Vectors_ the Dot and Cross Products_no Tensors

7/31/2019 Vectors_ the Dot and Cross Products_no Tensors

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THE DOT (INNER) AND CROSS (VECTOR) PRODUCTS

The Dot product

There are two important products involving vectors: the dot (inner or scalar) product and the

cross (or vector) productand both operations arise naturally in many applications. As their names

imply, the scalar product results in a real number whereas the cross product results in another vector.

We begin with the dot product:

The dot product at this time is a completely geometrical thing. It is determined be measuring lengths

and an angle, both geometrical concepts. Let us see how it naturally arises as aprojection problem.

Let

and

be vectors as shown below (in space or the plane). We are interested in finding the

lengthof theprojection of on (think of it as the shadow that casts along ), which in the figureis denoted by l.

Figure 1

Clearly, the lengthl is related to andby the cosine function: cos. We are done! Thatis all we wanted. If we knew the length of

and the angle

we would perform the operation above

and obtain the answer. However, what if all we are given are the vectors and ? Is there a way toobtain l if we only know the vectors components?Yes. Here is our first important result of this section:

This is a remarkable result because it tell us that there is no need to measure the angle between the

vectors - all we need to know is their components. We will show that these two versions of the dot

product yield the same number.

Example 1 Find if 1,1,3 and 0,1,4Solution 1 0 1 1 3 4 1 1Example 2 Find if 1,1,2 and 3,1,4Solution 1 3 1 1 2 4 10

cos

The dot product: Let uand be two vectors and let be the angle between them, defined to bethe angle generated when one vector is rotated toward the other through an angle in the range0,. Then the dot productof and , denoted by , is given by

l

If , , and , , Then


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Example 3 Find if 3,6 and 2,1Solution 3 2 6 1 0Example 4 Find the dot product of the vectors 2 0 . 5 and 0 . 2 5 Solution 2 0 0 . 5 1 0 0 . 2 5 0 . 5 So the dot product can be any real number, positive, negative, or zero.

Before establishing the equivalence of the two versions of the dot product, let us see how it helps us

with the projection problem.

We found that cos gives us the desired projection. Observe that we can get in thepicture as follows:

c o s cos ; We have a formula that does not rely on the measurement of an angle:

a completely algebraic formula! We call the component of along and denote it by :

So the dot product naturally arises in this context.

Example 5 find in the figure below given that 8, 10, and = 55o.

Solution

cos 8cos554.59

So the length of the shadow is approximately 4.59 linear units (cm for example). Notice that we didnot even bother to use since it cancels out anyway.Example6 find in the figure below given that 8, 10, and = 120o.

a

b

l


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Solution

this is very similar to the previous example but 9 0 so the shadow ofu lies over anextension of, as shown. It is tempting to use the angle instead of using , in which casethe projections length would be given by c o s 4 units.On the other hand, the definition tells us that

cos 810cos15010

cos 4

which is the correct answer and disagrees with our previous result. The key to mathematics

is to understand its definitions. They say precisely what is to be done and anything you do

outside their scope, even if you think you are simplifying the problem, can lead to trouble.

So unless you are certain that what you are doing is something logically equivalent to what a

definition states, stick to the definition.

Example 7 Find the component of on the vector Solution

We don not have the angle between the vectors, but we can use the algebraic version of

the dot product:

1,1,0 0,0,11

0This makes sense because

is a vector on the x-y plane whereas

is one the basis unit

vectors the one along the z-axis and is therefore perpendicular to .Now let us see if the two versions of the dot product are equivalent, first in . Consider thefollowing illustration:

Figure 2

Writing the vectors in trigonometric form:

c o s s i n c o s s i n

Where we use and for the magnitudes of and , respectively.


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The algebraic version of the dot product says that

Applying this formula to and:

c o s c o s s i n s i n cos cos sin s inThe term in parenthesis is cos , that is, the cosine of the angle between the vectors. Thus, cos

where . Notice that because the cosine is an even functioncos cos

so it does not matter in which order we take the difference.

The general case is treated a bit different. In the figure below, we use the vectors to form a triangle

which now may be in . By applying the law of cosines, we obtain an algebraic expression thatconnects the angle to the magnitude of the vectors.

Figure 3

Note:cis the vector that added toagives usb: a + c = b. Therefore, c = ba.

Now we apply the law of cosines:

2 cosSince we want a connection between the geometric formula and an algebraic one involving thecomponents of the vectors, let , , and , , . Then, , , , , , ,

2, , , , cosFrom which we obtain:

2 cos

Expanding,

2 2 2 2 cosCancelling from both sides of the equation:

2 2 2 2 cosFrom which it follows that cos

ab

c = b

a


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So there we have it. These two versions are equivalent and, depending on the information provided,

one can be a better option than the other.

This result has some interesting consequences. First observe that, solving forcos,

c o s

Hence, if we know the components of the vectors we can find the cosine of the angle between them

and therefore the angle itself.

Example 8 The figure below shows two vectors in space. Find the angle between them.

Figure 4

Solutionthe given vectors define a plane in space. The cosine of the angle between them is given by

c o s where 3,0,5 and 1,2,1. Substituting:

c o s 3 0 534 6

451To find , we use the cos1function: cos1 55.9We will see the dot product play an important role outside geometry soon.

In the previous example, would make any difference whether we let 3,0,5 and 1,2,1 asopposed to 3,0,5 and 1,2,1? No, the scalar product is commutative:

If two vectors are perpendicular, then the angle between them is

90o. A fancy word for

perpendicular is orthogonal. The reason for this word is because as we said in the previous lecture,not everything that is a vector is a directed line segment. It would not make sense for example to

say that two polynomials are perpendicular, but as we saw earlier certain sets of polynomials form

vector spaces so they are vectors. Hence the need for the new term.

Orthogonal Vectors

Two non-zero vectors and are orthogonal(perpendicular) if, and only if, 0The 0 vector is considered to be orthogonal to every vector.

3,0,51,2,1

For any two vectors and,


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This result follows immediately from the relation

c o s First, ifand are nonzero and 0, then c o s 0 9 0 .On the other hand, ifand are nonzero and 9 0 , then 0which implies that 0 or 0.Example 9 Determine if the vectors 1,4,2 and 6,0,3 are orthogonal.Solution 1 6 4 0 2 3 0

Yes, they are orthogonal.

Example 10 (application) A roof top has the shape shown in the figure below. Find the angle

between the two solid lines.

Figure 5Solution

Ok, my sense of perspective is not the best, but I think (or hope) you get the point. Let us

introduce a coordinate system as shown:

Figure 6

20

1544

20

15

z

y

x

20

15 44

20

15


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Observe that one of the vectors lies on the y-z plane and the other has a negative x-

component. The vector on the y-z plane can be described by 0 5 4 . The othervector, which we denote by , starts at the origin and ends at 7.5,5,4. Therefore, 7.5,5,4.The hard part is over. Now we simply apply the formula

c o s to obtain

c o s 0 7 . 5 5 5 4 46.409.86 0.6497Thus, cos1 0.649749.47

Direction Angles and Cosines

For vectors in it is sometimes convenient to describe their direction by means ofdirection anglesand direction cosines. The direction angles of a vector are the angles that the vector makes the

coordinate axes. The direction cosines are, of course, the cosine of these angles.

Figure 7

What happens when we take the dot product of a vectora and a basis vector? Say we compute ;since the angle between and is ,

c o s cosOn the other hand, , , 1,0,0 Thus,

cos

or

c o s Similarly,

c o s and cos These are called the direction cosines of the vector and the angles , , and are called itsdirection angles.

a

y

x

z


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Example 11 Find the direction cosines of the vector 5 3 Solution

c o s 535

c o s 335c o s 135The direction angles are 147.69, 59.53, and 80.27.The importance of these numbers is the following: they form the components of a unit vectorin the

direction of its vector , , . Observe:Since

c o s ; c o s ; c o s We have

cos c o s c o s 1

Therefore the vector cos,cos,cos is a unit vector. What is its direction? We have: cos ; cos ; cos

Hence, , , cos,cos,coswhich says that the vectorand the vector are parallel, and since is a unit vector we put a haton it and write: cos,cos,cosWork Done by a Force

Suppose that a constantforce F acts on an object as it moves through a distance s along a straight

line. A measure of the amount ofwork,W, done by the force is given by the product of theparallelcomponentof the force along the displacement and the magnitude of that displacement. This is

how work is defined in physics.

where is the component of along the direction of motion. This is not a dot product yet! Both and are scalars and the product is just a product of real numbers. However, we can make thedefinition more precise and convey more information with it if we introduce the vector whichrepresents the displacement vector relative to a pre-assigned origin. The figure below illustrates this.


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Figure 8

The force is decomposed into two components: which represents the component perpendicularto displacement, and which is the component parallel to the displacement. We have:

cos sin

Thus, c o s cosor equivalently,

Since 0 1 8 0 , the work done by a force is positive when the force acts in the direction ofmotion, zero when the force is perpendicular to the displacement and negative when the force acts in

the direction opposing motion (a retarding force).

Example 12 Suppose that a force 2,4,0 acts on an object as it moves along a straight linefrom point 1,0,1 to point 3,4,6. What is the work done by the force?Solution

The work done by the force is .F

is constant force ands

is the displacementvector from point 1,0,2 to point 3,4,6. Careful here, you must have the correctdisplacement vector, which in this case is given by 31,40,62 2,4,8.Thus, the work done is 2,4,0 2,4,8

4 1 6 012

The unit of work is the Joule when displacement is measured in meters (m) and force in Newtons

(Nt.) The Unit is theft-lb (foot-pound) when displacement is measured in feet and force in pounds.

In the previous example, in the metric system the work done by a force

2,4,0Nt. over a distance

of84 meters is 12 Joules. 84 meters is the distance from 1,0,2 to 3,4,6.Example 13 Of all the force with which we have to live, gravity if by far the most familiar one. Weall have fallen down and know that we cannot escape its influence.

Since gravity is a force, we can describe it by a vector, and if we choose up as the positive z

direction, then the force of gravity on a mass m is given byF= mgk where gis the accelerationdue to the force of gravity near the surface of the earth, whose magnitude is 32 ft/sec2 in the English

system or 9.8 m/sec2 in the metric system.

Which of the following would result in the largest value forW, the work done by gravity?

F

s

s


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A) an object moving horizontally a distance of 40 meters, B) an object moving vertically downwarda distance of4 meters, orC) an object moving straight up a distance of10 meters.Solution A) Since the force acts along the vertical direction, it does no work at all on an object that

moves horizontally (in this case 9 0 so c o s 0 and therefore 0).B) Vertically down means we can represent the displacement by the vectors =4k sothe

work done is 4 4, since 1.C) Here s = 10k so 10 10, a negative number which is, of

course, less than the positive number 4mg.

Thus, the largest value ofWoccurs when the object moves down, in the same direction as

gravity: Example 14 In the figure below, a 100-lb box is moved up a 20-ft ramp a distance of9 feet by an80-lb force that is always parallel to the incline. Find A) the work done by the force, and B) the

work done by the force of gravity.

SolutionA) The box moves a distance of9 feet up the ramp. The 80-lb force is parallel to thisdisplacement. Therefore,

8 0 9 c o s 0 720 ft-lbB) This is more tricky. We need to calculate the angle between the force of gravity and the

displacement vector. A free-body diagram will help:

cos120 1 0 0 9 0.5450 ft-lb

8030

30

60120 120

9

100


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The Cross Product

The dot product of two vectors gives rise to a scalar, a real number. It is also useful to define a

product that gives rise to a vector. This vector is called the cross product and, just like the dot

product, it has a geometric definition and an equivalent algebraic one.

The geometric point of view:

Ifa and bare vectors and is thesmallestangle between them, then theircross product, denoted by , is the vector sin where is a unit vector perpendicular to both and andin a direction such that the ordered set,, forms a right-handed system. Thus,

What we mean byordered set,, is that order matters; the sets ,, and ,, are equalas sets, but not as ordered sets. And what is meant by right-handed system is that they form a

right-handed system identical to the one defined by the basis vectors ,,.Example 15 Find the direction of and for the vectors shown below. Assume that bothlie on the plane of the page.

SolutionFirst . Point your right-hand fingers along the direction of and rotate them toward. Your right-hand thumb points in the direction of , which would be into the page.

Now for : point your right-hand fingers along the direction ofand curl them toward. Your right-hand thumb points in the direction of .

In both cases the magnitude of the cross product is sin, however the vectorspoint in opposite directions. Thus, the cross product is not commutative:

sin

We useanarrowsbutttorepresentvectorsthatgoesintothepage.

We use an arrows tip to representvectorsthatcomeoutofthepage.


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This looks like a very artificial and capricious definition, but there is a good reason for it; many

physical quantities behave in exactly the same way. For example, when a charged particle of charge enters a magnetic field with velocity , it experiences a force given by1

This is called the Lorentz force, in honor of the Dutch physicist H.A. Lorentz (1853 1928).

As it turns out, we can also obtain the cross product algebraically:

The algebraic point of view:

The cross product is also given by the determinant

Since determinants have the property that their sign is reversed by a row exchange,

As a refresher, recall that the determinant of the 2 2 matrix denoted by det isdefined to be det We also use the notation

Example 16 Find 0,1,2 2,1,0.Solution

0,1,2 2,1,0 0 1 22 1 0 1 21 0 0 22 0 0 12 1 1 0 1 2 0 0 2 2 0 1 2 1 2 4 2

Example 17 Find the cross product of two vectors in the y-z plane shown below.

1 When dealing with physical quantities one must always be mindful of the units being used.

x

60o

= 10 = 4y

z


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SolutionWe are given the magnitudes and the angle between the vectors so we know ; it is s in 1 0 4 s in 6 0 . What direction does it have? Cross a into bby pointingthe fingers of your right hand along a and curling them toward b, your thumb point in the

direction of .Since lies on the y-z plane, this direction is along the positive x-axis. Therefore, thedirection of the cross product is that ofi: s i n 104sin60

203Example 18 Find the magnitude of the cross product of the vectors shown below:

Solutionwe again have the magnitudes of the vectors, but we need the angle between them. The

angle shown might not necessarily be it; we need to figure it out. Let us slide so that itstail coincides with that of, as shown below:

What do you know! The angle given is, in fact, equal to the angle between the vectors by

the alternate interior angles result from geometry. Thus,

1 0 7 s in 4 5 702

What direction would this cross product have? If the vectors and are on the page, thecross product point out of the page.

Example 19 If 3,0, and 2,4,, find . Is there a value of for which ?Solution

3 0 2 4 0 4 3 2 1204 1 2

45o

7 10

7

1045o

45o


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Since the k-component is never zero, this vector will never be the zero vector.

Example 20 (area of parallelogram) Find a formula for the area of a parallelogram defined by twovectors

SolutionLet

, , and

, , be two non-collinear vectors. Then they define a

parallelogram:

The areaA of the parallelogram is . The problem is that, althoughwe can compute the length of vector

, we are not given

. All we are given are the vectors

and

. However, notice that sin

Thus, the area of the parallelogram defined by and is the norm of their cross product . Nice!Volume of a Parallelepiped

A parallelepiped is a six-faced solid the opposite faces of which are parallel. Think of it as a three

dimensional version of a parallelogram. Here is the picture:

The volume of this solid can be shown to be which is the area of the base time theheight. We can also obtain a formula using vectors; first, let us introduce the vectors as shown

below:

Figure 9

Viewedfromthetop

h

b

c

cb

c

a

h

bc


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The base of this solid is defined by the vectors band c. Looking at it from above, its base looks likethis:

Figure 10The area of this parallelogram is given by

sinwhich is just . Another look at figure 10 shows that h is nothing but cos. Thus, thevolume is given by

cos

Since volume must be a nonnegative number, we have the following result:The volume generated by the non-coplanar vectorsa,b,andcis given by

| |The quantity is called the triple productof the vectors ,,and and it can be shown that

Example 21 Find the volume of the solid whose edges are defined by the vectors 2,1,0, 0,4,0,and

1,1,5.

SolutionLet 2,1,0, 0,4,0 and 1,1,5. Then vectors b and c define the baseparallelogram, but any other order would have done just as well. First we compute the triple

product

1 1 52 1 00 4 0 4 0

The volume of the parallelepiped is |40| 40Torque

When a force acts on an object, it has two effects: it accelerates it along its line of action, and/or it

makes it rotate about a fixed point.

Figure 11

c

b

sin

F CM


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This force will accelerate the block along its line of action. On the other hand, the force in the figure belowwill produce a rotation about axis:

Figure 12

The rotational effect relative to a given point caused by a force is measured by its torque, which isa vector defined by

where is the position vector relative toP of the point at which the force is applied, and F is theforce. Graphically,

In the figure above,

points into the page

To appreciate the need to include in the definition of torque, try to open a door by pushing it nearits hinges; the closer you are to them the harder it will be. In fact, you might not even be able to

open the door.

The angle is equally important. Consider the three cases shown in the figure below where a dooris subjected to a force at the opposite en of the hinge side. The orientation of that force will have

very different effects:

Figure 13

P

Top View

F3Doorhinge

F2

F1

This force will produce arotation about linel.Fr

l


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Force 1 and 2 will tend to make the door swing counterclockwise while force 3 clockwise. Also,force 1 will be more effective that the other two in producing a rotation. Now move one of theseforces closer to the hinge and its turning effect will be diminished.

So from this discussion, it is clear that there are at least three factors we need to consider: how

powerful is the force, where is it applied, and at what angle is it applied. A convenient way to

incorporate all these elements is by the product sinWhich we immediately recognize as the magnitude of the cross product . This is the reasonwhy we defined the torque as we did.

Notice that this definition highlights the fact that the only component of the force responsible for

rotation is :

Figure 14

One can see that sin. Thus, the magnitude of the torque is also give by

If all we want is the magnitude of the torque, this formula will do.

Example 22 A 20 lb. force acts on a wheel barrel as shown below. What torque does this forceexert about point O? Assume that it is applied at a point

4.5feet from O.

SolutionFirst, let us simplify the picture by including only the pertinent information. This is thefree

body diagram the skeleton of the problem.

The position vector has length 4.5 ft. and acts at 18o to the horizontal. If you want to do thisproblem using components, then we have:

4.5cos184.5sin18 0 2 0

O18

o

20 lb.

Top ViewDoorhinge

F

r

O18o

20 lb.

4.5ft.


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The torque is given by

4.5 cos 18 4.5 sin 18 00 20 0 0 0 8 5 .6

The magnitude of the torque is approximately 85.6 ft-lb and its direction is out of the page.An alternate way of computing the torque about a point (mostly used in engineering textbooks) is bymultiplying the magnitude of the force times the moment arm, which is the distance from the pointabout which the torque is being calculated to the line of actionof the force:

The figure shows that sin. Therefore, sin

In the previous example, we can use this approach by identifying the moment arm:

You can clearly see that force times moment arm is given by 204.5cos1885.6 ft-lb.Example 23 Find the magnitude of the torque relative to point A in the figure shown below. Assumethe force acts horizontally.

SolutionThe free body diagram is shown to the right. In it, we see that the moment arm is 8ft.Therefore,

The magnitude of the torque is given by times M.

Moment armM

r

F

Line of action of the force

200 lb

8 ft12 ft

A

200 lb

8 ft12 ft A

Mr

Line of action of the force

O 18o20 lb.4.5ft.Forces line ofaction

M


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8 2 0 0 1 , 6 0 0 ft-lbExercise: Solve this problem by vectors.

Properties of the Dot and Cross product

The following properties are a direct consequence of the definitions of the dot and cross products:

The last one is particularly important and you will encounter it again if you study electromagnetism.

Exercise Why dont we have ?

Let , and be vectors and a scalar. Then,1) 2) 3) 4)

Vectors_ the Dot and Cross Products_no Tensors

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