For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

## Discussion

## Practice Questions

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## Table of Contents

## Transcription

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### More on Vectors & Norms

- An open disc centered at →a with radius b is written as || →x − →a ||< b.

- A unit vector has its norm equal to 1, so we check that || →b || = 1.

^{2}+ ( [(√3 )/2] )

^{2}} = √{[1/4] + [3/4]} = √1 = 1.

- First we find the norm of →a, so || →a || = √{( √3 )
^{2}+ ( − 2)^{2}} = √{3 + 4} = √7 . - Now we compute [(→a)/(|| →a ||)] = [(( √3 , − 2 ))/(√7 )] = [1/(√7 )]( √3 , − 2 ).

- To convert →b into a unit vector, we use the formula [(→b)/(|| →b ||)].
- Computing the norm of →b yields || →b || = √{1
^{2}+ 0^{2}+ 2^{2}} = √5 .

i) Find || →u + →v ||

^{2}

- Since →u and →v are perpendicular we can apply the pythagorean theorem for vectors so that || →u + →v ||
^{2}= || →u ||^{2}+ || →v ||^{2}. - Since →u and →v are unit vectors, their norms are 1. That is || →u ||
^{2}= 1 and || →v ||^{2}= 1.

^{2}= || →u ||

^{2}+ || →v ||

^{2}= 1 + 1 = 2.

ii) Find the angle q between →u and →v.

^{o}.

^{2}.

- Since →a and →b are not unit vectors, we compute the sume of the vectors first, followed by the norm.
- →a + →b = ( − 2,5, − 1) + (4, − 3,1) = (2,2,0)
- || →a + →b || = √{2
^{2}+ 2^{2}+ 0^{2}} = 2√2

^{2}= ( 2√2 )

^{2}= 8. Note that from our previous step, || →a + →b ||

^{2}= ( √{2

^{2}+ 2

^{2}+ 0

^{2}} )

^{2}= 2

^{2}+ 2

^{2}+ 0

^{2}= 8.

- The formula for finding the angle θ between two vectors is cosθ = [(→u ×→v)/(|| →u |||| →v ||)].
- Note that both u and v are unit vectors, so that our equation is now cosθ = →u ×→v.
- Computing the scalar product yields ( [1/(√2 )],[1/(√2 )] ) ×( [(√3 )/2], − [1/2] ) = [(√3 )/(2√2 )] − [1/(2√2 )] = [( − 1 + √3 )/(2√2 )].

^{ − 1}( [( − 1 + √3 )/(2√2 )] ) = 75

^{o}.

- Since →a and →b are not unit vectors, we use the general formula →P
_{→a→b}= [(→a ×→b)/(→b ×→b)]→b. - Computing we have, →a ×→b = ( [1/2], − [1/2] ) ×( [1/2], − [1/4] ) = [1/4] + [1/8] = [3/8]; →b ×→b = ( [1/2], − [1/4] ) ×( [1/2], − [1/4] ) = [1/4] + [1/16] = [5/16].
- So that [(→a ×→b)/(→b ×→b)] = [(3 \mathord/ phantom 3 8 8)/(5 \mathord/ phantom 5 16 16)] = [6/5].

_{→a→b}= [6/5]( [1/2], − [1/4] ) = ( [3/5], − [3/10] ). The image below displays how the projection (red vector) is perpendicular (blue line) to →a but and extension of →b.

_{1})/(√3 )],0,[(a

_{2})/(√3 )],0,[(a

_{3})/(√3 )] ) where a

_{1}

^{2}+ a

_{2}

^{2}+ a

_{3}

^{2}= 3. Show that →w is a unit vector.

- We show that || →w || = 1.
- || →w || = √{→w ×→w} = √{( [(a
_{1})/(√3 )] )^{2}+ 0^{2}+ ( [(a_{2})/(√3 )] )^{2}+ 0^{2}+ ( [(a_{3})/(√3 )] )^{2}} = √{[(a_{1}^{2}+ a_{2}^{2}+ a_{3}^{2})/3]} .

i) Find the first five terms of the sequence 1 + [1/n].

- By substituting n with 1,2,3,4 and 5 respectfully, we obtain the first five terms.

ii) Is the sequence 1 + [1/n] contained in the closed disc centered at →a with radius [1/4]? Explain.

iii) What is the minimum radius b that will contain the sequence 1 + [1/n] in a closed disc? Explain.

- Since our sequence is decreasing, we can see that as n reaches a high value, it will not exceed 1. For instance at n = 1000 we have [10001/1000] = 1.001
- We base our radius b on the highest term available from the sequence, in this case 2.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### More on Vectors & Norms

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- More on Vectors and Norms 0:38
- Open Disc
- Close Disc
- Open Ball, Closed Ball, and the Sphere
- Property and Definition of Unit Vector
- Example 1
- Three Special Unit Vectors
- General Pythagorean Theorem
- Projection
- Example 2
- Example 3

### Multivariable Calculus

### Transcription: More on Vectors & Norms

*Hello, and welcome back to educator.com.*0000

*Welcome back to multivariable Calculus.*0003

*Last lesson we introduced the scalar product, or dot product.*0005

*We also talked a little bit about norms.*0008

*Today, I am going to talk more about norms and vectors.*0010

*A lot of today, at least the first part of the lesson, is going to be basic properties.*0015

*Things that are pretty intuitively clear, but we want to formalize them.*0020

*Things that we are going to be talking about at any given moment.*0024

*We are going to be talking about open discs, open balls, and we want to make sure that we understand what these are, that we have been exposed to them so that we can proceed with our discussion of multivariable Calculus.*0028

*Okay, let us get started.*0037

*So let us let a be a point in the plane, remember a point is just a vector from the origin to that point.*0040

*A couple ways of thinking about it, let it be a point in the plane, let us see, then the set of points such that the norm, I should say the set of points x.*0052

*Let me do that.*0084

*It is the set of points x, such that the norm, in other words the distance between x and a, if it is less than some number, it is called the open disc.*0085

*Let me draw a picture of what this means.*0105

*Now, you know this notation here, so a norm is just a distance.*0111

*In the previous lesson, the norm between two points is the distance between those 2 points, that is all this is.*0116

*So this is saying that if there is a particular point in the plane, a, and if I want to take all of the points that are less than b units away from it -- let us say less than 2 units away from it -- I basically have all of the points in a circle, everything inside of that, here is what it looks like.*0125

*So here is our point a, let us go ahead and label that.*0145

*Then let us say b is some distance, b.*0151

*So all of the points in this disc, all of the points in here, all of the vectors, the points x such that the distance from x to , x to a, x to a, is less than sum b.*0157

*Basically it gives me a circular region in the plane that I am talking about.*0173

*Less than b, without this less than or equal to, we call it the open disc, so it is all of the points inside that circle of radius b.*0181

*It actually does not include all the points on the circle.*0190

*Now, if we say, the set of all points x such that the norm, in other words the distance from x to a,*0193

*Now if we say it is less than or equal to b, it is called the closed disc.*0217

*The closed disc actually includes, let me draw it separately up here, we have our a, then we have our b, it includes all points in here that are less than this length b.*0227

*But, it also includes all points on the boundary.*0246

*So closed, when we talk about an open set, we mean everything up to but not that boundary.*0250

*Closed is everything up to and including that boundary.*0256

*Now, let us see, this disc and a final one.*0264

*The set of all, actually I am not going to write it out since I have already written it out twice.*0270

*Let me just go ahead and give you the notation for it.*0277

*If we have the distance from the points x to a, which is the norm of x - a, if we say it equals b, well, this is the circle centered at a with radius b.*0278

*Now we are just talking about the boundary points, nothing outside, nothing inside.*0300

*So we have the open disc, we have the closed disc, and we have the circle.*0305

*We will often be talking about one of those or all of those.*0310

*It is very important that we specify which region in the plane we are talking about.*0314

*Now we are just going to generalize this, this is in 2 space.*0318

*Now we are going to generalize it to 3 space.*0324

*This is called the open disc, closed disc, the circle, we call it the... so for n=3, well let us say, for R3, let us be consistent, we have the closed ball and the sphere.*0326

*It is the same notation, the set of all x, either less than b, or less than or equal to b, or equal to b.*0358

*In 3 dimensions, we basically have that the picture we have is this.*0373

*Just do a quick picture of it, some point, and then instead of a circle, what you have is a ball.*0377

*That is it, just a ball, you have some center, you know this center right here is the point a, the vector a.*0388

*Then the radius of this ball is b.*0396

*So that is all it is.*0401

*This nomenclature of ball, closed ball, that actually applies to any number of dimensions.*0404

*So if I want to talk about the open ball in the 5th dimension, that is what it is.*0412

*It tells me there is some vector a in the center, and then all the points that are less than some specified value away from that point, we call it the open ball, the closed ball, the open sphere, depending on which set of points we are talking about.*0415

*Alright, let us move on.*0435

*We are just talking about things that we are going to be using over and over again.*0436

*So this is going to be a property, again this particular property is intuitive, but we might as well formalize it.*0440

*Let x be a number, then the norm of x × a, a vector, = x × norm of a.*0450

*What this says is that if I have a vector a and I multiply it by some number, let us say 15, then I take the norm of what I get.*0470

*It is the same as taking the norm of the vector first, and then multiplying by the number 15 afterward.*0480

*This is intuitively clear, but then I will formalize it.*0485

*Okay, now, let us give a definition.*0489

*Now I am going to define something called a unit vector.*0496

*It is exactly what you think it is, a unit vector is a vector whose norm is one.*0498

*In other words, whose length is 1.*0507

*Formally, that is, if the norm is 1, it is a unit vector.*0513

*That is it, it is just a unit vector whose length is 1.*0520

*1 is nice because it is one of those numbers that you can use as a standard and you can multiply it by any number you want to make it as big as you want, or to make it as small as you want.*0523

*In mathematics it is called the identity, in mathematics it is a very important number.*0535

*Okay, given any vector, b, just a random vector -- let me do this notation a little differently.*0537

*1/norm(b), and the norm is a number, × the vector b itself = a unit vector in the direction of b.*0555

*This is very important, so basically if I am given some vector b of any specified length, I can actually take that vector.*0576

*And, if I divide that vector by its norm, or more specifically, if I multiply it by the reciprocal of its norm as a constant in the front, remember we can do that, we can multiply vectors by constants, lengthen them, shorten them -- I can actually create a unit vector, specifically a vector of length 1 in the direction b.*0586

*That is what makes this very nice.*0609

*You have any vector, you just go ahead and you essentially divide it by its norm, and you have created a unit vector.*0611

*That is standard procedure for creating a unit vector in any given direction.*0616

*It is going to be very important when we talk about the direction of derivatives later on.*0621

*Now a little bit more, we say 2 vectors have the same direction.*0630

*Let me see, let me specify what they are, actually.*0645

*We say that 2 vectors a and b, have the same direction if there is some number C, such that C × a = b, or a = c ×b,*0653

*Where you put the c does not matter.*0685

*All that says if I have 2 vectors, let us say that, and that, I can find some number since they are in the same direction.*0687

*Basically we know that when you multiply a vector by a scalar, by some number, all you do is you lengthen it or shorten it, you scale it out.*0695

*You are scaling it, that is where the word scalar comes from.*0702

*If this is a, and this is b, well I know that if I start with a and it looks like here I maybe have to multiply it by 3 to get b.*0706

*If there is such a number that I can multiply by a to get b, that automatically tells me that a and b are in the same direction.*0717

*I do not necessarily know that, but this property allows me to say the two vectors happen to be in the same direction.*0724

*If I need to check they are in the same direction, I need to find some number c that satisfies this property.*0732

*Let us say I start with some vector b and I need to see if a happens to be in the same direction,*0737

*If I can take b and multiply it by 1/3, I can get a, sure enough there is some number b that if I multiply by 1/3, I end up getting the vector a.*0745

*That means they are in the same direction.*0755

*So this is just a formal definition of when some things are in the same direction.*0757

*Again, what mathematics does, it takes things that are generally intuitive and it actually formalizes them.*0763

*It gives them specific algebraic definitions so that we can actually do something with them.*0770

*Geometry helps, we want to be able to draw a picture and talk about two vectors in the same direction.*0775

*But geometry is not math, geometry is just some sort of pictorial representation that helps you visualize something.*0780

*It is algebra that is math, that is where the power of mathematics comes in.*0787

*Now, let us move on with some more unit vectors.*0795

*Unit vectors are fantastic because we can basically take any vector and we can convert it to a unit vector.*0799

*Once we convert it to a unit vector, then we can multiply that unit vector by any number c to make it any length that we want.*0807

*That is the power of unit vectors, it gives us a standard, an identity vector, if you will,*0814

*To multiply by, that we can always use over and over again.*0822

*I can get any length I want by just choosing the proper c.*0825

*If I have a unit vector and I want a length of 15, I can just multiply it by 15.*0829

*If I want it to be a vector in the opposite direction to have a length of 446, I take -446 × my unit vector.*0833

*Let us do an example.*0845

*Okay, so example number 1.*0850

*Let the vector a = (3,2,-1), so we are working in R3, 3 dimensional space.*0855

*Now the norm, let us go ahead and calculate the norm, we are going to be finding the unit vector in this direction.*0863

*Again, we need to find the norm so that we can actually multiply by the reciprocal of the norm or divide by the norm to create the unit vector.*0870

*So, the norm of a, you remember from Calculus that things started to get a little notationally intensive.*0878

*In multivariable calculus, things are going to get even more notationally intensive.*0888

*Do not let the tedium of the notation throw you off, it is just notation.*0890

*You want to be very, very clear about what you are doing.*0894

*Whatever you do, do not do things in your head, write out everything.*0897

*This is very important.*0900

*When you are doing things in your head, I think it is really nice to be able to do that.*0903

*But the idea in mathematics is to be correct, to know what is going on, not to show somebody that you can do something in your head.*0908

*If you write things out, especially because this is not arithmetic anymore, this is serious upper level math, this is real mathematics.*0913

*You want to be able to notate everything properly.*0924

*I know it is tedious, but write out the notation, it equals sqrt(9) + 4 + 1, and we end up with sqrt(14).*0927

*So, the unit vector in the direction of a which I will symbolize like that.*0945

*We are often going to be throwing symbols in here that are a little less than standard, that you may not see in your book.*0952

*I will certainly try to be as standard as possible, I will try to use notation that you will see all the time.*0960

*But again, a lot of times when we are talking about a concept, we need to be able to use notations that make sense at the moment.*0965

*You are free to do that yourself.*0972

*What is important is not the symbolism, but the underlying mathematics.*0975

*This is just a symbol for a unit vector in the direction of a, you can symbolize it any way you want.*0977

*= 1/sqrt(14) × the vector (3,2,-1).*0984

*Right? Because we said a unit vector is just 1/reciprocal of the norm × the vector itself.*0993

*Well the vector is this.*1001

*When we multiply a constant by a vector, we multiply everything by that constant,*1005

*So we get 3/sqrt(14), 2/sqrt(14) and -1/(sqrt14), this is a perfectly good vector, you can leave it like that.*1009

*You do not need to rationalize the denominator like they made you do in high school.*1020

*It is a perfectly valid number, it is a unit vector in the direction of a.*1025

*If you were to take the norm of this number, you would find that its norm is 1, you have created a unit vector.*1033

*You might want to do that yourself just to convince yourself that this is true.*1039

*Okay, now we are going to talk about 3 very special unit vectors.*1044

*So, 3 special unit vectors.*1050

*And we are going to be talking about our 3 dimensional space, so we have the z axis, the x-axis, and the y-axis.*1060

*Let me go ahead and label those.*1069

*Again, this is the standard right handed representation of three dimensional space, x, y, z.*1074

*Well, a vector in the direction of z, a vector in the direction of x, and a vector in the direction of y -- each having a length of 1 -- each of those have special names.*1080

*We call them, we can go ahead and say e _{1}, or we can go ahead and say, you know what, I am going to use e_{x}.*1096

*So this vector right here, the point, remember, it is a point that a unit vector represents, is (1,0,0).*1106

*My notation is not doing too well today, my apologies, (1,0,0), that is it.*1125

*Now the unit vector in the direction of y is notated as e _{y}, sometimes notated as e_{2}.*1130

*It equals (0,1,0), there is nothing in the x direction, nothing in the z direction, but a length of 1 in the y direction.*1139

*Of course e _{z} is equal to (0,0,1).*1150

*These are mutually perpendicular, they are orthogonal.*1155

*In other words if you took the dot product of this, it would be 0, this and this, 0, this and this, 0.*1157

*That is the whole idea.*1163

*This is called an ortho-normal-basis, you do not have to know that.*1164

*Just know that they are perpendicular.*1168

*Because we often talk about 3 space, we will often be using 3 space and 2 space.*1171

*We just wanted to introduce this special collection of unit vectors in the direction of the coordinate axes.*1176

*Okay, now let us go ahead and do our general Pythagorean theorem.*1184

*You know our Pythagorean theorem, it is of course if you have a right triangle in the plane, a ^{2} + b^{2} = c^{2}.*1188

*The sum of the squares of the two shorter sides is equal to the square of the hypotenuse.*1197

*As it turns out this is generally true in all dimensions.*1200

*I am not going to go ahead and do a proof of the theorem.*1205

*Again, in these set of courses, we want to be able to list the theorems, and we want to be able to use them, but we are not going to go through the proofs.*1207

*We want to do it in an intuitive sense, we want to get a feeling of it by doing problems and by discussing it, not necessarily by doing proofs.*1218

*If you want the proofs, by all means they are in your book.*1225

*I would certainly suggest looking at them at least occasionally.*1228

*At least to see what is going on if something seems like it fell out of the sky.*1230

*Most of these things are reasonably intuitively clear.*1232

*The general Pythagorean Theorem, we are going to be using norm notation, so again a norm is a length -- if a and b are orthogonal, in other words if their dot product is equal to 0, if they are perpendicular, then the norm of a + b ^{2} = norm(a^{2}) + norm(b^{2}).*1237

*Again, a norm is a length.*1282

*It tells me that if I have a and b that are orthogonal to each other, a and b orthogonal to each other, the length of a ^{2} + the length of b^{2} is equal to the length of a+b^{2}.*1284

*That is that length, that is all that is going on, this is just the standard Pythagorean theorem expressed for any number of dimensions.*1300

*You can have R3, R5, R15, it does not matter.*1308

*It holds in all dimensions, that is actually pretty extraordinary that it does so.*1312

*Let me go ahead and draw this out a little bit more formally here.*1319

*If this is a and this is b, well, so this is vector a and this is the vector b, the vector a+b is this vector.*1325

*Remember when we said if we add vectors, we go from tail to head of one, then we take b, that direction, and we just go this way.*1335

*That is it, this is a + b, you start here and end up here, then you start here and you end up here.*1347

*So this vector right here actually is a+b.*1351

*Of course this is perpendicular therefore this is perpendicular, so this vector here is our a+b.*1360

*Sure enough, this squared + this squared = that squared, that is all this is saying.*1368

*The norm of this + the norm of that = the norm of this.*1372

*Now we are going to introduce a very very important notion, something called a projection and we will do an example of that.*1380

*A projection, let us go ahead and define what we mean by that.*1390

*First of all I am going to draw a picture.*1392

*Let us say I have a vector a, and another vector b, so this is a and this is b, the order is very important here.*1400

*If I imagine shining a light straight down this way, this a is actually going to cast a shadow on this vector b.*1410

*That shadow is of course, if you shine the light straight down, it is going to be perpendicular to that.*1421

*This thing right here, this vector, it is a vector along b, but its length is the actual length of the shadow that is cast by the vector a.*1427

*Obviously if this angle were bigger, then you would have a shorter shadow.*1442

*This thing right here, that is the projection.*1447

*It is the projection of a onto b.*1450

*I am going to notate it this way.*1455

*The projection is an actual vector.*1457

*The projection of a onto b.*1460

*This is going to be the notation that I use for my projection, it is a slightly different than you are going to use in your book but I think it makes everything clear.*1464

*You are saying that you are projecting the vector a onto the vector b, not the vector b onto the vector a.*1472

*That actually exists, you can project any vector onto another vector, but it is an entirely different vector.*1480

*The order matters.*1484

*The algebraic definition of the projection, this is the geometric, we just wanted to see what we mean by that in 2 space and 3 space, so algebraically, the project of a onto b = a · b/b · b ×, so this is a number, × the vector b itself.*1487

*a · b/b · b is just some number, so basically what we are doing is, again, we want to find what this vector is.*1518

*It is in the direction of b, so we know that this vector is some number × b, right?*1528

*It is basically just you are scaling out b, you either make it bigger or smaller, in this case it ends up being smaller.*1535

*That is all you are doing, is we want to know what vector = the projection of this vector on top of this vector.*1545

*The projection geometrically is what shadow does this cast on that.*1554

*Again, this is just some number, dot products are just numbers.*1560

*A number × a number, we are scaling b to find out what this is.*1564

*Now, let us move on to the next set.*1570

*The component of a along b, let me actually, so c = a · b/ b · b, all this is is just the component of the vector a along the vector b.*1576

*In other words, if I take a look at this vector a, it has this component this way, and another component this way.*1605

*It is the component of a in the direction of b, that is all it is.*1613

*The best way to think about it is this whole idea of shining a light and casting a shadow, that is what is going on.*1618

*If b happens to be a unit vector, then since the norm of b = 1, b · b, that denominator = 1.*1628

*Remember the definition of the norm is the square root of the dot product.*1657

*So if the norm is 1, and I take the square of that which is also 1, it means that that product is 1.*1661

*The projection of a onto b = just a · b × vector b.*1672

*In other words the denominator disappears, because b · b = 1.*1683

*So if you happen to be projecting something onto a unit vector, which looks like this, let us say that is a vector of unit 1 and this is b.*1685

*It is not a unit vector of 1, it is a unit vector, so it has a length of 1, my apologies.*1697

*If that is a, well again, if I shine a light, drop a perpendicular, that is all I get.*1701

*I drop a perpendicular from the end of a down onto the vector b, well, the vector b extends this way, right?*1708

*But this is a unit vector, so what I am looking for is that vector.*1717

*It has a length and it has a direction.*1720

*That length and that direction is given by this expression right here, if b happens to be a unit vector.*1723

*If not, then you just use a general vector addition.*1730

*So let us do an example.*1735

*Example number 2.*1740

*We will let a = (2,4,2) and b = (-3,1,3).*1743

*Okay, now we want to find the projection of a onto b.*1755

*Well we use our formula, so the projection of a onto b = a · b/ b · b × vector b itself.*1763

*Okay, well, a · b = 2 × -3 is -6, 4 × 1 is 4, 2 ×3 is 6, -6 + 6 is 0, so that is 4.*1783

*And b · b = -3 × 3 is 9, and 1 × 1 is 1, and 3 × 3 is 9, so we get 19, if I am not mistaken.*1799

*Therefore the projection equals 4/19 × vector b itself.*1814

*Well the vector b is, I am going to write it vertically, (-3,1,3).*1823

*Of course we have a constant times a vector, so we multiply everything right through.*1834

*So that is (-12/19, 4/19, 12/19), this vector (-12/19, 4/19,12/19) it is the projection of the vector (2,4,2) on (-3,1,3) in that direction.*1838

*That is what is going on here.*1863

*That is it, this is algebraic.*1865

*Very, very important.*1868

*Let us go ahead and finish it off with a nice further geometric interpretation, so let us go ahead and take our standard drawing of, this is vector a and this is vector b, and we know that this is the projection right perpendicular.*1873

*So this right here happens to be, this length is the norm(a), that is what the length is, what the norm is, it is a length.*1890

*This length right here, well it happens to be, remember c × the norm(b).*1900

*Basically I have taken the length of b and I have multiplied by some number c, and that c = that a · b/b · b.*1908

*This would be our definition.*1920

*So if I have the norm of b, and I multiply by this a · b / b · b, from the projection, that is the length of this thing.*1923

*If I, now look, I have some angle θ in between those 2 vectors, in between this vector and this vector or this vector and this vector, it actually does not really matter.*1934

*I can actually write cos(θ) = the adjacent/the hypotenuse of this right triangle.*1945

*So I end up with c × norm(b)/the norm(a).*1950

*C happens to be a · b/b · b, and I am multiplying that by the norm(b).*1960

*Then I divide all of that by the norm(a).*1978

*Remember the definition of the norm, remember the norm(b), or any vector, is the sqrt(b · b).*1983

*So b · b, if I square both sides, I get the norm squared.*1997

*So I am going to take b · b, and I am going to replace this with that.*2004

*So that equals a · b/the norm of b ^{2} × the norm divided by the norm of a.*2010

*Okay, now this is squared, this is on top, this cancels this, and of course when I rearrange this because this is actually equal to a · b/norm of b × norm of a.*2027

*Look at that.*2045

*I have the cosine of the angle θ between those two vectors = the dot product of those vectors/the product of the norms of those vectors.*2048

*Or, another way of writing it is a · b = norm(a) × norm(b) × cos(θ).*2063

*In some books, possibly in your own, this is actually taken to be the definition of the dot product.*2079

*That is fine, it is not really a problem in and of itself.*2085

*However, notice we have introduced some angle θ.*2093

*Well the angle θ makes sense in 2 and 3 dimensions.*2096

*The algebraic definition of dot product which we gave is the product of the, you know, corresponding entries added together.*2100

*That strictly involves a and b, it does not involve something else, and it does not involve something else called the norm.*2111

*I think this is fine, I think it is better to use the algebraic definition and derive this as an identity as opposed to taking this as a definition.*2120

*The nice thing about it is all of the other properties of the scalar product fall out easily when you use the algebraic definition as opposed to this definition.*2129

*Starting out with this definition, it is okay, but I think it is just better mathematically to work the other way around.*2138

*Notice we had just a nice straight basic definition of the dot product, and then we ended up with this identity.*2149

*Let us go ahead and do our final example for this lesson.*2155

*Example 3.*2162

*We will let a = (2,-2,6) and we will let b = (3,0,1).*2165

*We want to find θ, the angle between those vectors.*2175

*Well, we know that cos(θ) = a · b/norm(a) × norm(b).*2190

*a · b = 6 + 0 + 6,*2205

*The norm of a, if I do that out is going to be sqrt(44).*2211

*The norm(b) is going to be sqrt(10).*2215

*I end up with 12/sqrt(440), which is equal to 0.5721.*2218

*When I take the inverse cos, or arccos, I get θ = 55.1 degrees.*2228

*That is it, so you have 2 vectors in 3 space, the angle between them is 55.1 degrees.*2240

*We used the norm, and we used the dot product.*2245

*It is fine to think about this geometrically when you think about our angle in R2, or an angle in R3.*2250

*We do not really talk about angles in R4 or R5, they do not really make sense.*2257

*We can still certainly talk about them, this value still exists, this identity is still valid, but again, we would like to make sure that we work algebraically, not just geometrically.*2260

*Do not try to figure intuition and make it fit the mathematics, do not try to make the mathematics fit your intuition.*2275

*Let the mathematics guide you.*2283

*At the end you have to defer to the mathematics, not your intuition.*2288

*Now that we are getting into higher mathematics your intuition will not always lead you in the right direction.*2291

*Thank you for joining us here at educator.com, we will see you next time.*2295

1 answer

Last reply by: Professor Hovasapian

Mon Dec 21, 2015 7:42 PM

Post by Jamal Tischler on December 21, 2015

We could say at the projection of a vector just, ||a|| cos(tetha)* b/||b||. Where ||a|| cos (tetha) is the legth of the projection( in the right triangle, I applied cos) and b/||b|| is the init vector along b.

1 answer

Last reply by: Professor Hovasapian

Wed Jan 28, 2015 12:38 PM

Post by Jacob Kalbfleisch on January 25, 2015

Prof. Hovasapian is hands down the best lecturer on Educator. So clear and succinct, and offers fantastic advice very bluntly. I've so much respect for this man!

4 answers

Last reply by: Professor Hovasapian

Fri Jun 20, 2014 4:50 PM

Post by abc def on June 11, 2014

Good morning Professor Hovasapian,

I was playing around with some Angle-Difference Identities to wrap my mind around the dot product in 2-3D. I observed that A*B = ||A||||B||cos(@) can be proven by using the Angle-Difference Identity in respect of cos(@),

A*B=||A||||B||cos(@a - @b) = ||A||||B||[cos(@a)cos(@b)+sin(@a)sin(@b)] = [||A||cos(@a)||B||cos(@b)+||A||sin(@a)||B||sin(@b)] = (Ax)(Bx)+(Ay)(By) QED.

Is there a operation that _____=||A||||B||sin(@) would work?

1 answer

Last reply by: Professor Hovasapian

Wed Oct 2, 2013 2:50 PM

Post by Said Sabir on October 2, 2013

how to calculate the angel if I know already the Cos of that angel ?