The Dot Product

If we apply a force to an object so that the object moves, we say that work is done by the force. In Introduction to Applications of Integration on integration applications, we looked at a constant force and we assumed the force was applied in the direction of motion of the object. Under those conditions, work can be expressed as the product of the force acting on an object and the distance the object moves. In this chapter, however, we have seen that both force and the motion of an object can be represented by vectors.

In this section, we develop an operation called the dot product, which allows us to calculate work in the case when the force vector and the motion vector have different directions. The dot product essentially tells us how much of the force vector is applied in the direction of the motion vector. The dot product can also help us measure the angle formed by a pair of vectors and the position of a vector relative to the coordinate axes. It even provides a simple test to determine whether two vectors meet at a right angle.

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 u and v are two-dimensional vectors, we calculate the dot product in a similar fashion. Thus, if $\text{u}=\langle u_1,u_2\rangle$

and $\text{v}=\langle v_1,v_2\rangle ,$

then

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.

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

Proof

Let $\text{u}=\langle u_1,u_2,u_3\rangle$

and $\text{v}=\langle v_1,v_2,v_3\rangle .$

Then

The associative property looks like the associative property for real-number multiplication, but pay close attention to the difference between scalar and vector objects:

The proof that $c\left(\text{u}·\text{v}\right)=\text{u}·\left(c\text{v}\right)$

is similar.

The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:

□

Note that the definition of the dot product yields $0·\text{v}=0.$

By property iv., if $\text{v}·\text{v}=0,$

then $\text{v}=0.$

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 ([link]). 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.

Proof

Place vectors u and v in standard position and consider the vector $\text{v}-\text{u}$

([link]). These three vectors form a triangle with side lengths $\Vert \text{u}\Vert ,\Vert \text{v}\Vert ,\text{and}\Vert \text{v}-\text{u}\Vert .$

Recall from trigonometry that the law of cosines describes the relationship among the side lengths of the triangle and the angle θ. Applying the law of cosines here gives

The dot product provides a way to rewrite the left side of this equation:

Substituting into the law of cosines yields

□

We can use this form of the dot product to find the measure of the angle between two nonzero vectors. The following equation rearranges [link] to solve for the cosine of the angle:

Using this equation, we can find the cosine of the angle between two nonzero vectors. Since we are considering the smallest angle between the vectors, we assume $0\text{°}\le \theta \le 180\text{°}$

(or $0\le \theta \le \pi$

if we are working in radians). The inverse cosine is unique over this range, so we are then able to determine the measure of the angle $\theta .$

The angle between two vectors can be acute $\left(0<\text{cos}\theta <1\right),$

obtuse $\left(\mathrm{-1}<\text{cos}\theta <0\right),$

or straight $\left(\text{cos}\theta =\mathrm{-1}\right).$

If $\text{cos}\theta =1,$

then both vectors have the same direction. If $\text{cos}\theta =0,$

then the vectors, when placed in standard position, form a right angle ([link]). We can formalize this result into a theorem regarding orthogonal (perpendicular) vectors.

Proof

Let u and v be nonzero vectors, and let $\theta$

denote the angle between them. First, assume $\text{u}·\text{v}=0.$

Then

However, $\Vert \text{u}\Vert \ne 0$

and $\Vert \text{v}\Vert \ne 0,$

so we must have $\text{cos}\theta =0.$

Hence, $\theta =90\text{°},$

and the vectors are orthogonal.

Now assume u and v are orthogonal. Then $\theta =90\text{°}$

and we have

□

The terms orthogonal, perpendicular, and normal each indicate that mathematical objects are intersecting at right angles. The use of each term is determined mainly by its context. We say that vectors are orthogonal and lines are perpendicular. The term normal is used most often when measuring the angle made with a plane or other surface.

The angle a vector makes with each of the coordinate axes, called a direction angle, is very important in practical computations, especially in a field such as engineering. For example, in astronautical engineering, the angle at which a rocket is launched must be determined very precisely. A very small error in the angle can lead to the rocket going hundreds of miles off course. Direction angles are often calculated by using the dot product and the cosines of the angles, called the direction cosines. Therefore, we define both these angles and their cosines.

In [link], the direction cosines of $\text{v}=\langle 2,3,3\rangle$

are $\text{cos}\alpha =\frac{2}{\sqrt{22}},$

and $\text{cos}\gamma =\frac{3}{\sqrt{22}}.$

The direction angles of v are $\alpha =1.130\text{rad},$

and $\gamma =0.877\text{rad}.$

So far, we have focused mainly on vectors related to force, movement, and position in three-dimensional physical space. However, vectors are often used in more abstract ways. For example, suppose a fruit vendor sells apples, bananas, and oranges. On a given day, he sells 30 apples, 12 bananas, and 18 oranges. He might use a quantity vector, $\text{q}=\langle 30,12,18\rangle ,$

to represent the quantity of fruit he sold that day. Similarly, he might want to use a price vector, $\text{p}=\langle 0.50,0.25,1\rangle ,$

to indicate that he sells his apples for 50¢ each, bananas for 25¢ each, and oranges for $1 apiece. In this example, although we could still graph these vectors, we do not interpret them as literal representations of position in the physical world. We are simply using vectors to keep track of particular pieces of information about apples, bananas, and oranges.

This idea might seem a little strange, but if we simply regard vectors as a way to order and store data, we find they can be quite a powerful tool. Going back to the fruit vendor, let’s think about the dot product, $\text{q}·\text{p}.$

We compute it by multiplying the number of apples sold (30) by the price per apple (50¢), the number of bananas sold by the price per banana, and the number of oranges sold by the price per orange. We then add all these values together. So, in this example, the dot product tells us how much money the fruit vendor had in sales on that particular day.

When we use vectors in this more general way, there is no reason to limit the number of components to three. What if the fruit vendor decides to start selling grapefruit? In that case, he would want to use four-dimensional quantity and price vectors to represent the number of apples, bananas, oranges, and grapefruit sold, and their unit prices. As you might expect, to calculate the dot product of four-dimensional vectors, we simply add the products of the components as before, but the sum has four terms instead of three.

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 ([link]). We return to this example and learn how to solve it after we see how to calculate projections.

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 $\text{v}=\langle 6,\mathrm{-4}\rangle$

and let $\text{u}=\langle 3,1\rangle .$

We want to decompose the vector v into orthogonal components such that one of the component vectors has the same direction as u.

We first find the component that has the same direction as u by projecting v onto u. Let $\text{p}={\text{proj}}_{\text{u}}\text{v}.$

Then, we have

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

We have

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

and

Therefore, q and p are orthogonal.

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 ([link]).

We have

In U.S. standard units, we measure the magnitude of force $\Vert \text{F}\Vert$

in pounds. The magnitude of the displacement vector $\Vert \overrightarrow{PQ}\Vert$

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).

Key Concepts

• The dot product, or scalar product, of two vectors $\text{u}=\langle u_1,u_2,u_3\rangle$

  and

  $\text{v}=\langle v_1,v_2,v_3\rangle$

  is

  $\text{u}·\text{v}=u_1{v}_1+u_2{v}_2+u_3{v}_3.$
• The dot product satisfies the following properties:
  • $\text{u}·\text{v}=\text{v}·\text{u}$
  • $\text{u}·\left(\text{v}+\text{w}\right)=\text{u}·\text{v}+\text{u}·\text{w}$
  • $c\left(\text{u}·\text{v}\right)=\left(c\text{u}\right)·\text{v}=\text{u}·\left(c\text{v}\right)$
  • $\text{v}·\text{v}={\Vert \text{v}\Vert }^2$
• The dot product of two vectors can be expressed, alternatively, as $\text{u}·\text{v}=\Vert \text{u}\Vert \Vert \text{v}\Vert \text{cos}\theta .$

  This form of the dot product is useful for finding the measure of the angle formed by two vectors.

• Vectors u and v are orthogonal if $\text{u}·\text{v}=0.$
• 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 v onto u is the vector ${\text{proj}}_{\text{u}}\text{v}=\frac{\text{u}·\text{v}}{{\Vert \text{u}\Vert }^2}\text{u}.$

  The magnitude of this vector is known as the scalar projection of v onto u, given by

  ${\text{comp}}_{\text{u}}\text{v}=\frac{\text{u}·\text{v}}{\Vert \text{u}\Vert }.$
• 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 $W=\text{F}·s=\Vert \text{F}\Vert \Vert \text{s}\Vert \text{cos}\theta .$

Key Equations

• Dot product of u and v

  ---

  $\begin{array}{cc}\hfill \text{u}·\text{v}& =u_1{v}_1+u_2{v}_2+u_3{v}_3\hfill \\ & =\Vert \text{u}\Vert \Vert \text{v}\Vert \text{cos}\theta \hfill \end{array}$
• Cosine of the angle formed by u and v

  ---

  $\text{cos}\theta =\frac{\text{u}·\text{v}}{\Vert \text{u}\Vert \Vert \text{v}\Vert }$
• Vector projection of v onto u

  ---

  ${\text{proj}}_{\text{u}}\text{v}=\frac{\text{u}·\text{v}}{{\Vert \text{u}\Vert }^2}\text{u}$
• Scalar projection of v onto u

  ---

  ${\text{comp}}_{\text{u}}\text{v}=\frac{\text{u}·\text{v}}{\Vert \text{u}\Vert }$
• Work done by a force F to move an object through displacement vector $\overrightarrow{PQ}$