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

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

### Scalar Product & Norm

- The scalar product is defined as →a ×→b = a
_{1}a_{2}+ b_{1}b_{2}for →a = (a_{1},b_{1}) and →b = (a_{2},b_{2}). - Substituting yields →a ×→b = ( − 2)(5) + (7)( − 4) = − 10 + ( − 28).

- The scalar product is defined as →a ×→b = a
_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}for →a = (a_{1},b_{1},c_{1}) and →b = (a_{2},b_{2},c_{2}). - Substituting yields →a ×→b = (1)( − [1/2] ) + (0)(8) + ( − 3)(6) = − [1/2] + 0 + ( − 18).

- 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 ).

- 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.

- 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.

- We must check that →i→j = 0, →i→k = 0 and →j→k = 0.
- Note that →i→j = (1,0,0) ×(0,1,0) = 0 because no two components have a nonzero product.

i) Find − →a

- The opposite of a vector is the product of − 1 to each component.

ii) Find − →a + →a

- The sum of two vector is the addition of each component.

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 .

^{2}= ( √3 )

^{2}= 3.

iv) Find || − →a || + || →a ||

- Recall that || − →a || = || →a ||, so that our problem is to find || →a || + || →a || = 2|| →a ||.

- The norm of a vector is defined as || →u || = √{→u ×→u} .
- Substituting gives || →u || = √{( − √5 ,7) ×( − √5 ,7)} = √{( − √5 )
^{2}+ 7^{2}} = √{5 + 49} = √{54} .

- 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)} = √{1
^{2}+ ( − 2)^{2}} = √5 .

- 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) || = √{1.5
^{2}+ 0.2^{2}+ 1.9^{2}+ 4.6^{2}} = √{27.06} .

*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

### 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

### Multivariable Calculus

### Transcription: Scalar Product & Norm

*Hello, and welcome back to educator.com, 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 course*0925

*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 x ^{2}+y^{2} 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 = a1 ^{2} + a2^{2} 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 x ^{2} + y^{2} = z^{2}, 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 educator.com, we will join you next time, bye-bye.*1822

1 answer

Last reply by: Professor Hovasapian

Fri Aug 26, 2016 7:28 PM

Post by Kaye Lim on August 25, 2016

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, 2016

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?

Thanks

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 â†’a = (12, âˆ’ 5) and â†’b = (3, âˆ’ 13). Find || â†’a âˆ’ â†’b ||.

Step 1. First compute the difference between the vectors, â†’a âˆ’ â†’b = (12, âˆ’ 5) âˆ’ (3,13) = (12, âˆ’ 5) + ( âˆ’ 3, âˆ’ 13) = (9, âˆ’ 18) = 9(1, âˆ’ 2).

Shouldn't the vector of â†’a âˆ’ â†’b be (9,8)since â†’a + (-â†’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 â†’a = ( âˆ’ 2,5) and â†’b = (7, âˆ’ 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 â†’a Ã—â†’b = a1a2 + b1b2 for â†’a = (a1,b1) and â†’b = (a2,b2).

Step 2. Substituting yields â†’a Ã—â†’b = ( âˆ’ 2)(5) + (7)( âˆ’ 4) = âˆ’ 10 + ( âˆ’ 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

Raffi,

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