression is a straightforward application of the dot product:
  \[{‖\text{b}‖}^{2}=\text{b}\cdot \text{b}=〈0,2,4〉\cdot 〈0,2,4〉={0}^{2}+{2}^{2}+{4}^{2}=0+4+16=20.\]
Find the following products for $\text{p}=〈7,0,2〉,$ $\text{q}=〈-2,2,-2〉,$ and $\text{r}=〈0,2,-3〉.$
1. $(\text{r}\cdot \text{p})\text{q}$
2. ${‖\text{p}‖}^{2}$
Vis svaret

a. $(\text{r}\cdot \text{p})\text{q}=〈12,-12,12〉;$ b. ${‖\text{p}‖}^{2}=53$
Find the measure of the angle between each pair of vectors.
1. i + j + k and 2i – j – 3k
2. $〈2,5,6〉$ and $〈-2,-4,4〉$
Vis svaret
1. To find the cosine of the angle formed by the two vectors, substitute the components of the vectors into :
  \[\begin{array}{ll}\text{cos}\ \theta & =\frac{(\text{i}+\text{j}+\text{k})\cdot (2\text{i}-\text{j}-3\text{k})}{‖\text{i}+\text{j}+\text{k}‖\cdot ‖2\text{i}-\text{j}-3\text{k}‖} \\ & =\frac{1(2)+(1)(-1)+(1)(-3)}{\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\ \sqrt{{2}^{2}+{(-1)}^{2}+{(-3)}^{2}}} \\ & =\frac{-2}{\sqrt{3}\ \sqrt{14}}=\frac{-2}{\sqrt{42}}.\end{array}\]
  Therefore, $\theta =\text{arccos}\ \frac{-2}{\sqrt{42}}$.
2. Start by finding the value of the cosine of the angle between the vectors:
  \[\begin{array}{ll}\text{cos}\ \theta & =\frac{〈2,5,6〉\cdot 〈-2,-4,4〉}{‖〈2,5,6〉‖\cdot ‖〈-2,-4,4〉‖} \\ & =\frac{2(-2)+(5)(-4)+(6)(4)}{\sqrt{{2}^{2}+{5}^{2}+{6}^{2}}\ \sqrt{{(-2)}^{2}+{(-4)}^{2}+{4}^{2}}} \\ & =\frac{0}{\sqrt{65}\ \sqrt{36}}=0.\end{array}\]
  Now, $\text{cos}\ \theta =0$ and $0\le \theta \le \pi ,$ so $\theta =\pi \text{/}2.$
Find the measure of the angle, in radians, formed by vectors $\text{a}=〈1,2,0〉$ and $\text{b}=〈2,4,1〉.$ Round to the nearest hundredth.

Vis svaret

$\theta \approx 0.22$ rad
Determine whether $\text{p}=〈1,0,5〉$ and $\text{q}=〈10,3,-2〉$ are orthogonal vectors.

Vis svaret

Using the definition, we need only check the dot product of the vectors:
\[\text{p}\cdot \text{q}=1(10)+(0)(3)+(5)(-2)=10+0-10=0.\]
Because $\text{p}\cdot \text{q}=0,$ the vectors are orthogonal ().
For which value of x is $\text{p}=〈2,8,-1〉$ orthogonal to $\text{q}=〈x,-1,2〉?$

Vis svaret

$x=5$
Let $\text{v}=〈2,3,3〉.$ Find the measures of the angles formed by the following vectors.
1. $\text{v}$ and i
2. $\text{v}$ and j
3. $\text{v}$ and k
Vis svaret
1. Let α be the angle formed by $\text{v}$ and i:
  \[\begin{array}{ll}\text{cos}\ \alpha & =\frac{\text{v}\cdot \text{i}}{‖\text{v}‖\cdot ‖\text{i}‖} \\ & =\frac{〈2,3,3〉\cdot 〈1,0,0〉}{\sqrt{{2}^{2}+{3}^{2}+{3}^{2}}\ \sqrt{1}} \\ & =\frac{2}{\sqrt{22}}.\end{array}\]
  \[\alpha =\text{arccos}\ \frac{2}{\sqrt{22}}\approx 1.130\ \text{rad}.\]
2. Let β represent the angle formed by $\text{v}$ and j:
  \[\begin{array}{ll}\text{cos}\ \beta & =\frac{\text{v}\cdot \text{j}}{‖\text{v}‖\cdot ‖\text{j}‖} \\ & =\frac{〈2,3,3〉\cdot 〈0,1,0〉}{\sqrt{{2}^{2}+{3}^{2}+{3}^{2}}\ \sqrt{1}} \\ & =\frac{3}{\sqrt{22}}.\end{array}\]
  \[\beta =\text{arccos}\ \frac{3}{\sqrt{22}}\approx 0.877\ \text{rad.}\]
3. Let γ represent the angle formed by $\text{v}$ and k:
  \[\begin{array}{ll}\text{cos}\ \gamma & =\frac{\text{v}\cdot \text{k}}{‖\text{v}‖\cdot ‖\text{k}‖} \\ & =\frac{〈2,3,3〉\cdot 〈0,0,1〉}{\sqrt{{2}^{2}+{3}^{2}+{3}^{2}}\ \sqrt{1}} \\ & =\frac{3}{\sqrt{22}}.\end{array}\]
  \[\gamma =\text{arccos}\ \frac{3}{\sqrt{22}}\approx 0.877\ \text{rad.}\]
Let $\text{v}=〈3,-5,1〉.$ Find the measure of the angles formed by each pair of vectors.
1. $\text{v}$ and i
2. $\text{v}$ and j
3. $\text{v}$ and k
Vis svaret

a. $\alpha \approx 1.04$ rad; b. $\beta \approx 2.58$ rad; c. $\gamma \approx 1.40$ rad
AAA Party Supply Store sells invitations, party favors, decorations, and food service items such as paper plates and napkins. When AAA buys its inventory, it pays 25¢ per package for invitations and party favors. Decorations cost AAA 50¢ each, and food service items cost 20¢ per package. AAA sells invitations for $2.50 per package and party favors for $1.50 per package. Decorations sell for $4.50 each and food service items for $1.25 per package.
During the month of May, AAA Party Supply Store sells 1258 invitations, 342 party favors, 2426 decorations, and 1354 food service items. Use vectors and dot products to calculate how much money AAA made in sales during the month of May. How much did the store make in profit?

Vis svaret

The cost, price, and quantity vectors are
\[\begin{array}{l}\text{c}=〈0.25,0.25,0.50,0.20〉 \\ \text{p}=〈2.50,1.50,4.50,1.25〉 \\ \text{q}=〈1258,342,2426,1354〉.\end{array}\]
AAA sales for the month of May can be calculated using the dot product $\text{p}\cdot \text{q}.$ We have
\[\begin{array}{ll}\text{p}\cdot \text{q} & =〈2.50,1.50,4.50,1.25〉\cdot 〈1258,342,2426,1354〉 \\ & =3145+513+10917+1692.5 \\ & =16267.5.\end{array}\]
So, AAA took in $16,267.50 during the month of May.
To calculate the profit, we must first calculate how much AAA paid for the items sold. We use the dot product $\text{c}\cdot \text{q}$ to get
\[\begin{array}{ll}\text{c}\cdot \text{q} & =〈0.25,0.25,0.50,0.20〉\cdot 〈1258,342,2426,1354〉 \\ & =314.5+85.5+1213+270.8 \\ & =1883.8.\end{array}\]
So, AAA paid $1,883.80 for the items they sold. Their profit, then, is given by
\[\begin{array}{ll}\text{p}\cdot \text{q}-\text{c}\cdot \text{q} & =16267.5-1883.8 \\ & =14383.7.\end{array}\]
Therefore, AAA Party Supply Store made $14,383.70 in May.
On June 1, AAA Party Supply Store decided to increase the price they charge for party favors to $2 per package. They also changed suppliers for their invitations, and are now able to purchase invitations for only 10¢ per package. All their other costs and prices remain the same. If AAA sells 1408 invitations, 147 party favors, 2112 decorations, and 1894 food service items in the month of June, use vectors and dot products to calculate their total sales and profit for June.

Vis svaret

Sales = $15,685.50; profit = $14,073.15
Find the projection of $\text{v}$ onto u.
1. $\text{v}=〈3,5,1〉$ and $\text{u}=〈-1,4,3〉$
2. $\text{v}=3\text{i}-2\text{j}$ and $\text{u}=\text{i}+6\text{j}$
Vis svaret
1. Substitute the components of $\text{v}$ and $\text{u}$ into the formula for the projection:
  \[\begin{array}{ll}{\text{proj}}_{\text{u}}\text{v} & =\frac{\text{u}\cdot \text{v}}{{‖\text{u}‖}^{2}}\text{u} \\ & =\frac{〈-1,4,3〉\cdot 〈3,5,1〉}{{‖〈-1,4,3〉‖}^{2}}〈-1,4,3〉 \\ & =\frac{-3+20+3}{{(-1)}^{2}+{4}^{2}+{3}^{2}}〈-1,4,3〉 \\ & =\frac{20}{26}〈-1,4,3〉 \\ & =〈-\frac{10}{13},\frac{40}{13},\frac{30}{13}〉.\end{array}\]
2. To find the two-dimensional projection, simply adapt the formula to the two-dimensional case:
  \[\begin{array}{ll}{\text{proj}}_{\text{u}}\text{v} & =\frac{\text{u}\cdot \text{v}}{{‖\text{u}‖}^{2}}\text{u} \\ & =\frac{(\text{i}+6\text{j})\cdot (3\text{i}-2\text{j})}{{‖\text{i}+6\text{j}‖}^{2}}(\text{i}+6\text{j}) \\ & =\frac{1(3)+6(-2)}{{1}^{2}+{6}^{2}}(\text{i}+6\text{j}) \\ & =-\frac{9}{37}(\text{i}+6\text{j}) \\ & =-\frac{9}{37}\text{i}-\frac{54}{37}\text{j}.\end{array}\]
Express $\text{v}=〈8,-3,-3〉$ as a sum of orthogonal vectors such that one of the vectors has the same direction as $\text{u}=〈2,3,2〉.$

Vis svaret

Let p represent the projection of $\text{v}$ onto $\text{u}$:
\[\begin{array}{ll}\text{p} & ={\text{proj}}_{\text{u}}\text{v} \\ & =\frac{\text{u}\cdot \text{v}}{{‖\text{u}‖}^{2}}\text{u} \\ & =\frac{〈2,3,2〉\cdot 〈8,-3,-3〉}{{‖〈2,3,2〉‖}^{2}}〈2,3,2〉 \\ & =\frac{16-9-6}{{2}^{2}+{3}^{2}+{2}^{2}}〈2,3,2〉 \\ & =\frac{1}{17}〈2,3,2〉 \\ & =〈\frac{2}{17},\frac{3}{17},\frac{2}{17}〉.\end{array}\]
Then,
\[\text{q}=\text{v}-\text{p}=〈8,-3,-3〉-〈\frac{2}{17},\frac{3}{17},\frac{2}{17}〉=〈\frac{134}{17},-\frac{54}{17},-\frac{53}{17}〉.\]
To check our work, we can use the dot product to verify that p and $\text{q}$ are orthogonal vectors:
\[\text{p}\cdot \text{q}=〈\frac{2}{17},\frac{3}{17},\frac{2}{17}〉\cdot 〈\frac{134}{17},-\frac{54}{17},-\frac{53}{17}〉=\frac{268}{289}-\frac{162}{289}-\frac{106}{289}=0.\]
Then,
\[\text{v}=\text{p}+\text{q}=〈\frac{2}{17},\frac{3}{17},\frac{2}{17}〉+〈\frac{134}{17},-\frac{54}{17},-\frac{53}{17}〉.\]
Express $\text{v}=5\text{i}-\text{j}$ as a sum of orthogonal vectors such that one of the vectors has the same direction as $\text{u}=4\text{i}+2\text{j}.$

Vis svaret

$\text{v}=\text{p}+\text{q},$ where $\text{p}=\frac{18}{5}\text{i}+\frac{9}{5}\text{j}$ and $\text{q}=\frac{7}{5}\text{i}-\frac{14}{5}\text{j}$
A container ship leaves port traveling $15\text{^{\circ}}$ north of east. Its engine generates a speed of 20 knots along that path (see the following figure). In addition, the ocean current moves the ship northeast at a speed of 2 knots. Considering both the engine and the current, how fast is the ship moving in the direction $15\text{^{\circ}}$ north of east? Round the answer to two decimal places.

Vis svaret

Let $\text{v}$ be the velocity vector generated by the engine, and let $\text{w}$ be the velocity vector of the current. We already know $‖\text{v}‖=20$ along the desired route. We just need to add in the scalar projection of $\text{w}$ onto $\text{v}$. We get
\[\begin{array}{ll}{\text{comp}}_{\text{v}}\text{w} & =\frac{\text{v}\cdot \text{w}}{‖\text{v}‖} \\ & =\frac{‖\text{v}‖‖\text{w}‖\text{cos}(30\text{^{\circ}})}{‖\text{v}‖} \\ & =‖\text{w}‖\text{cos}(30\text{^{\circ}}) \\ & =2\frac{\sqrt{3}}{2}=\sqrt{3}\approx 1.73\ \text{knots}.\end{array}\]
The ship is moving at 21.73 knots in the direction $15\text{^{\circ}}$ north of east.
Repeat the previous example, but assume the ocean current is moving southeast instead of northeast, as shown in the following figure.

Vis svaret

21 knots
A conveyor belt generates a force $\text{F}=5\text{i}-3\text{j}+\text{k}$ that moves a suitcase from point $(1,1,1)$ to point $(9,4,7)$ along a straight line. Find the work done by the conveyor belt. The distance is measured in meters and the force is measured in newtons.

Vis svaret

The displacement vector $\overset{\to}{PQ}$ has initial point $(1,1,1)$ and terminal point $(9,4,7)\text{:}$
\[\overset{\to}{PQ}=〈9-1,4-1,7-1〉=〈8,3,6〉=8\text{i}+3\text{j}+6\text{k}.\]
Work is the dot product of force and displacement:
\[\begin{array}{ll}W & =\text{F}\cdot \overset{\to}{PQ} \\ & =(5\text{i}-3\text{j}+\text{k})\cdot (8\text{i}+3\text{j}+6\text{k}) \\ & =5(8)+(-3)(3)+1(6) \\ & =37\text{N}\cdot \text{m} \\ & =37\ \text{J}.\end{array}\]
A constant force of 30 lb is applied at an angle of 60° to pull a handcart 10 ft across the ground (). What is the work done by this force?

Vis svaret

150 ft-lb
$\text{u}=〈3,0〉,$ $\text{v}=〈2,2〉$

Vis svaret

6
$\text{u}=〈3,-4〉,$ $\text{v}=〈4,3〉$
$\text{u}=〈2,2,-1〉,$ $\text{v}=〈-1,2,2〉$

Vis svaret

0
$\text{u}=〈4,5,-6〉,$ $\text{v}=〈0,-2,-3〉$
$\text{a}=〈2,0,-3〉,$ $\text{b}=〈-4,-7,1〉,$ $\text{c}=〈1,1,-1〉$

Vis svaret

$(\text{a}\cdot \text{b})\text{c}=〈-11,-11,11〉;$ $(\text{a}\cdot \text{c})\text{b}=〈-20,-35,5〉$
$\text{a}=〈0,1,2〉,$ $\text{b}=〈-1,0,1〉,$ $\text{c}=〈1,0,-1〉$
$\text{a}=\text{i}+\text{j},$ $\text{b}=\text{i}-\text{k},$ $\text{c}=\text{i}-2\text{k}$

Vis svaret

$(\text{a}\cdot \text{b})\text{c}=〈1,0,-2〉;$ $(\text{a}\cdot \text{c})\text{b}=〈1,0,-1〉$
$\text{a}=\text{i}-\text{j}+\text{k},$ $\text{b}=\text{j}+3\text{k},$ $\text{c}=\text{-}\text{i}+2\text{j}-4\text{k}$
[T] $\text{a}=〈3,-1〉,$ $\text{b}=〈-4,0〉$

Vis svaret

a. $\theta =2.82$ rad; b. $\theta$ is not acute.
[T] $\text{a}=〈2,1〉,$ $\text{b}=〈-1,3〉$
$\text{u}=3\text{i},$ $\text{v}=4\text{i}+4\text{j}$

Vis svaret

a. $\theta =\frac{\pi }{4}$ rad; b. $\theta$ is acute.
$\text{u}=5\text{i},$ $\text{v}=-6\text{i}+6\text{j}$
$\text{a}=〈3,-1,2〉,$ $\text{b}=〈1,-1,-2〉$

Vis svaret

$\theta =\frac{\pi }{2}$
$\text{a}=〈0,-1,-3〉,$ $\text{b}=〈2,3,-1〉$
$\text{a}=\text{i}+\text{j},$ $\text{b}=\text{j}-\text{k}$

Vis svaret

$\theta =\frac{\pi }{3}$
$\text{a}=\text{i}-2\text{j}+\text{k},$ $\text{b}=\text{i}+\text{j}-2\text{k}$
[T] $\text{a}=3\text{i}-\text{j}-2\text{k},$ $\text{b}=\text{v}+\text{w},$ where $\text{v}=-2\text{i}-3\text{j}+2\text{k}$ and $\text{w}=\text{i}+2\text{k}$

Vis svaret

$\theta =2$ rad
[T] $\text{a}=3\text{i}-\text{j}+2\text{k},$ $\text{b}=\text{v}-\text{w},$ where $\text{v}=2\text{i}+\text{j}+4\text{k}$ and $\text{w}=6\text{i}+\text{j}+2\text{k}$
$\text{a}=〈x,y〉,$ $\text{b}=〈\text{-}y,x〉,$ where x and y are nonzero real numbers

Vis svaret

Orthogonal

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

imaginary unit
i² = −1.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: The Dot Product

Calculate the dot product of two given vectors.
Determine whether two given vectors are perpendicular.
Find the direction cosines of a given vector.
Explain what is meant by the vector projection of one vector onto another vector, and describe how to compute it.
Calculate the work done by a given force.
Find the dot product of
Find the scalar product of
Substitute the vector components into the formula for the dot product:

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

Prøv din egen

Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Mer i Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems

Dot product of \(\)u and v	\(\begin{array}{ll}\text{u}\cdot \text{v} & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3} \\ & =‖\text{u}‖‖\text{v}‖\text{cos}\ \theta \end{array}\)
Cosine of the angle formed by \(\text{u}\) and \(\text{v}\)	\(\text{cos}\ \theta =\frac{\text{u}\cdot \text{v}}{‖\text{u}‖‖\text{v}‖}\)
Vector projection of \(\text{v}\) onto \(\text{u}\)	\({\text{proj}}_{\text{u}}\text{v}=\frac{\text{u}\cdot \text{v}}{{‖\text{u}‖}^{2}}\text{u}\)
Scalar projection of \(\text{v}\) onto \(\text{u}\)	\({\text{comp}}_{\text{u}}\text{v}=\frac{\text{u}\cdot \text{v}}{‖\text{u}‖}\)
Work done by a force F to move an object through displacement vector \(\overset{\to}{PQ}\)	\(W=\text{F}\cdot \overset{\to}{PQ}=‖\text{F}‖‖\overset{\to}{PQ}‖\text{cos}\ \theta\)