0:00:01.950,0:00:06.214 In this unit, we're going to[br]have a look at a method for 0:00:06.214,0:00:10.204 combining two vectors. This[br]method is called the scalar 0:00:10.204,0:00:14.076 product. And it's called the[br]scalar product because when we 0:00:14.076,0:00:18.332 calculate it, the result that we[br]get or the answer is a scalar as 0:00:18.332,0:00:19.548 opposed to a vector. 0:00:20.550,0:00:23.820 So let's look at two vectors.[br]Here's the first vector. 0:00:27.610,0:00:29.050 That's cool that vector A. 0:00:32.520,0:00:36.953 And here's the second vector.[br]Let's call this vector B and 0:00:36.953,0:00:41.386 you'll notice that I've drawn[br]these two vectors so that the 0:00:41.386,0:00:43.804 tales of the two vectors[br]coincide. 0:00:46.210,0:00:47.788 Now that they coincide, we can 0:00:47.788,0:00:52.090 measure this angle. Between the[br]two vectors and I'm going to 0:00:52.090,0:00:53.470 label the angle theater. 0:00:54.590,0:00:56.410 So there are two vectors. 0:00:57.340,0:01:00.993 Now when we calculate the scalar[br]product, we do it like this. We 0:01:00.993,0:01:05.013 find the length. Or the[br]magnitude, or the modulus of 0:01:05.013,0:01:06.000 the first vector. 0:01:07.050,0:01:09.420 Which is the modulus of a. 0:01:11.460,0:01:13.338 We multiply it by the length. 0:01:14.530,0:01:17.880 Of the second vector, which is[br]the modulus of beat. 0:01:19.600,0:01:25.072 And we multiplied by the cosine[br]of this angle between the two 0:01:25.072,0:01:28.720 vectors. We multiplied by the[br]cosine of Theta. 0:01:29.970,0:01:33.413 And that's how we calculate the[br]scalar product. The length of 0:01:33.413,0:01:36.543 the first vector times the[br]length of the second vector 0:01:36.543,0:01:39.986 multiplied by the cosine of the[br]angle in between the two 0:01:39.986,0:01:44.400 vectors. And we have a[br]notation for the scalar 0:01:44.400,0:01:49.380 product and we write it like[br]this vector A dot vector be. 0:01:50.720,0:01:54.640 And this is the definition.[br]Sometimes this scalar product is 0:01:54.640,0:01:59.344 also referred to as a dot[br]product because of that dot in 0:01:59.344,0:02:04.048 there. So sometimes you'll hear[br]it referred to as a dot product. 0:02:06.970,0:02:10.652 OK, so let's have an example and[br]see if we can calculate the dot 0:02:10.652,0:02:11.704 product of two vectors. 0:02:14.350,0:02:16.180 Let's suppose our vector a here. 0:02:16.740,0:02:19.220 Has got length for units. 0:02:23.200,0:02:28.960 And let's suppose we have a[br]vector B up here and let's 0:02:28.960,0:02:32.320 suppose vector B has length[br]5 units. 0:02:33.420,0:02:37.644 And let's suppose that the angle[br]for the sake of argument between 0:02:37.644,0:02:39.756 the two vectors is 60 degrees. 0:02:41.220,0:02:45.367 And let's use this formula to[br]calculate the scalar product A 0:02:45.367,0:02:47.629 dot B. It's a vector there. 0:02:49.250,0:02:51.990 A dot B will be. 0:02:52.870,0:02:55.686 Well, it's the modulus or the[br]length of the first vector. 0:02:57.120,0:02:58.050 Which is 4. 0:03:01.730,0:03:04.907 Multiplied by the modulus[br]or the length of the 0:03:04.907,0:03:07.731 second vector and the[br]length of the second 0:03:07.731,0:03:08.790 vector is 5. 0:03:11.950,0:03:16.435 Multiplied by the cosine of the[br]angle between the two, and so we 0:03:16.435,0:03:18.505 want the cosine of 60 degrees. 0:03:20.950,0:03:25.066 Now the cosine of 60 degrees. If[br]you don't know, you can work it 0:03:25.066,0:03:28.888 out on your Calculator, but it's[br]a common angle an the cosine of 0:03:28.888,0:03:33.886 60 degrees is 1/2, so we've got[br]4 * 5 * 1/2, so if we work all 0:03:33.886,0:03:38.296 that out, we get 4 files, or 20[br]and a half of 20 is 10. 0:03:40.350,0:03:43.620 So that's our first example of[br]calculating a scalar product, 0:03:43.620,0:03:47.544 and let me remind you of[br]something that I said Very early 0:03:47.544,0:03:51.468 on that when we calculate the[br]scalar product, the answer is a 0:03:51.468,0:03:55.719 scalar and you'll see the answer[br]here 10 is just a single number 0:03:55.719,0:03:59.316 Whilst we started with two[br]vectors A&B, we finish up with 0:03:59.316,0:04:02.913 just a single number, a scalar,[br]and that's the scalar product. 0:04:05.640,0:04:09.400 Let me illustrate some other[br]features of this scalar product 0:04:09.400,0:04:12.784 by doing the calculation the[br]opposite way round, supposing 0:04:12.784,0:04:17.296 I'd try to calculate the dotted[br]with a instead. Let's work that 0:04:17.296,0:04:19.176 through and see what happens. 0:04:20.350,0:04:23.998 So I've done the operation in a[br]different order instead of a 0:04:23.998,0:04:25.822 baby. I've now got dot A. 0:04:27.350,0:04:30.610 Well, again, using the[br]definition we want to say that 0:04:30.610,0:04:32.240 the dot product of DNA. 0:04:32.800,0:04:36.220 Is the modulus or length of the[br]first vector which is the 0:04:36.220,0:04:40.495 modulus or the length of be this[br]time and the modulus of B is 5? 0:04:43.310,0:04:46.401 Multiply by the modulus or the[br]length of the second vector. 0:04:47.890,0:04:49.078 Which is now 4. 0:04:50.670,0:04:53.946 Multiplied by the cosine of the[br]angle between the two and the 0:04:53.946,0:04:55.038 angle still 60 degrees. 0:04:56.340,0:05:01.395 The cosine of 60 degrees is 1/2,[br]so if we work this out this time 0:05:01.395,0:05:06.113 will get five 420 1/2 of 20 is[br]10, so you'll see that whichever 0:05:06.113,0:05:10.494 way we did the calculation[br]whether we worked out a dot B or 0:05:10.494,0:05:14.875 whether we worked out B dotted[br]with a, we get the same answer. 0:05:15.650,0:05:20.859 10 and that's true in general[br]for any two vectors that we 0:05:20.859,0:05:23.302 choose. If we workout a dot B. 0:05:26.980,0:05:31.171 That will be the same as working[br]out B Dot A. 0:05:32.110,0:05:36.160 And that's another important[br]property of the scalar product, 0:05:36.160,0:05:38.410 and we call this property 0:05:38.410,0:05:42.540 commutativity. We say that the[br]dot product or the scalar 0:05:42.540,0:05:43.560 product is commutative. 0:05:56.490,0:05:59.250 Now, as well as the[br]commutativity property, I 0:05:59.250,0:06:02.355 want to tell you about[br]another property which is 0:06:02.355,0:06:03.390 known as distributivity. 0:06:04.960,0:06:11.470 Suppose we have a dotted with[br]the sum of two vectors B Plus C. 0:06:13.420,0:06:16.228 The distributivity property[br]tells us to expand these 0:06:16.228,0:06:20.089 brackets in the normal sort of[br]algebraic way that you would 0:06:20.089,0:06:23.248 expect. We work this out by[br]saying it's a. 0:06:23.980,0:06:25.219 Dotted with B. 0:06:28.580,0:06:29.348 Add it to. 0:06:31.210,0:06:33.970 A dotted with C. 0:06:38.320,0:06:41.580 This also works the other[br]way round, so suppose we 0:06:41.580,0:06:42.884 have B Plus C. 0:06:45.390,0:06:50.556 And we want to dot it with the[br]vector a As you might predict 0:06:50.556,0:06:54.246 the result that will get is B[br]dotted with a. 0:06:57.600,0:06:58.390 Attitude. 0:07:00.580,0:07:01.768 See dotted with a. 0:07:03.450,0:07:08.733 And these properties are said to[br]be the distributivity. 0:07:10.780,0:07:11.480 Rules. 0:07:16.140,0:07:19.400 I will need those rules along[br]with the commutativity rules 0:07:19.400,0:07:21.030 later on in this unit. 0:07:27.050,0:07:30.002 Now there's another property[br]very important property of the 0:07:30.002,0:07:33.938 scalar product that I like to[br]tell you about as well, and 0:07:33.938,0:07:36.562 it's the scalar product of two[br]perpendicular vectors. 0:07:49.190,0:07:53.315 Let's see what happens when the[br]two vectors that we're dealing 0:07:53.315,0:07:57.440 with A&B are perpendicular. That[br]means there at right angles. So 0:07:57.440,0:08:01.940 we have a situation like this.[br]There's a vector B. There's a 0:08:01.940,0:08:06.440 vector A and they're separated[br]by an angle of 90 degrees, so 0:08:06.440,0:08:07.940 the vectors are perpendicular. 0:08:09.510,0:08:12.880 Let's use our formula for the[br]dot product. The scalar 0:08:12.880,0:08:16.587 product, the modulus of the[br]first one with the modulus of 0:08:16.587,0:08:19.957 the second one times the[br]cosine of the angle between 0:08:19.957,0:08:23.327 the two. Let's use that[br]formula in this specific case, 0:08:23.327,0:08:25.012 when the vectors are[br]perpendicular. 0:08:27.930,0:08:30.757 Well, we want the length of[br]the first one. Still the 0:08:30.757,0:08:31.528 modulus of A. 0:08:32.680,0:08:36.424 The length of the second one[br]will be the modulus of B and 0:08:36.424,0:08:40.168 this time we want the cosine of[br]Theta, but theater is 90 degrees 0:08:40.168,0:08:41.896 from the cosine of 90 degrees. 0:08:43.830,0:08:46.888 Now you either should know[br]or you can easily check on 0:08:46.888,0:08:49.668 your Calculator that the[br]cosine of 90 degrees is 0. 0:08:51.600,0:08:54.048 So these two vectors, these two 0:08:54.048,0:08:58.088 moduli? Whatever they are[br]multiplied by zero. So the 0:08:58.088,0:09:00.440 answer that will get will be 0. 0:09:01.480,0:09:05.848 This is a very important result[br]that if you have two vectors 0:09:05.848,0:09:08.760 which are perpendicular, their[br]scalar product is 0. 0:09:21.920,0:09:24.530 So for two perpendicular[br]vectors a. 0:09:26.070,0:09:26.540 The. 0:09:28.080,0:09:30.964 Important result which[br]you should learn that 0:09:30.964,0:09:34.260 their dot product their[br]scalar product is 0. 0:09:37.880,0:09:42.425 Now the converse of this is also[br]true, and by that I mean that if 0:09:42.425,0:09:46.061 we choose any two vectors at[br]all, and we find their dot 0:09:46.061,0:09:50.000 product and we find that the[br]answer is 0, that will mean that 0:09:50.000,0:09:53.333 provided that neither a norby[br]was zero, those two vectors must 0:09:53.333,0:09:57.272 be perpendicular and will use[br]this as a test later on in this 0:09:57.272,0:09:59.393 unit to test whether or not to 0:09:59.393,0:10:01.970 given vector. Are[br]perpendicular or not? 0:10:08.840,0:10:13.880 I want to move on now to look at[br]how we calculate the scalar 0:10:13.880,0:10:17.480 product of two vectors when[br]they're given in Cartesian form. 0:10:17.480,0:10:20.720 Let me remind you about[br]cartesian form when we're 0:10:20.720,0:10:23.960 dealing with vectors in[br]Cartesian form, where we express 0:10:23.960,0:10:28.280 the vector in terms of unit[br]vectors IJ&K. So we might be 0:10:28.280,0:10:30.080 dealing with vectors of this 0:10:30.080,0:10:33.260 form 3I. Minus two J. 0:10:34.440,0:10:35.950 Plus 7K. 0:10:37.700,0:10:41.426 Expect to be might be something[br]like minus 5I. 0:10:42.730,0:10:48.760 Plus 4J. Minus 3K. So when[br]we see the vectors written down 0:10:48.760,0:10:51.448 like this with eyes and Jays and 0:10:51.448,0:10:55.186 kays in. These vectors are now[br]given in Cartesian form. 0:10:55.800,0:10:59.921 And we're going to learn how[br]to do now is figure out how 0:10:59.921,0:11:02.457 to calculate the dot[br]product. The scalar product 0:11:02.457,0:11:03.408 vectors like these. 0:11:04.470,0:11:08.331 Before we do that, I'd like to[br]introduce some other important 0:11:08.331,0:11:11.841 results. Let me remind you a[br]little bit about IJ&K. 0:11:15.830,0:11:19.232 In three dimensions, where we[br]have an X axis. 0:11:20.410,0:11:23.310 AY axis. And as Ed Access. 0:11:24.540,0:11:27.969 We did note unit vectors along[br]the X axis. 0:11:29.020,0:11:36.466 I I along the Y axis by J and[br]along the Z Axis by K and you'll 0:11:36.466,0:11:38.656 notice that the vectors IJ&K. 0:11:39.570,0:11:43.730 US are all it's 90 degrees to[br]each other, so the angle between 0:11:43.730,0:11:47.890 I&J is 90 degrees. The angle[br]between Jane K is 90 degrees and 0:11:47.890,0:11:51.730 the angle between I&K is 90[br]degrees and all of these vectors 0:11:51.730,0:11:55.570 I Jane KA unit vectors. That[br]means that their length is one. 0:11:57.370,0:12:02.639 Now let's just calculate the dot[br]product of I with J. 0:12:08.300,0:12:11.600 We use the result we had before[br]the definition of the dot 0:12:11.600,0:12:14.625 product, which says that we[br]write down the length of the 0:12:14.625,0:12:18.475 first vector and the length of I[br]being a unit vector is just one. 0:12:20.880,0:12:24.389 The length of J being a unit[br]vector is also one. 0:12:25.800,0:12:29.687 And we want the cosine of the[br]angle between the two and the 0:12:29.687,0:12:33.275 angle between the two is 90[br]degrees, so we want the cosine 0:12:33.275,0:12:38.500 of 90 degrees. And again, the[br]cosine of 90 degrees is 0, so 0:12:38.500,0:12:42.964 we find that I got Jay is[br]simply zero. We could have 0:12:42.964,0:12:45.940 deduced that from the[br]previous result that I 0:12:45.940,0:12:49.660 obtained for you, which was[br]that for any two vectors 0:12:49.660,0:12:52.636 which are perpendicular,[br]their dot product is 0. 0:12:54.900,0:13:00.234 So I got Jays 0. Now the same[br]argument tells us also that Jay 0:13:00.234,0:13:02.520 dotted with K must be 0. 0:13:07.400,0:13:09.296 Be cause J&K are[br]perpendicular vectors 0:13:09.296,0:13:12.456 there at right angles and[br]the angle between them is 0:13:12.456,0:13:15.300 90 degrees and the cosine[br]of 90 is 0. 0:13:16.770,0:13:23.610 What about INK? Well, I dotted[br]with K must also be 0. 0:13:25.900,0:13:28.958 What about dotting the vector[br]with itself? Let's just have a 0:13:28.958,0:13:31.738 look at that. Suppose we wanted[br]I dotted with I. 0:13:34.330,0:13:36.877 Well, the length of the[br]first vector is one. 0:13:38.360,0:13:42.248 The length of the second vector[br]is still one, and if we're 0:13:42.248,0:13:45.488 dotting a vector with itself,[br]the angle between the two 0:13:45.488,0:13:49.700 vectors must be 0. So this time[br]we want the cosine of 0. 0:13:50.680,0:13:53.859 And the cosine of 0. The cosine[br]of note is one. 0:13:54.540,0:13:59.556 So we have 1 * 1 * 1, which[br]is just one. 0:14:00.170,0:14:03.686 So if we're dotting the vector I[br]with itself, we get one. 0:14:04.690,0:14:09.201 A similar argument will tell us[br]that if we got the vector Jay 0:14:09.201,0:14:14.059 with itself, we also get one,[br]and if we drop the vector K with 0:14:14.059,0:14:15.447 itself, we get one. 0:14:16.400,0:14:19.344 Let me summarize those[br]important results here and 0:14:19.344,0:14:23.392 I've results that you should[br]remember that if you took the 0:14:23.392,0:14:27.808 vector I and dotted it with[br]itself, that would be the same 0:14:27.808,0:14:29.648 as dotting Jay with itself. 0:14:31.060,0:14:35.785 And it's the same as Dot Inc. A[br]with itself, and in any case we 0:14:35.785,0:14:37.045 get the answer 1. 0:14:38.850,0:14:46.007 If we don't, I with J we get[br]zero if we got I with K if we 0:14:46.007,0:14:51.480 get zero and if we got J with K[br]we also get 0. 0:14:52.480,0:14:55.312 And all those results are[br]particularly important, and 0:14:55.312,0:14:58.852 once that you should become[br]familiar with as you work 0:14:58.852,0:15:00.976 through the exercises[br]accompanying this unit. 0:15:05.890,0:15:10.674 Now I want to use those results[br]about the dot products of those 0:15:10.674,0:15:15.458 unit vectors to find a way to[br]calculate the dot product of two 0:15:15.458,0:15:19.138 arbitrary vectors that are given[br]in Cartesian form. So let's 0:15:19.138,0:15:22.450 choose two arbitrary vectors.[br]Let's suppose the first one. 0:15:23.750,0:15:26.060 Is a one. 0:15:27.430,0:15:28.010 I. 0:15:29.830,0:15:35.120 A2, J and[br]a 3K. 0:15:36.520,0:15:38.062 So this is now an arbitrary 0:15:38.062,0:15:44.094 vector. A where a one A2 and a[br]three are the three cartesian 0:15:44.094,0:15:46.154 components of the vector A. 0:15:47.270,0:15:52.926 And let's suppose vector B[br]similarly is B1I. 0:15:54.500,0:16:00.593 Plus B2J Plus V3K and[br]again B1B2B3 arbitrary numbers, 0:16:00.593,0:16:06.686 so this is an arbitrary[br]vector in three dimensions. 0:16:08.260,0:16:12.280 What I want to do is workout the[br]dot product of A&B. 0:16:13.510,0:16:17.434 So let's work it through a[br]dot B equals. 0:16:18.900,0:16:21.268 It's the first vector[br]which is all this. 0:16:31.180,0:16:36.060 Dotted with the second vector[br]be, which is all this. 0:16:46.100,0:16:51.170 OK. Now what we want to do[br]now is use our 0:16:51.170,0:16:54.329 distributivity rule to be[br]able to workout to expand 0:16:54.329,0:16:57.137 these brackets and workout[br]the individual dot products. 0:16:58.310,0:17:02.524 And we can do that using the[br]rules that we know that in turn 0:17:02.524,0:17:04.029 each one of these vectors. 0:17:04.850,0:17:07.775 In the first bracket,[br]multiplies each of these in 0:17:07.775,0:17:11.025 the SEC bracket and just the[br]normal algebraic way. So 0:17:11.025,0:17:15.575 first of all we want a one I[br]dotted with B1 I those terms. 0:17:24.480,0:17:26.196 Then we want a one I. 0:17:26.880,0:17:28.800 Dotted with B2J 0:17:37.020,0:17:41.388 and finally a one[br]I dotted with B3K. 0:17:47.490,0:17:50.290 So that takes care of this first 0:17:50.290,0:17:54.730 vector here. And we moved to[br]the second vector. So now we 0:17:54.730,0:17:57.470 want plus A2, J, dotted with[br]each of these three. 0:18:01.200,0:18:04.024 So it's a 2 J dotted with B1I. 0:18:09.120,0:18:11.450 A2 J dotted with B2J. 0:18:17.180,0:18:19.686 And a two J dotted with B3K. 0:18:27.000,0:18:30.443 And finally, this loss vector in[br]the first bracket dotted with 0:18:30.443,0:18:34.199 each of these vectors in the SEC[br]bracket will be a 3K. 0:18:37.350,0:18:38.940 Dotted with B1I 0:18:40.860,0:18:43.500 plus a 3K dotted with B2J. 0:18:48.500,0:18:51.524 And finally, a 3K[br]dotted with B3K. 0:18:56.540,0:19:00.016 Now this looks horrendous. With[br]all these nine terms in here, 0:19:00.016,0:19:04.124 but a lot of this is going to[br]cancel out now. You'll remember 0:19:04.124,0:19:06.968 that any vectors which are[br]perpendicular to each other. 0:19:08.000,0:19:10.065 Have a dot product which is 0. 0:19:11.590,0:19:12.640 Now this vector. 0:19:14.320,0:19:16.742 Is in the eye direction and this 0:19:16.742,0:19:21.792 vector. Is in the J direction,[br]so a one I is perpendicular to 0:19:21.792,0:19:22.830 be 2 J. 0:19:23.890,0:19:28.420 So where we have the item dotted[br]with the J term, this must be 0. 0:19:31.840,0:19:36.076 Similarly, any vector in the[br]direction of I any multiple of I 0:19:36.076,0:19:38.900 must be perpendicular to any[br]multiple of K. 0:19:39.900,0:19:42.147 So this dot product is[br]also zero. 0:19:43.890,0:19:47.850 And similarly for the A2JB1I,[br]those vectors are perpendicular. 0:19:47.850,0:19:50.050 Their dot product is 0. 0:19:52.620,0:19:55.955 A2J dot B3K. A perpendicular 0:19:55.955,0:20:01.200 that's zero. Similarly, that[br]will be 0 and that will be 0. 0:20:01.940,0:20:05.117 And we'll be left with[br]three non zero terms. 0:20:07.310,0:20:08.610 What about this term here? 0:20:10.170,0:20:15.386 A1 I be one I now this is a[br]multiple of I and this is a 0:20:15.386,0:20:17.342 multiple of I. So these two 0:20:17.342,0:20:22.251 vectors are parallel. The angle[br]between them is 0, so when we 0:20:22.251,0:20:26.319 work out the dot product will[br]want the length of the first 0:20:26.319,0:20:28.014 one, which is a one. 0:20:28.810,0:20:30.310 The length of the second one, 0:20:30.310,0:20:31.510 which is. He wants. 0:20:33.170,0:20:36.052 And the cosine of the angle[br]between the two. The angle 0:20:36.052,0:20:41.649 between the two. Is zero and the[br]cosine of 0 is one, so we get a 0:20:41.649,0:20:43.264 one B 1 * 1. 0:20:44.340,0:20:47.740 So this first term, whilst it[br]looked quite complicated, just 0:20:47.740,0:20:49.440 simplifies to a 1B one. 0:20:52.090,0:20:54.884 Similarly, these two vectors[br]here. This is a vector in the 0:20:54.884,0:20:58.186 direction of J and this is a[br]vector in the direction of Jay. 0:20:58.830,0:21:03.000 So the angle between the two of[br]them is 0, the cosine of 0 is 0:21:03.000,0:21:04.946 one, so when we workout this dot 0:21:04.946,0:21:09.453 product we want. The length of[br]the first one, which is a 2. The 0:21:09.453,0:21:10.911 length of the second one which 0:21:10.911,0:21:16.373 is B2. Multiplied by the cosine[br]of 0. Cosine of 0 is one, so we 0:21:16.373,0:21:17.657 just get a 2B2. 0:21:19.150,0:21:22.102 And finally, same argument[br]applies here. This is a 0:21:22.102,0:21:26.038 multiple of K. This is a[br]multiple of K, so the vectors 0:21:26.038,0:21:29.974 are parallel. The angle between[br]them is zero and the cosine of 0:21:29.974,0:21:34.894 0 is one. So we get the length[br]of the first being A3 length of 0:21:34.894,0:21:39.486 the second being B3 and the[br]cosine of 0 cosine of 0 is one, 0:21:39.486,0:21:42.766 so the whole lot. All of this[br]complicated stuff just 0:21:42.766,0:21:46.702 simplifies down at the end of[br]the day to this very important 0:21:46.702,0:21:47.358 result here. 0:21:49.810,0:21:55.036 And what this result says is[br]if you want to find the scalar 0:21:55.036,0:21:59.056 product of two vectors A&B[br]which are given in this 0:21:59.056,0:22:03.880 cartesian form, like that, all[br]we need to do is multiply the 0:22:03.880,0:22:08.302 I components together. You see[br]the A1B one is the eye 0:22:08.302,0:22:09.508 components multiplied[br]together. 0:22:11.410,0:22:13.374 Multiply the J components 0:22:13.374,0:22:18.458 together A2B2. Multiply the K[br]components together A3B3. 0:22:19.370,0:22:21.860 And then finally add up these 0:22:21.860,0:22:24.400 results. So the dot product. 0:22:25.420,0:22:26.488 Is the sum. 0:22:27.170,0:22:30.579 Of the products of the[br]corresponding components. 0:22:32.220,0:22:35.751 And you'll notice that the[br]result we get here on the 0:22:35.751,0:22:39.603 right hand side doesn't have[br]any eyes or Jays orkez in the 0:22:39.603,0:22:43.134 answer anymore. This is purely[br]a number. This is purely a 0:22:43.134,0:22:46.344 scalar, as we'd expect when we[br]calculate a scalar product. 0:22:48.460,0:22:52.660 Point out also that this[br]expression on the right, which 0:22:52.660,0:22:57.280 is the sum of the products of[br]the corresponding components is 0:22:57.280,0:23:00.640 also sometimes referred to as[br]the inner product. 0:23:02.330,0:23:07.530 Of A&B, so on occasions you may[br]hear some staff or reading some 0:23:07.530,0:23:11.530 textbooks that this is called[br]the inner product of Bambi. 0:23:19.520,0:23:22.424 OK, so we have a formula[br]now for the dot product. 0:23:24.800,0:23:31.807 It's a one B 1 +[br]8, two B2 plus A3B3. 0:23:35.210,0:23:38.928 Can very important results and[br]this will allow us to calculate 0:23:38.928,0:23:42.308 the scalar product of two[br]vectors given in cartesian form. 0:23:42.308,0:23:44.336 Let me give you an example. 0:23:49.030,0:23:56.950 Suppose the first vector a is[br]this one 4I Plus 3J minus. 0:23:56.950,0:23:58.930 Sorry plus 7K. 0:24:00.620,0:24:03.952 And Suppose B is the vector two 0:24:03.952,0:24:06.480 I. +5 J. 0:24:07.980,0:24:08.910 Plus 4K. 0:24:11.300,0:24:16.459 Suppose we want to calculate the[br]scalar product A dot B. 0:24:18.390,0:24:22.636 Well, this formula tells us that[br]we multiply the corresponding I 0:24:22.636,0:24:27.268 components together A1B one.[br]This is a one for this is B1 0:24:27.268,0:24:31.514 which is 2. So we multiply the[br]four by the two. 0:24:35.290,0:24:38.530 We multiply the corresponding J[br]components together. A2B2 well, 0:24:38.530,0:24:43.570 this is a two and this is B2, so[br]we want 3 * 5. 0:24:47.080,0:24:50.280 And then we multiply the[br]corresponding K components 0:24:50.280,0:24:53.080 together. A3B3, which is 7 * 4. 0:24:56.050,0:25:03.415 When we do that and we add up[br]all the results and will get 8 0:25:03.415,0:25:10.289 + 3, five, 15, Seven, 428. And[br]if we work those out we shall 0:25:10.289,0:25:17.163 get 2815 and eight to add up[br]to 51. So the dot product of 0:25:17.163,0:25:19.618 A&B is the scalar 51. 0:25:22.420,0:25:25.940 Now, sometimes, instead of[br]writing it all out like this, we 0:25:25.940,0:25:29.780 can use column vector notation[br]and some people find it a bit 0:25:29.780,0:25:33.620 easier to work with column[br]vectors. So just let me show you 0:25:33.620,0:25:37.460 how we do the same calculation.[br]If we had these vectors given 0:25:37.460,0:25:41.300 not in this form, but as column[br]vectors, suppose we had a 0:25:41.300,0:25:47.258 written as 437. And[br]be written AS254. 0:25:49.020,0:25:51.306 Then the a dotted with B. 0:25:52.170,0:25:57.786 Would be 437 dotted[br]with 254. 0:26:00.350,0:26:02.886 Now to multiply corresponding[br]components together is quite 0:26:02.886,0:26:07.641 easy to see what they are now[br]because we want 4 * 2, which is 0:26:07.641,0:26:13.820 8. 3 * 5 which is 15 and 7 *[br]4, which is 28. And if we had 0:26:13.820,0:26:17.850 those that will get 51 like we[br]did over here. So some people 0:26:17.850,0:26:21.880 might find it a bit easier just[br]to write them as column vectors 0:26:21.880,0:26:24.360 and then just read off the[br]corresponding components, 0:26:24.360,0:26:27.770 multiply them together, and add[br]them up to get the result. 0:26:32.200,0:26:36.088 Now it's so important that you[br]know how to calculate a scalar 0:26:36.088,0:26:39.976 product that I'm going to give[br]you one more example, just to 0:26:39.976,0:26:42.244 make sure that you understand[br]the process. 0:26:43.420,0:26:48.060 And will go straight into column[br]vector notation. Suppose the 0:26:48.060,0:26:51.308 vector A is the vector minus 6. 0:26:51.870,0:26:54.420 3 - 11. 0:26:55.710,0:26:59.014 That's minus six I plus 3J minus 0:26:59.014,0:27:06.532 11 K. And let's suppose[br]the vector B is the 0:27:06.532,0:27:11.600 vector 12:04 that's twelve[br]IOJ plus 4K. 0:27:14.670,0:27:20.354 So the dot product A dot B will[br]be minus six 3 - 11. 0:27:21.520,0:27:24.348 Dotted with twelve 04. 0:27:26.050,0:27:29.504 And in the column vector[br]notation, it's so easy to work 0:27:29.504,0:27:33.272 this out. We want minus 6 * 12,[br]which is minus 72. 0:27:35.410,0:27:37.680 3 * 0 is 0. 0:27:40.100,0:27:43.604 And minus 11 * 4 is minus 44. 0:27:44.980,0:27:46.764 And if we add these up, we get. 0:27:48.180,0:27:51.210 Minus 116 0:27:53.080,0:27:57.030 so the dot product of A&B is[br]just the sum. 0:27:57.850,0:28:00.986 Of the products of the[br]corresponding components. 0:28:06.430,0:28:11.022 I want to move on now to look at[br]some applications of the scalar 0:28:11.022,0:28:13.974 product and the first[br]application is one that I 0:28:13.974,0:28:17.910 mentioned very early on and it's[br]to do with testing whether or 0:28:17.910,0:28:19.550 not two given vectors are 0:28:19.550,0:28:23.799 perpendicular or not. Now you[br]remember from the definition of 0:28:23.799,0:28:27.506 the scalar product of two[br]vectors, 8 be that is the 0:28:27.506,0:28:31.550 modulus of the first one times[br]the modulus of the second one 0:28:31.550,0:28:35.257 times the cosine of the angle in[br]between the two vectors. 0:28:36.280,0:28:39.460 Now, if we find that[br]when we work this out 0:28:39.460,0:28:41.050 that the result is 0. 0:28:42.330,0:28:47.048 Then either the modulus of a[br]must be 0, or the modulus of be 0:28:47.048,0:28:50.418 must be 0. Or Alternatively,[br]cosine theater must be 0. 0:28:51.540,0:28:55.740 Now, if we know that the two[br]given vectors do not have a zero 0:28:55.740,0:28:59.640 length, In other words, the[br]modulus of a can't be 0, and the 0:28:59.640,0:29:01.440 modulus of B can't be 0. 0:29:02.070,0:29:06.714 Then the only conclusion we can[br]draw is that cosine theater must 0:29:06.714,0:29:12.132 be 0, and from that we did use[br]that theater must be 90 degrees. 0:29:13.290,0:29:16.865 In other words, if we take the[br]dot product of two vectors and 0:29:16.865,0:29:18.790 we find we get the answer 0. 0:29:19.800,0:29:24.510 Then, provided the two factors[br]were nonzero vectors, we can 0:29:24.510,0:29:28.278 deduce the amb must be[br]perpendicular vectors, so. 0:29:30.360,0:29:32.110 If A&B. 0:29:34.090,0:29:35.839 A nonzero vectors. 0:29:41.660,0:29:42.550 Search that. 0:29:46.000,0:29:49.575 Hey, don't be when we work it[br]out. Turns out to be 0. 0:29:50.420,0:29:53.640 Then A&B. 0:29:55.780,0:29:56.710 Are perpendicular. 0:30:02.530,0:30:03.630 There at right angles. 0:30:04.300,0:30:08.216 Another word we sometimes use is[br]orthogonal. You may hear that, 0:30:08.216,0:30:11.420 particularly in more advanced[br]work, two vectors at right 0:30:11.420,0:30:14.624 angles to vectors which are[br]perpendicular are sometimes said 0:30:14.624,0:30:15.692 to be orthogonal. 0:30:16.660,0:30:20.566 So we have a test here we take[br]the two vectors, we find their 0:30:20.566,0:30:24.193 dot product and if the answer[br]that we get zero then we deduce 0:30:24.193,0:30:27.262 that the two vectors must be[br]perpendicular. So that's a very 0:30:27.262,0:30:30.889 easy test we can apply to see if[br]two vectors are Purple ****. 0:30:33.560,0:30:34.946 Let me give you an example. 0:30:39.840,0:30:46.522 Suppose we have the Vector A,[br]which is the vector three 2 - 0:30:46.522,0:30:52.690 1 and the Vector B which is[br]1 - 2 - 1. 0:30:54.580,0:30:58.701 Now I think you would agree that[br]just by looking at these vectors 0:30:58.701,0:31:00.603 three, I +2 J Minus K. 0:31:01.530,0:31:05.618 And one I minus two J minus K.[br]You'd have no idea just from 0:31:05.618,0:31:08.830 looking at them written down[br]like that. Whether or not these 0:31:08.830,0:31:10.582 vectors where at right angles to 0:31:10.582,0:31:15.800 each other. We can apply this[br]dot product test and workout a 0:31:15.800,0:31:19.870 baby to see what happens. So[br]let's workout a dot B. 0:31:21.390,0:31:27.166 I want the dot product of three[br]2 - 1 with 1 - 2 - 1. 0:31:28.550,0:31:30.020 And we'll get 3 ones or three. 0:31:30.950,0:31:33.380 Two times minus 2 - 4. 0:31:33.940,0:31:38.168 And minus one times minus one is[br]plus one. So we've got three and 0:31:38.168,0:31:43.000 one which is 4 - 4 which is 0.[br]So you'll see that when we work 0:31:43.000,0:31:46.624 out the dot product of these two[br]nonzero vectors, the answer that 0:31:46.624,0:31:47.832 we get is 0. 0:31:48.660,0:31:52.296 So we can deduce from this[br]result that this vector a must 0:31:52.296,0:31:55.629 be at right angles to this[br]vector B and that's something 0:31:55.629,0:31:59.265 that's not obvious just from[br]looking at it. So we've got a 0:31:59.265,0:32:02.598 very useful test there to test[br]for. The two vectors are 0:32:02.598,0:32:03.507 perpendicular or not. 0:32:09.320,0:32:11.980 Now another important[br]application of the scalar 0:32:11.980,0:32:16.160 product is to finding the angle[br]between two given vectors. Let 0:32:16.160,0:32:20.340 me remind you of the definition[br]of the scalar product again 0:32:20.340,0:32:24.900 because we will need that the[br]scalar product of A&B is found 0:32:24.900,0:32:29.460 by taking the length of the[br]first vector times the length of 0:32:29.460,0:32:33.640 the second vector times the[br]cosine of the angle in between 0:32:33.640,0:32:35.160 the two vectors so. 0:32:35.270,0:32:38.286 If we get a vector, if we get[br]two vectors in Cartesian form. 0:32:39.190,0:32:40.665 And we calculate their dot 0:32:40.665,0:32:45.238 product. And if we know how to[br]calculate the modulus of each of 0:32:45.238,0:32:48.826 those two vectors, then the only[br]thing in this expression that we 0:32:48.826,0:32:52.414 don't know is cosine Theta, so[br]will be able to use this 0:32:52.414,0:32:55.404 definition to calculate cosine[br]theater the cosine of the angle 0:32:55.404,0:32:58.693 between the two vectors, from[br]which we can deduce the angle 0:32:58.693,0:33:01.476 itself. OK, let's rearrange 0:33:01.476,0:33:06.540 this. If we divide both sides by[br]the modular surveying, the 0:33:06.540,0:33:11.860 modulus of B will be able to get[br]cosine theater is a dot B 0:33:11.860,0:33:16.420 divided by the modulus of a[br]modulus of B and that's the 0:33:16.420,0:33:20.980 formula that we're going to use[br]in a minute to find cosine, 0:33:20.980,0:33:22.500 Theta and hence theater. 0:33:23.940,0:33:27.252 Now we've done a lot of work[br]already in this unit on 0:33:27.252,0:33:30.564 calculating the dot product. Let[br]me just remind you a little bit 0:33:30.564,0:33:34.152 about how you find the modulus[br]of a vector when it's given in 0:33:34.152,0:33:41.640 cartesian form. If we have a[br]Vector A, which is a one 0:33:41.640,0:33:45.270 I plus A2J plus a 3K. 0:33:46.470,0:33:51.370 Then the modulus of this[br]vector is found by squaring 0:33:51.370,0:33:53.820 the individual components,[br]adding them. 0:33:55.890,0:33:59.894 And finally, taking the square[br]root of the result. So this is a 0:33:59.894,0:34:03.590 formula that we're going to use[br]for finding the modulus of a 0:34:03.590,0:34:06.978 vector in Cartesian form, and[br]this has been covered in an 0:34:06.978,0:34:10.366 early you unit. If you need to[br]look back at that. 0:34:14.830,0:34:17.701 Let's look at a specific[br]example where we want to find 0:34:17.701,0:34:19.006 the angle between two vectors. 0:34:23.760,0:34:26.140 So the problem is[br]find the angle. 0:34:34.840,0:34:41.128 Between the two vectors, I'm[br]going to choose are A which is 0:34:41.128,0:34:43.748 4I Plus 3J plus 7K. 0:34:45.320,0:34:46.410 And B. 0:34:47.740,0:34:54.149 Which is 2 I plus 5J Plus[br]4K, so two vectors given in 0:34:54.149,0:34:59.079 cartesian form and the[br]problem is to find the angle 0:34:59.079,0:35:00.065 between them. 0:35:01.300,0:35:06.370 And we have this result that we[br]just reduced the cosine of the 0:35:06.370,0:35:10.660 required angle. Is the dot[br]product 8B divided by the length 0:35:10.660,0:35:15.730 of a times the length of be? So[br]that's the formula that we're 0:35:15.730,0:35:16.900 going to use. 0:35:18.470,0:35:23.926 OK, a dot B now. In fact a dot B[br]has already been evaluated in an 0:35:23.926,0:35:28.359 earlier example, but I'll just[br]remind you of that a dot B. If 0:35:28.359,0:35:31.428 we use the column vector[br]notation will be 437. 0:35:32.460,0:35:34.359 Dotted with 254 0:35:37.210,0:35:44.347 which is 428-3515 and Seven 428.[br]And if you work that out, you 0:35:44.347,0:35:46.543 find you get 51. 0:35:48.130,0:35:50.715 So all we need to do is[br]calculate now the modular 0:35:50.715,0:35:53.535 survey in the modulus of B and[br]then we've got everything we 0:35:53.535,0:35:56.355 need to know to put in the[br]right hand side of this 0:35:56.355,0:35:56.590 formula. 0:35:58.010,0:35:59.518 OK, the modular survey. 0:36:02.030,0:36:06.650 Now the modulus of a is found by[br]taking the square root of the 0:36:06.650,0:36:10.280 squares of these components[br]added up. So we want 4 squared. 0:36:11.470,0:36:12.570 3 squared 0:36:13.620,0:36:14.700 7 squared. 0:36:16.700,0:36:19.730 Which is going to be[br]the square root of 16. 0:36:21.030,0:36:26.670 330977's of 49 and if you work[br]that out, you'll find that 0:36:26.670,0:36:29.490 that's the square root of 74. 0:36:31.860,0:36:37.827 Similarly, the modulus of B will[br]be the square root of 2 squared. 0:36:39.350,0:36:42.298 +5 squared, +4 squared. 0:36:44.400,0:36:51.862 Which is the square root of 4[br]+ 25 + 16 and 25 + 0:36:51.862,0:36:54.527 4 + 16 is 45. 0:36:55.890,0:36:58.740 So we've got all the[br]ingredients we need now to 0:36:58.740,0:36:59.880 pop into this formula. 0:37:01.190,0:37:06.572 Cosine theater will be a dotted[br]with B, which we found was 51 0:37:06.572,0:37:11.540 divided by the modular survey,[br]which is the square root of 74. 0:37:12.810,0:37:16.982 The modulus of B, which is the[br]square root of 45 now will need 0:37:16.982,0:37:18.770 a Calculator to work this out. 0:37:21.650,0:37:22.949 We want 51. 0:37:24.980,0:37:27.745 Divided by the square[br]root of 74. 0:37:29.090,0:37:30.390 Divided by the square root. 0:37:31.370,0:37:38.560 45 Which is not .8838[br]to 4 decimal places. That's the 0:37:38.560,0:37:43.609 cosine of the angle between the[br]two vectors A&B. 0:37:44.440,0:37:46.306 If we find the inverse cosine. 0:37:49.340,0:37:51.610 Of .8838. 0:37:54.160,0:37:58.228 Which is 27.90[br]degrees. 0:38:00.980,0:38:04.566 So there we've seen how we can[br]use the scalar product. 0:38:06.040,0:38:09.883 To find the angle between[br]two vectors when they're 0:38:09.883,0:38:11.591 given in Cartesian form. 0:38:17.270,0:38:21.314 Now there's one more application[br]I'd like to tell you about, and 0:38:21.314,0:38:25.021 it's concerned with finding the[br]components of one vector in the 0:38:25.021,0:38:28.728 direction of a second vector.[br]Let me give you an example. 0:38:30.860,0:38:33.308 Suppose we have a vector A. 0:38:35.170,0:38:36.556 Let me have a second vector. 0:38:38.300,0:38:38.990 B. 0:38:42.280,0:38:46.988 And I've drawn these vectors so[br]that their tails coincide and. 0:38:47.580,0:38:51.280 As usual, the angle between the[br]two vectors is theater. 0:38:52.440,0:38:54.260 Let's call this .0. 0:38:54.850,0:38:58.546 And what I'm going to do is[br]I'm going to drop a 0:38:58.546,0:39:01.010 perpendicular from this[br]point, which I'll call B 0:39:01.010,0:39:01.318 down. 0:39:04.330,0:39:06.100 To meet the vector A. 0:39:09.510,0:39:11.640 Let's call this point here a. 0:39:13.100,0:39:15.347 And we can regard the vector OB. 0:39:16.810,0:39:21.046 As the sum of the vector[br]from O2 Capital A and then 0:39:21.046,0:39:25.282 the vector from A to B. In[br]other words, the vector OB. 0:39:27.190,0:39:28.940 Is the sum of OA. 0:39:31.680,0:39:32.988 Plus a bee. 0:39:36.550,0:39:40.598 Now this vector here, which[br]we've called OK, that vector is 0:39:40.598,0:39:45.750 said to be the component of B in[br]the direction of a. You'll see 0:39:45.750,0:39:50.902 we can regard be as being made[br]up of two components, one in the 0:39:50.902,0:39:56.054 direction of A and one which is[br]at right angles to that. So this 0:39:56.054,0:39:57.158 component, oh A. 0:39:57.960,0:40:00.910 Is the component of B[br]in the direction of a. 0:40:18.340,0:40:21.980 And what we're going to do is[br]we're going to see how we can 0:40:21.980,0:40:24.060 use the scalar product to[br]find this component. 0:40:26.070,0:40:30.282 Let's call this length here the[br]length from O to a. Let's call 0:40:30.282,0:40:34.930 that little. And then let's[br]focus our attention on this 0:40:34.930,0:40:36.070 right angled triangle. 0:40:36.830,0:40:38.300 Oba. 0:40:39.470,0:40:41.514 So there's a right angle[br]in there. 0:40:43.320,0:40:47.390 Now, using our knowledge of[br]trigonometry, we can write down 0:40:47.390,0:40:51.867 the cosine of this angle is[br]adjacent over hypotenuse, so the 0:40:51.867,0:40:53.088 cosine of Theta. 0:40:54.640,0:40:57.874 Is this length L[br]the adjacent side? 0:40:59.350,0:41:01.878 Over the hypotenuse, now[br]the hypotenuse is the 0:41:01.878,0:41:03.142 length of this side. 0:41:04.650,0:41:07.410 And this is the length of[br]this side is the modulus of 0:41:07.410,0:41:07.870 vector be. 0:41:12.020,0:41:16.299 Now, In other words, if we[br]rearrange this L, the component 0:41:16.299,0:41:22.134 of B in the direction of a week[br]and right as the modulus of be 0:41:22.134,0:41:23.690 times, the cosine Theta. 0:41:27.750,0:41:31.290 Now we've already got an[br]expression for the cosine of 0:41:31.290,0:41:35.538 this angle in the previous work[br]that we've done, we already know 0:41:35.538,0:41:40.140 that the cosine of feta can be[br]written as the dot product for 0:41:40.140,0:41:45.070 amb. Divided by the modular[br]survey times the modulus of be, 0:41:45.070,0:41:47.344 that's the result we just had. 0:41:48.470,0:41:51.200 So if you look at this[br]expression now, you'll see 0:41:51.200,0:41:54.476 with the modulus B in the[br]numerator and a modulus of be 0:41:54.476,0:41:55.841 in the denominator, they[br]cancel. 0:41:56.900,0:41:58.420 So we can write this. 0:41:59.170,0:42:03.535 As a not be over[br]the modulus of A. 0:42:06.760,0:42:08.730 Let me write that down[br]again on the next page. 0:42:20.970,0:42:24.546 So this remember L is the[br]component of be in the direction 0:42:24.546,0:42:28.718 of a. We can write it like this.[br]I'm going to Rearrange this a 0:42:28.718,0:42:32.592 little bit to write it as a[br]divided by the modulus of A. 0:42:34.000,0:42:35.908 Dotted with effective be. 0:42:37.240,0:42:40.771 And because of the result we[br]got very early on the 0:42:40.771,0:42:43.981 commutativity of the dot[br]product. We can write this as 0:42:43.981,0:42:46.870 B dot A divided by the modulus[br]of A. 0:42:49.400,0:42:53.313 Now when you take any vector and[br]you divide it by its modulus, 0:42:53.313,0:42:57.226 what you get is a unit vector in[br]the direction of that vector, 0:42:57.226,0:43:01.440 and we write a unit vector as[br]the vector with a little hat on 0:43:01.440,0:43:05.353 like that. So this is a unit[br]vector in the direction of A. 0:43:14.770,0:43:18.502 So this is an important result[br]we've got. We've got L, which 0:43:18.502,0:43:22.856 is the component of be in the[br]direction of a can be found by 0:43:22.856,0:43:26.588 taking the dot product of be[br]with a unit vector in the 0:43:26.588,0:43:27.521 direction of A. 0:43:31.340,0:43:37.794 So suppose we wish to find the[br]component of Vector B which is 3 0:43:37.794,0:43:43.787 I plus J Plus 4K in the[br]direction of the Vector A, which 0:43:43.787,0:43:46.553 is I minus J Plus K. 0:43:48.180,0:43:53.306 What we want to do is[br]evaluate B dotted with a 0:43:53.306,0:43:55.636 divided by the modular[br]survey. 0:43:59.190,0:44:03.026 Now we can do that as follows.[br]We can workout the dot product B 0:44:03.026,0:44:07.136 dot A and then divide by the[br]modulus of a. So B Dot A is 0:44:07.136,0:44:09.602 going to be 3 * 1, which is 3. 0:44:10.190,0:44:15.970 Plus one times minus one which[br]is minus 1 + 4 * 1 which is 4 or 0:44:15.970,0:44:20.390 divided by the modulus of a[br]which is the square root of 1 0:44:20.390,0:44:24.130 squared plus minus one squared[br]plus one squared, which is the 0:44:24.130,0:44:25.490 square root of 3. 0:44:27.570,0:44:32.988 If we work that out, will get 3[br]- 1 is 2 + 4 six six over Route 0:44:32.988,0:44:37.503 3. This means that the component[br]of B in the direction of a is 6. 0:44:37.503,0:44:41.115 Over Route 3. We might want to[br]write that in an alternative 0:44:41.115,0:44:45.329 form just to tidy it up so we[br]don't have the square root in 0:44:45.329,0:44:48.640 the denominator and we can do[br]that by multiplying top and 0:44:48.640,0:44:52.553 bottom by the square root of 3,[br]which will produce Route 3 times 0:44:52.553,0:44:56.165 route 3 in the denominator,[br]which is 3 and three into six 0:44:56.165,0:44:58.884 goes twice. So that[br]would just simplify to 0:44:58.884,0:45:00.404 two square root of 3. 0:45:02.560,0:45:06.169 So in this unit we've introduced[br]the scalar product. 0:45:06.760,0:45:10.225 Learn how to calculate it and[br]looked at some of its 0:45:10.225,0:45:11.170 properties and applications.