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Lecture Comments (22)

1 answer

Last reply by: Professor Hovasapian
Fri Aug 26, 2016 7:28 PM

Post by Kaye Lim on August 25 at 09:40:41 PM

Dear Prof. Hovasapian,

-How do I know if the distance formula between 2 vectors held true for 2 vectors in a much higher dimension?

-How was the formula for distance between 2 vectors (a-b)developed? Was it found to be true in one case, thus later tested and found to be true for all 2 vectors in 2D and 3D space, thus, it was concluded to be true and applicable for any dimension?

Thank you!

1 answer

Last reply by: Professor Hovasapian
Sat Mar 26, 2016 4:41 AM

Post by elgendy ahmed on March 6 at 08:01:18 PM

Thanks Professor Raffi,

Now, the product of a scalar multiplication is a number and not a vector as you mentioned. However, isn't the number a one-dimension space vector. Or the number in this case is just a number an not a vectorized object? Could you please elaborate on this?


1 answer

Last reply by: Professor Hovasapian
Wed Aug 13, 2014 11:30 PM

Post by Denny Yang on August 10, 2014

Quick Question.

Q. Let →a = (12, − 5) and →b = (3, − 13). Find || →a − →b ||.

Step 1. First compute the difference between the vectors, →a − →b = (12, − 5) − (3,13) = (12, − 5) + ( − 3, − 13) = (9, − 18) = 9(1, − 2).

Shouldn't the vector of →a − →b be (9,8)since →a + (-→b) = (12-3,-5-(-13)) = (9,8).

1 answer

Last reply by: Professor Hovasapian
Thu Apr 24, 2014 5:00 PM

Post by Patrick Gleason on April 24, 2014

I am looking at the first practice question:

Q. Find the scalar product of →a = ( − 2,5) and →b = (7, − 4).

*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.

   Step 1. The scalar product is defined as →a ×→b = a1a2 + b1b2 for →a = (a1,b1) and →b = (a2,b2).
   Step 2. Substituting yields →a ×→b = ( − 2)(5) + (7)( − 4) = − 10 + ( − 28).

Shouldn't the answer to this be -34 not -38? The solution shows a1a2 + b1b2 but per the example slide it should be a1b1 + a2b2. Or am I completely misunderstanding something?

1 answer

Last reply by: Professor Hovasapian
Sun Apr 27, 2014 4:27 PM

Post by Juan Castro on April 17, 2014

I think that the answer to question 4 is wrong (i.e.orthogonality of (7,3,2,-1) and (-1,2,1,0)). I get 1 instead of zero.

1 answer

Last reply by: Professor Hovasapian
Sat Apr 12, 2014 5:26 PM

Post by Vinu Gowda on April 11, 2014

I like the way you teach, and I like the most that you give extra information on the topics.. That's really good. Thanks,,,  

2 answers

Last reply by: Professor Hovasapian
Sun Jan 26, 2014 3:08 PM

Post by Eddie Chan on January 26, 2014

The answer of practice question 1 is wrong

1 answer

Last reply by: Professor Hovasapian
Sun Sep 16, 2012 4:17 PM

Post by Nishaan Brahmananthan on September 16, 2012

You are very much help full in helping me on Maths

2 answers

Last reply by: Jonathan Bello
Thu Jul 26, 2012 7:39 PM

Post by Jonathan Bello on July 24, 2012

Prof Raffi,

May I ask what University or College do you teach at? They are lucky to have you.

1 answer

Last reply by: Professor Hovasapian
Sat Jul 14, 2012 7:00 PM

Post by Senghuot Lim on July 10, 2012

my hero

Scalar Product & Norm

Find the scalar product of a = ( − 2,5) and b = (7, − 4).
  • The scalar product is defined as a ×b = a1a2 + b1b2 for a = (a1,b1) and b = (a2,b2).
  • Substituting yields a ×b = ( − 2)(5) + (7)( − 4) = − 10 + ( − 28).
The solution is a ×b = − 38. Note that this number is a scalar, not a vector.
Find the scalar product of a = (1,0, − 3) and b = ( − [1/2],8,6 ).
  • The scalar product is defined as a ×b = a1a2 + b1b2 + c1c2 for a = (a1,b1,c1) and b = (a2,b2,c2).
  • Substituting yields a ×b = (1)( − [1/2] ) + (0)(8) + ( − 3)(6) = − [1/2] + 0 + ( − 18).
The solution is a ×b = − [37/2]. Note that this number is a scalar, not a vector.
Let a = (1,0), b = ( √3 , − √2 ) and c = ( − √2 ,√3 ). Verify that a ×(b + c) = (a ×b) + (a ×c).
  • Substitute the left hand side and right hand side of the equation with the given vectors.
  • Computing the left hand side yields a ×(b + c) = (1,0) ×[ ( √3 , − √2 ) + ( − √2 ,√3 ) ] = (1,0) ×( √3 − √2 , − √2 + √3 ) = √3 − √2 .
  • Computing the right hand side yields (a ×b) + (a ×c) = (1,0) ×( √3 , − √2 ) + (1,0) ×( − √2 ,√3 ) = √3 + ( − √2 ).
Since both sides of our equation equal √3 − √2 we have verified that the property a ×(b + c) = (a ×b) + (a ×c) is true.
Are u = (7,3,2, − 1) and v = ( − 1,2,1,0) orthogonal?
  • Recall that two vectors are orthogonal (or perpendicular) when their scalar product equals zero.
  • The scalar product of any two vectors is the sum of the product of each component so u ×v = 7( − 1) + 3(2) + 2(1) + ( − 1)(0) = − 7 + 6 + 2 − 1.
Our scalar product yeilds u ×v = 0, hence u and v are orthogonal.
Let a = (x,6) and b = (3, − 1). Find an integer value of x that makes a and b perpendicular.
  • Two vectors are perpendicular if a ×b = 0.
  • The scalar product of our vectors is (x,6) ×(3, − 1) = 3x − 6.
  • Equating to zero gives 3x − 6 = 0.
Solving for x yields our solution, x = 2.
Verify that i = (1,0,0), j = (0,1,0) and k = (0,0,1) are orthogonal to each other.
  • We must check that ij = 0, ik = 0 and jk = 0.
  • Note that ij = (1,0,0) ×(0,1,0) = 0 because no two components have a nonzero product.
We can also conclude that ik = 0 and jk = 0 for the same reason. Hence the vectors i, j and k are orthogonal.
Let a = (1,1,1).
i) Find − a
  • The opposite of a vector is the product of − 1 to each component.
Hence − a = ( − 1(1), − 1(1), − 1(1)) = ( − 1, − 1, − 1).
Let a = (1,1,1).
ii) Find − a + a
  • The sum of two vector is the addition of each component.
So − a + a = ( − 1, − 1, − 1) + (1,1,1) = (0,0,0).
Let a = (1,1,1).
iii) Find || a ||2
  • The norm of a vector is defined as || a || = √{a ×a} .
  • Substituting our vector gives || a || = √{(1,1,1) ×(1,1,1)} = √{1 + 1 + 1} = √3 .
So that || a ||2 = ( √3 )2 = 3.
Let a = (1,1,1).
iv) Find || − a || + || a ||
  • Recall that || − a || = || a ||, so that our problem is to find || a || + || a || = 2|| a ||.
Note that || a || = √3 , so that our solution is 2√3 .
Let u = ( − √5 ,7 ), find || u ||.
  • The norm of a vector is defined as || u || = √{u ×u} .
  • Substituting gives || u || = √{( − √5 ,7) ×( − √5 ,7)} = √{( − √5 )2 + 72} = √{5 + 49} = √{54} .
Simplifying resutlts in √{54} = √{9 ×6} = 3√6 .
Let a = (12, − 5) and b = (3, − 13). Find || a − b ||.
  • First compute the difference between the vectors, a − b = (12, − 5) − (3,13) = (12, − 5) + ( − 3, − 13) = (9, − 18) = 9(1, − 2).
  • Now we find the norm of the vector 9(1, − 2), note that || 9(1, − 2) || = | 9 ||| (1, − 2) ||.
  • So || (1, − 2) || = √{(1, − 2) ×(1, − 2)} = √{12 + ( − 2)2} = √5 .
Our solution is 9√5 .
Find the distance between u = (1.0,1.5,2.7,3.5) and v = ( − 0.5,1.3,0.8, − 1.1).
  • Finding the distance bewteen two vectors is analogous to finding || u − v ||.
  • Now, u − v = (1.0,1.5,2.7,3.5) + (0.5, − 1.3, − 0.8,1.1) = (1.5,0.2,1.9,4.6).
  • Computing || (1.5,0.2,1.9,4.6) || = √{1.52 + 0.22 + 1.92 + 4.62} = √{27.06} .
So the distance between u and v is √{27.06} ≈ 5.20

*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.


Scalar Product & Norm

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
  • Scalar Product and Norm 1:05
    • Introduction to Scalar Product
    • Example 1
    • Properties of Scalar Product
    • Definition: Orthogonal
    • Example 2: Orthogonal
    • Definition: Norm of a Vector
    • Example 3
    • Distance Between Two Vectors
    • Example 4

Transcription: Scalar Product & Norm

Hello, and welcome back to, and welcome back to multivariable calculus.0000

Last lesson we introduced vectors, points, and space.0005

Now, we are going to start operating with vectors, we are going to start doing things with vectors.0007

We are going to start doing things to them, multiplying them and seeing how long they are, so that we can really start getting into more of the deeper mathematics and the basic properties that we are going to be using over and over again throughout this course.0011

Today, we are going to talk about something called the scalar product, also known as the dot product, and the norm.0025

The norm is just a fancy word for length, the distance.0030

So, when we are talking about the norm of the vector, we are just saying "how long is that vector?"0034

We are accustomed to dealing with 2 and 3 space, but remember this is true in any dimension.0037

This is why when we say scalar product norm, we are talking about a vector in any number of dimensions, it turns out.0043

A vector in 15 space does have a length.0051

We cannot visualize that length, or a drawing of any kind, but it does exist.0054

It is a real thing.0060

Having said that, let us just jump right on in.0062

Again, so that we can continue to use our intuition, which we want to be able to use geometrically to help our algebra, we are mostly going to be working in 2 and 3 space in our examples.0067

Let us start off by defining a couple of vectors here.0080

Let a = (a1, a2, a3) and the vector b = (b1, b2, b3).0086

Okay, so define their scalar product or dot product, a · b, this is the symbol for the dot product or scalar product, = a1 × b1 + a2 × b2 + a3 × b3.0106

Now, let us just talk a little before we do the example. 0138

First of all, a product, you know that if I take two numbers, like 14 and 2, the product is 28.0142

As it turns out, I am with vectors, so now we have turned on to bigger more complicated objects in mathematics.0150

Vectors have their components as numbers, but they are treated as individual objects.0158

I can take a vector × a vector.0162

As it turns out, there is more than 1 way to multiply 2 vectors.0165

In fact there are any number of ways to multiply two vectors if you like.0169

One of the ways is the scalar product, the dot product.0173

Later on, we will learn something called the vector product, or the cross product.0176

So, now the operations start to become a little bit more complex because the objects themselves start to become a little bit more complex.0181

But, you can see, because vectors are represented by components, the definition of the products is slightly more complicated, but it is still just a multiplication of numbers added together. 0191

So, let us just do an example.0202

Again, the scalar product, given two vectors, (a1, a2, a3) and (b1, b2, b3), you are just going to take the product of the first components + the product of the second components + the product of the third components.0207

By product, we just mean numbers multiplied by numbers.0218

So, example 1.0223

We will let a = (-3,4,2), and we will let b = (4,2,1).0227

Well, the dot product, a · b is equal to -- I will write this out explicitly -- -3 × 4 + 4 × 2 + 2 × 1.0239

Then, I go ahead and equal them over here, equals -12 + 8 + 2, that is going to equal -2.0258

Okay, so we have two vectors which are points in space. 0268

We multiply those vectors by some multiplication that we call the scalar product or dot product.0272

We end up with a number.0279

Notice what we did, this is very important.0281

We took two vectors, which are not numbers, and we ended up with a number.0285

That is why in fact we call it the scalar product.0291

A scalar is just a fancy word for a number.0295

It is something that just has a magnitude, in other words, a value. 0299

A vector in of itself has a magnitude, a length, and it has a direction, but this is a scalar.0303

So, we took two vectors, we multiplied them and we ended up with a scalar.0312

That is really extraordinary.0315

Let me write that out formally.0319

Notice that this is a number, not another vector.0321

When we added 2 vectors together, we took a vector, and we took another vector, we added them together, and we ended up with a vector.0336

In other words we stayed in the space of vectors. 0341

Here we took a vector, we multiplied it by a vector, this procedure, and we ended up with a number. 0345

We did not end up with another vector, we jumped to another space, the space of real numbers.0350

I would say that that is really extraordinary. 0356

I have these in mind, I mean it is not ultimately important when we do the specific problems, but it is really sort of extraordinary to think about things globally -- that we are jumping around from space to space sometimes. 0360

Okay, so the scalar product also has some properties, let us go ahead and list those.0374

These I will number.0383

1) a·b = b·a, so if I want to take the scalar product, it is commutative, I can actually do the product in any order.0384

As it turns out later, it is not true for other vector products.0394

2) a·(b + c) = a·b + a·c, so as it turns out, the dot product is actually distributive over addition.0402

3) If I take some constant c and multiply it by vector a, and of course get another vector and I dot that with another vector, I can actually pull the constant out, dot the vectors first, then multiply it by a constant, it = c × a · b.0420

That is all that means, it means I can switch the order of constants and vector multiplication.0440

4) Okay, the 0 vector dotted by the 0 vector is equal to 0, notice the notation, 0 vector, 0 vector, 0 number, because the dot product is equal to another number, not another vector.0446

Last but not least, this is actually related to number 4, if a does not equal the 0 vector, then a dotted with itself is greater than 0 -- this is a very important thing.0460

A vector dotted with itself is a positive number. 0477

Now, there is a, well, you know what, I do not need to write this down, I can go ahead and tell you.0485

I do not want to Waste space and waste time here.0497

There is a geometric interpretation for what the dot product is.0499

In other words we can draw something out in the plane and tell you what the dot product means visually.0503

We will get to that, actually in not too far from now.0510

However, again what we want to do is we want to concentrate and we want to develop our algebraic abilities.0513

When we give a definition, which we did for scalar product, the definition was very clear. 0518

It tells you to follow this procedure to get this thing, this scalar product.0522

We want to develop algebraic capacities -- that is where the real mathematics takes place.0529

We use geometry to guide some of our intuition, obviously we do not want to throw out our intuition, we want to use it.0534

But again, for those of you who have taken a look at the linear algebra, intuition does not always guide you to mathematical truth.0542

You have to trust the mathematics.0548

At this level, that is often the case.0551

So we will talk about the geometric meaning of dot product but we will deal with it algebraically.0553

That is very important.0560

Now, let us talk about an interesting little property here.0564

You know that if I have x × y = 0, in other words two numbers equal to 0,0566

Then you know that either x or y = 0.0576

In other words you cannot have two numbers that multiplied together equal 0, one of them has to equal 0.0580

That is the whole foundation for factoring.0585

That is the whole foundation for finding the roots of polynomial equations.0589

It is the idea that you can set each factor equal to 0. 0591

You know that either x or y is 0.0595

This is not true for vectors, as it turns out.0610

In other words, this is not true for vectors.0616

In other words, there can be the vector a, which is not equal to 0, and a vector b, which is also not equal to 0,0627

In other words, a non-zero vector,0641

But, when I take a · b, I actually get 0.0646

As it turns out, there are certain properties.0652

As you can see, vectors share many of the properties of real numbers but they do not share this property. 0654

I can actually take the dot product of two vectors and get 0, without either a or b being equal to 0. 0659

Let us just do a quick demonstration of that.0667

If the vector a = (6,-2,4), and b = (2,2,-2),0670

If we do a · b, that is equal to 6 × 2 is 12, -2 × 2 is -4, and 4 × -2 is -8,0683

Look at that, we end up with 0 and neither one of these is the 0 vector.0697

Now, we are going to define and give a very important definition.0703

It will be a very simple thing to do, a very simple definition.0708

Do not let the simplicity of the definition make you think that it is not relevant.0713

It is actually profoundly relevant.0717

Define 2 vectors, a and b are said to be orthogonal, or perpendicular.0722

I am going to put perpendicular in parentheses. 0746

We definitely want to start to use the word orthogonal, because it is an algebraic definition and not a geometric definition. 0748

I will explain what that means in a minute.0754

a and b are said to be orthogonal, if a · b = 0.0757

Okay, so this is an algebraic definition.0767

It says that I have two vectors and if I take their scalar product, those two vectors are said to be orthogonal.0769

Orthogonal in 2 space and 3 space means that the vectors are perpendicular.0777

Perpendicular is a geometric notion.0782

Again from geometry, from trigonometry, when we think of things that are perpendicular, it just means that they make a right angle with each other.0785

Well, what if I am in 17 space.0793

A vector in 17 space, I have no way of picturing that.0795

If I have another vector in 17 space, if I take the dot product of those, the scalar product, and they it ends up being 0, how can I think about perpendicularity in that space? 0798

They do not actually make a 90 degree angle. 0807

A 90 degree angle is a geometric notion that only happens in the plane in 2 space.0810

As it turns out, orthogonal is the general term for perpendicularity.0815

It is true, you can go ahead and say perpendicular and orthogonal interchangeably, it is not a problem,0819

As long as you understand that it is the algebra that defines this.0825

Now we are moving away from just geometric notions of perpendicularity, of length.0828

We will still use some of these terms, but now we are generalizing those terms.0835

Algebraically if I take the dot product, if I multiply corresponding components of vectors and I end up with 0, those vectors are orthogonal.0840

Now, in 3 space, that just means they are perpendicular, but we will not be limiting ourselves to 3 space.0849

Let us do a quick example.0858

Example 2, we will let vector a = (1,1,-1) and vector b = (2,1,3).0860

The question is, are a and b orthogonal?0880

Well, let us take their dot product. a · b = 1 × 2 which is 2, 1 × 1 which is 1, -1 × 3 is -3. 0890

2 + 1 is 3, 3 - 3 is 0. 0904

The answer is yes, these vectors are orthogonal, they are perpendicular.0909

Okay, that is it, nice and easy.0911

The simplicity belies its importance. 0916

This notion of orthogonally is going to be very useful.0918

It is going to show up everywhere throughout the rest of the course0925

Now, let us talk about something called the norm.0930

So, we will start with a definition.0935

We said at the beginning of the lesson that the norm is just a fancy word for the length.0938

Again, we think of the length as a physical thing. 0946

The length of the pen that I am holding, the length of a vector in 2 space, the length of a vector in 3 space, these things we can generalize. 0949

We can visualize them, we can draw them.0957

But what if we had a vector in 10 space?0962

How do you talk about the length of a vector in 10 space? 0963

Well this definition that I am about to give you, which is algebraic, is that.0968

The norm of a vector symbolized by a vertical line, a double vertical line, a vector surrounded by double vertical lines,0975

That is just a symbol, okay?0995

That is defined as the vector a dotted with itself, and then you take the square root of that.1001

That is it, this is a very important definition, it is the norm of a vector, the length of the vector, the magnitude of the vector.1020

All of those terms are going to be used.1029

You take a vector and you dot it with itself, so a · a, then you get a number, then you take the square root of that number. 1031

Remember, we said a · a, as long as a · a is positive, so you can take the square root of that positive number.1040

This is actually nothing more than the pythagorean theorem in any number of dimensions. 1048

That is all this is.1052

You know that if I have some triangle sort of facing the vector like this, let us just draw this out.1054

You know that if this is the x-coordinate and this is the y-coordinate, and this is (x,y), a vector in 2 space.1061

Well you know that this length here is nothing more than x2+y2 under a radical.1068

This is the exact same thing, except we generalized it to any number of dimensions, it is this, I promise.1075

Okay, let us see, I am going to actually elaborate just a little bit more on this.1086

What I did regarding the pythagorean theorem. 1098

Let a = (a1,a2).1101

Let a = the vector (a1,a2), then the norm of a = sqrt(a · a).1106

Let us see, let us draw this out, if that is our vector a.1120

This is a1, this is a2, right? This is a1, this is a2, the first coordinate, the second coordinate, along the x, along the y.1128

What is a · a, well, a · a = a12 + a22 and square root of that.1138

This squared and that squared under a square root.1157

So you can see it is a vector way of describing the Pythagorean Theorem. 1160

Now that we give a vector definition of it, it generalizes into any number of dimensions.1166

We are no longer locked into just x2 + y2 = z2, no more.1170

Let us do an example.1175

I think this is example number 3.1186

We will let a = (4,2,6), and the norm of a, I know that the notation can be a bit tedious, but definitely take the time to write it out.1190

Whatever you do in mathematics, do not do it in your head, write it out.1205

Being able to do something in your head is not a measure of anything, it is just a measure of being able to do something in your head.1210

This stuff tends to be complex, you want to be able to see what you are doing, write it out.1216

It takes a little extra time, but I promise it will pay huge dividends, do not write things out in your head.1220

Okay, we said it is again, a · a under the radical,1226

a · a = this times itself, so it is 4 × 4 = 16 + 2 × 2 = 4, + 6 × 6 = 36, all under the radical equals sqrt(56).1234

So for this vector, (4,2,6), the length of this vector is sqrt(56).1251

In other words, the distance from the origin to this point is sqrt(56), that is all that is.1257

Now, observe the following.1264

The norm of a vector a = the norm of the vector -a.1275

Well, this makes sense just in terms of the definition, because again, we are squaring things.1283

When we square things the negative sign goes away.1288

If you want the geometric interpretation of this, it is just this.1292

If this is our vector a, we know that -1 is the vector of the same length in the other direction, so this is -a.1295

Well the length is the same, they have the same length.1304

As it turns out, this is true, so if we negate a vector, the norm is the same.1308

If you want, you can recalculate the norm, but it is just good to sort of observe that is the case.1314

Now, let me get a clean page here.1326

Talk about the last thing that we want to talk about today.1329

Let a and b be 2 vectors.1333

In other words 2 individual vectors, not 2 vectors, not a vector in 2 space.1342

Now, we know the vectors are just another way of to specify points in space.1348

The distance between these points is the vector a - b.1381

We will stop and think about this for a second.1395

This is a vector representation of the distance between two points.1398

We speak about a vector, a mathematical object, a directed line segment, one going this way, one going this way, they are actually points in space.1401

The distance between those two points in space is the vector a - the vector b.1410

Let us show you what that looks like geometrically, this is actually important to understand geometrically.1415

Let us just take our one vector a, and put it over here.1421

That is vector a, we will do that.1427

Now let us put vector b this way.1430

That represents this point in space, and that represents this point in space -- how do I find the distance between these two points in space? 1435

Well it is true, you are going to use the distance formula in 2 and 3 space.1444

However, as it turns out, the definition that I am about to give you for the distance between 2 points in space, or 2 vectors,1446

It is again, a generalization of the distance formula for any number of dimensions.1455

Now we are working in any number of dimensions, we are no longer limited.1459

Here is how it happens, a - b, when we are adding vectors, in this case a + -b,1462

This is just equivalent to vector a + the vector -b.1470

When we add vectors geometrically, we just take the first vector,1476

Then we take the other vector, and we can put the tail of that vector on the head of the first one, and we go some place else.1480

So the vector a - b looks like this.1490

It means do a first, then take b, which is this vector here.1494

But, since it is -b, go in the other direction.1499

That is all you are doing, if we said take the vector a + b, it would be a + b.1505

If we had another vector c, let us say we had another vector c here, if we said, what is a + b + c?1515

We would go a first, then from there you would go b in the same length and the same direction, and then you would go c, which is like that.1520

Head to tail, head to tail, head to tail. 1530

Well, in the case of a + a - b, you have a + a negative b, so you are going to be moving in the opposite direction of b in the same length,1534

That is going to give you a new vector, I will do this one with a dotted line.1545

That is the vector a - b, as it turns out, look at how long this is, that turns out to be exactly as long as this.1550

What you end up with is this parallelogram geometrically, so the distance between this point and this point, is the same as the distance between this point and this point.1560

Well, this point and this point marks the vector a - b.1575

That is what is going on here geometrically.1580

Let us go ahead and define this distance.1583

Define the distance between vector a and vector b as the following: It is the norm of the vector a - b.1590

This is the symbolism, it equals exactly what a norm is, a - b is a vector.1613

We said that the norm of the vector is the vector dotted by itself, then you take the square root.1620

So we have a - b · a - b under the square root.1627

Let us just do an example, that is what is important.1637

Example 4.1645

Let a = (-1,1,6) and b = (1,2,3,).1648

Again these are 3 vectors, points in 3 space.1660

The vector a - b equals, I am going to write all of this out explicitly, (-1,1,6) - (1,2,3). 1664

Well, that equals (-1-1,1-2,6-3) = (-2,-1,3), the vector (-2,-1,3) that is the vector a - b.1674

Since that is the vector a - b based on the definition that we just wrote, let us dot this with itself.1696

a - b · a - b, dot product with itself, again we are going to write this out explicitly.1706

(-2,-1,3) · (-2,-1,3) = -2 × -2 is 4, -1 × -1 is 1, 3 × 3 is 9, square root of that 4,1,9, 14,1719

So we end up with radical 14.1740

Therefore the norm of a-b = sqrt(14).1743

In other words, the distance between the vector a and the vector b is sqrt(14).1750

The distance between the point that a represents, and the point that b represents is the distance of 14.1756

Believe it or not, this is just the formula that I gave which is a generalization of the distance formula in any number of dimensions, expressed in vectors.1767

Again, trust the definition.1778

Hopefully the geometry, or the geometric interpretation that we gave you helps a little bit.1780

It is fine to use that geometric intuition at this point, think about it, look at it, if you want break to it up into components.1786

Write everything out until you are comfortable with this notion.1795

It is just a straight application of basic algebra.1800

The norm is this, and this distance between two points is the norm.1802

It is a norm because that is what a norm is, it is a distance, you have that definition.1808

Okay, so we have introduced scalar product, we have introduced norm, 1814

We will go ahead and stop here for today, and next time we will continue on with more vectors.1817

Thank you for joining us here at, we will join you next time, bye-bye.1822