1 00:00:01,950 --> 00:00:06,214 In this unit, we're going to have a look at a method for 2 00:00:06,214 --> 00:00:10,204 combining two vectors. This method is called the scalar 3 00:00:10,204 --> 00:00:14,076 product. And it's called the scalar product because when we 4 00:00:14,076 --> 00:00:18,332 calculate it, the result that we get or the answer is a scalar as 5 00:00:18,332 --> 00:00:19,548 opposed to a vector. 6 00:00:20,550 --> 00:00:23,820 So let's look at two vectors. Here's the first vector. 7 00:00:27,610 --> 00:00:29,050 That's cool that vector A. 8 00:00:32,520 --> 00:00:36,953 And here's the second vector. Let's call this vector B and 9 00:00:36,953 --> 00:00:41,386 you'll notice that I've drawn these two vectors so that the 10 00:00:41,386 --> 00:00:43,804 tales of the two vectors coincide. 11 00:00:46,210 --> 00:00:47,788 Now that they coincide, we can 12 00:00:47,788 --> 00:00:52,090 measure this angle. Between the two vectors and I'm going to 13 00:00:52,090 --> 00:00:53,470 label the angle theater. 14 00:00:54,590 --> 00:00:56,410 So there are two vectors. 15 00:00:57,340 --> 00:01:00,993 Now when we calculate the scalar product, we do it like this. We 16 00:01:00,993 --> 00:01:05,013 find the length. Or the magnitude, or the modulus of 17 00:01:05,013 --> 00:01:06,000 the first vector. 18 00:01:07,050 --> 00:01:09,420 Which is the modulus of a. 19 00:01:11,460 --> 00:01:13,338 We multiply it by the length. 20 00:01:14,530 --> 00:01:17,880 Of the second vector, which is the modulus of beat. 21 00:01:19,600 --> 00:01:25,072 And we multiplied by the cosine of this angle between the two 22 00:01:25,072 --> 00:01:28,720 vectors. We multiplied by the cosine of Theta. 23 00:01:29,970 --> 00:01:33,413 And that's how we calculate the scalar product. The length of 24 00:01:33,413 --> 00:01:36,543 the first vector times the length of the second vector 25 00:01:36,543 --> 00:01:39,986 multiplied by the cosine of the angle in between the two 26 00:01:39,986 --> 00:01:44,400 vectors. And we have a notation for the scalar 27 00:01:44,400 --> 00:01:49,380 product and we write it like this vector A dot vector be. 28 00:01:50,720 --> 00:01:54,640 And this is the definition. Sometimes this scalar product is 29 00:01:54,640 --> 00:01:59,344 also referred to as a dot product because of that dot in 30 00:01:59,344 --> 00:02:04,048 there. So sometimes you'll hear it referred to as a dot product. 31 00:02:06,970 --> 00:02:10,652 OK, so let's have an example and see if we can calculate the dot 32 00:02:10,652 --> 00:02:11,704 product of two vectors. 33 00:02:14,350 --> 00:02:16,180 Let's suppose our vector a here. 34 00:02:16,740 --> 00:02:19,220 Has got length for units. 35 00:02:23,200 --> 00:02:28,960 And let's suppose we have a vector B up here and let's 36 00:02:28,960 --> 00:02:32,320 suppose vector B has length 5 units. 37 00:02:33,420 --> 00:02:37,644 And let's suppose that the angle for the sake of argument between 38 00:02:37,644 --> 00:02:39,756 the two vectors is 60 degrees. 39 00:02:41,220 --> 00:02:45,367 And let's use this formula to calculate the scalar product A 40 00:02:45,367 --> 00:02:47,629 dot B. It's a vector there. 41 00:02:49,250 --> 00:02:51,990 A dot B will be. 42 00:02:52,870 --> 00:02:55,686 Well, it's the modulus or the length of the first vector. 43 00:02:57,120 --> 00:02:58,050 Which is 4. 44 00:03:01,730 --> 00:03:04,907 Multiplied by the modulus or the length of the 45 00:03:04,907 --> 00:03:07,731 second vector and the length of the second 46 00:03:07,731 --> 00:03:08,790 vector is 5. 47 00:03:11,950 --> 00:03:16,435 Multiplied by the cosine of the angle between the two, and so we 48 00:03:16,435 --> 00:03:18,505 want the cosine of 60 degrees. 49 00:03:20,950 --> 00:03:25,066 Now the cosine of 60 degrees. If you don't know, you can work it 50 00:03:25,066 --> 00:03:28,888 out on your Calculator, but it's a common angle an the cosine of 51 00:03:28,888 --> 00:03:33,886 60 degrees is 1/2, so we've got 4 * 5 * 1/2, so if we work all 52 00:03:33,886 --> 00:03:38,296 that out, we get 4 files, or 20 and a half of 20 is 10. 53 00:03:40,350 --> 00:03:43,620 So that's our first example of calculating a scalar product, 54 00:03:43,620 --> 00:03:47,544 and let me remind you of something that I said Very early 55 00:03:47,544 --> 00:03:51,468 on that when we calculate the scalar product, the answer is a 56 00:03:51,468 --> 00:03:55,719 scalar and you'll see the answer here 10 is just a single number 57 00:03:55,719 --> 00:03:59,316 Whilst we started with two vectors A&B, we finish up with 58 00:03:59,316 --> 00:04:02,913 just a single number, a scalar, and that's the scalar product. 59 00:04:05,640 --> 00:04:09,400 Let me illustrate some other features of this scalar product 60 00:04:09,400 --> 00:04:12,784 by doing the calculation the opposite way round, supposing 61 00:04:12,784 --> 00:04:17,296 I'd try to calculate the dotted with a instead. Let's work that 62 00:04:17,296 --> 00:04:19,176 through and see what happens. 63 00:04:20,350 --> 00:04:23,998 So I've done the operation in a different order instead of a 64 00:04:23,998 --> 00:04:25,822 baby. I've now got dot A. 65 00:04:27,350 --> 00:04:30,610 Well, again, using the definition we want to say that 66 00:04:30,610 --> 00:04:32,240 the dot product of DNA. 67 00:04:32,800 --> 00:04:36,220 Is the modulus or length of the first vector which is the 68 00:04:36,220 --> 00:04:40,495 modulus or the length of be this time and the modulus of B is 5? 69 00:04:43,310 --> 00:04:46,401 Multiply by the modulus or the length of the second vector. 70 00:04:47,890 --> 00:04:49,078 Which is now 4. 71 00:04:50,670 --> 00:04:53,946 Multiplied by the cosine of the angle between the two and the 72 00:04:53,946 --> 00:04:55,038 angle still 60 degrees. 73 00:04:56,340 --> 00:05:01,395 The cosine of 60 degrees is 1/2, so if we work this out this time 74 00:05:01,395 --> 00:05:06,113 will get five 420 1/2 of 20 is 10, so you'll see that whichever 75 00:05:06,113 --> 00:05:10,494 way we did the calculation whether we worked out a dot B or 76 00:05:10,494 --> 00:05:14,875 whether we worked out B dotted with a, we get the same answer. 77 00:05:15,650 --> 00:05:20,859 10 and that's true in general for any two vectors that we 78 00:05:20,859 --> 00:05:23,302 choose. If we workout a dot B. 79 00:05:26,980 --> 00:05:31,171 That will be the same as working out B Dot A. 80 00:05:32,110 --> 00:05:36,160 And that's another important property of the scalar product, 81 00:05:36,160 --> 00:05:38,410 and we call this property 82 00:05:38,410 --> 00:05:42,540 commutativity. We say that the dot product or the scalar 83 00:05:42,540 --> 00:05:43,560 product is commutative. 84 00:05:56,490 --> 00:05:59,250 Now, as well as the commutativity property, I 85 00:05:59,250 --> 00:06:02,355 want to tell you about another property which is 86 00:06:02,355 --> 00:06:03,390 known as distributivity. 87 00:06:04,960 --> 00:06:11,470 Suppose we have a dotted with the sum of two vectors B Plus C. 88 00:06:13,420 --> 00:06:16,228 The distributivity property tells us to expand these 89 00:06:16,228 --> 00:06:20,089 brackets in the normal sort of algebraic way that you would 90 00:06:20,089 --> 00:06:23,248 expect. We work this out by saying it's a. 91 00:06:23,980 --> 00:06:25,219 Dotted with B. 92 00:06:28,580 --> 00:06:29,348 Add it to. 93 00:06:31,210 --> 00:06:33,970 A dotted with C. 94 00:06:38,320 --> 00:06:41,580 This also works the other way round, so suppose we 95 00:06:41,580 --> 00:06:42,884 have B Plus C. 96 00:06:45,390 --> 00:06:50,556 And we want to dot it with the vector a As you might predict 97 00:06:50,556 --> 00:06:54,246 the result that will get is B dotted with a. 98 00:06:57,600 --> 00:06:58,390 Attitude. 99 00:07:00,580 --> 00:07:01,768 See dotted with a. 100 00:07:03,450 --> 00:07:08,733 And these properties are said to be the distributivity. 101 00:07:10,780 --> 00:07:11,480 Rules. 102 00:07:16,140 --> 00:07:19,400 I will need those rules along with the commutativity rules 103 00:07:19,400 --> 00:07:21,030 later on in this unit. 104 00:07:27,050 --> 00:07:30,002 Now there's another property very important property of the 105 00:07:30,002 --> 00:07:33,938 scalar product that I like to tell you about as well, and 106 00:07:33,938 --> 00:07:36,562 it's the scalar product of two perpendicular vectors. 107 00:07:49,190 --> 00:07:53,315 Let's see what happens when the two vectors that we're dealing 108 00:07:53,315 --> 00:07:57,440 with A&B are perpendicular. That means there at right angles. So 109 00:07:57,440 --> 00:08:01,940 we have a situation like this. There's a vector B. There's a 110 00:08:01,940 --> 00:08:06,440 vector A and they're separated by an angle of 90 degrees, so 111 00:08:06,440 --> 00:08:07,940 the vectors are perpendicular. 112 00:08:09,510 --> 00:08:12,880 Let's use our formula for the dot product. The scalar 113 00:08:12,880 --> 00:08:16,587 product, the modulus of the first one with the modulus of 114 00:08:16,587 --> 00:08:19,957 the second one times the cosine of the angle between 115 00:08:19,957 --> 00:08:23,327 the two. Let's use that formula in this specific case, 116 00:08:23,327 --> 00:08:25,012 when the vectors are perpendicular. 117 00:08:27,930 --> 00:08:30,757 Well, we want the length of the first one. Still the 118 00:08:30,757 --> 00:08:31,528 modulus of A. 119 00:08:32,680 --> 00:08:36,424 The length of the second one will be the modulus of B and 120 00:08:36,424 --> 00:08:40,168 this time we want the cosine of Theta, but theater is 90 degrees 121 00:08:40,168 --> 00:08:41,896 from the cosine of 90 degrees. 122 00:08:43,830 --> 00:08:46,888 Now you either should know or you can easily check on 123 00:08:46,888 --> 00:08:49,668 your Calculator that the cosine of 90 degrees is 0. 124 00:08:51,600 --> 00:08:54,048 So these two vectors, these two 125 00:08:54,048 --> 00:08:58,088 moduli? Whatever they are multiplied by zero. So the 126 00:08:58,088 --> 00:09:00,440 answer that will get will be 0. 127 00:09:01,480 --> 00:09:05,848 This is a very important result that if you have two vectors 128 00:09:05,848 --> 00:09:08,760 which are perpendicular, their scalar product is 0. 129 00:09:21,920 --> 00:09:24,530 So for two perpendicular vectors a. 130 00:09:26,070 --> 00:09:26,540 The. 131 00:09:28,080 --> 00:09:30,964 Important result which you should learn that 132 00:09:30,964 --> 00:09:34,260 their dot product their scalar product is 0. 133 00:09:37,880 --> 00:09:42,425 Now the converse of this is also true, and by that I mean that if 134 00:09:42,425 --> 00:09:46,061 we choose any two vectors at all, and we find their dot 135 00:09:46,061 --> 00:09:50,000 product and we find that the answer is 0, that will mean that 136 00:09:50,000 --> 00:09:53,333 provided that neither a norby was zero, those two vectors must 137 00:09:53,333 --> 00:09:57,272 be perpendicular and will use this as a test later on in this 138 00:09:57,272 --> 00:09:59,393 unit to test whether or not to 139 00:09:59,393 --> 00:10:01,970 given vector. Are perpendicular or not? 140 00:10:08,840 --> 00:10:13,880 I want to move on now to look at how we calculate the scalar 141 00:10:13,880 --> 00:10:17,480 product of two vectors when they're given in Cartesian form. 142 00:10:17,480 --> 00:10:20,720 Let me remind you about cartesian form when we're 143 00:10:20,720 --> 00:10:23,960 dealing with vectors in Cartesian form, where we express 144 00:10:23,960 --> 00:10:28,280 the vector in terms of unit vectors IJ&K. So we might be 145 00:10:28,280 --> 00:10:30,080 dealing with vectors of this 146 00:10:30,080 --> 00:10:33,260 form 3I. Minus two J. 147 00:10:34,440 --> 00:10:35,950 Plus 7K. 148 00:10:37,700 --> 00:10:41,426 Expect to be might be something like minus 5I. 149 00:10:42,730 --> 00:10:48,760 Plus 4J. Minus 3K. So when we see the vectors written down 150 00:10:48,760 --> 00:10:51,448 like this with eyes and Jays and 151 00:10:51,448 --> 00:10:55,186 kays in. These vectors are now given in Cartesian form. 152 00:10:55,800 --> 00:10:59,921 And we're going to learn how to do now is figure out how 153 00:10:59,921 --> 00:11:02,457 to calculate the dot product. The scalar product 154 00:11:02,457 --> 00:11:03,408 vectors like these. 155 00:11:04,470 --> 00:11:08,331 Before we do that, I'd like to introduce some other important 156 00:11:08,331 --> 00:11:11,841 results. Let me remind you a little bit about IJ&K. 157 00:11:15,830 --> 00:11:19,232 In three dimensions, where we have an X axis. 158 00:11:20,410 --> 00:11:23,310 AY axis. And as Ed Access. 159 00:11:24,540 --> 00:11:27,969 We did note unit vectors along the X axis. 160 00:11:29,020 --> 00:11:36,466 I I along the Y axis by J and along the Z Axis by K and you'll 161 00:11:36,466 --> 00:11:38,656 notice that the vectors IJ&K. 162 00:11:39,570 --> 00:11:43,730 US are all it's 90 degrees to each other, so the angle between 163 00:11:43,730 --> 00:11:47,890 I&J is 90 degrees. The angle between Jane K is 90 degrees and 164 00:11:47,890 --> 00:11:51,730 the angle between I&K is 90 degrees and all of these vectors 165 00:11:51,730 --> 00:11:55,570 I Jane KA unit vectors. That means that their length is one. 166 00:11:57,370 --> 00:12:02,639 Now let's just calculate the dot product of I with J. 167 00:12:08,300 --> 00:12:11,600 We use the result we had before the definition of the dot 168 00:12:11,600 --> 00:12:14,625 product, which says that we write down the length of the 169 00:12:14,625 --> 00:12:18,475 first vector and the length of I being a unit vector is just one. 170 00:12:20,880 --> 00:12:24,389 The length of J being a unit vector is also one. 171 00:12:25,800 --> 00:12:29,687 And we want the cosine of the angle between the two and the 172 00:12:29,687 --> 00:12:33,275 angle between the two is 90 degrees, so we want the cosine 173 00:12:33,275 --> 00:12:38,500 of 90 degrees. And again, the cosine of 90 degrees is 0, so 174 00:12:38,500 --> 00:12:42,964 we find that I got Jay is simply zero. We could have 175 00:12:42,964 --> 00:12:45,940 deduced that from the previous result that I 176 00:12:45,940 --> 00:12:49,660 obtained for you, which was that for any two vectors 177 00:12:49,660 --> 00:12:52,636 which are perpendicular, their dot product is 0. 178 00:12:54,900 --> 00:13:00,234 So I got Jays 0. Now the same argument tells us also that Jay 179 00:13:00,234 --> 00:13:02,520 dotted with K must be 0. 180 00:13:07,400 --> 00:13:09,296 Be cause J&K are perpendicular vectors 181 00:13:09,296 --> 00:13:12,456 there at right angles and the angle between them is 182 00:13:12,456 --> 00:13:15,300 90 degrees and the cosine of 90 is 0. 183 00:13:16,770 --> 00:13:23,610 What about INK? Well, I dotted with K must also be 0. 184 00:13:25,900 --> 00:13:28,958 What about dotting the vector with itself? Let's just have a 185 00:13:28,958 --> 00:13:31,738 look at that. Suppose we wanted I dotted with I. 186 00:13:34,330 --> 00:13:36,877 Well, the length of the first vector is one. 187 00:13:38,360 --> 00:13:42,248 The length of the second vector is still one, and if we're 188 00:13:42,248 --> 00:13:45,488 dotting a vector with itself, the angle between the two 189 00:13:45,488 --> 00:13:49,700 vectors must be 0. So this time we want the cosine of 0. 190 00:13:50,680 --> 00:13:53,859 And the cosine of 0. The cosine of note is one. 191 00:13:54,540 --> 00:13:59,556 So we have 1 * 1 * 1, which is just one. 192 00:14:00,170 --> 00:14:03,686 So if we're dotting the vector I with itself, we get one. 193 00:14:04,690 --> 00:14:09,201 A similar argument will tell us that if we got the vector Jay 194 00:14:09,201 --> 00:14:14,059 with itself, we also get one, and if we drop the vector K with 195 00:14:14,059 --> 00:14:15,447 itself, we get one. 196 00:14:16,400 --> 00:14:19,344 Let me summarize those important results here and 197 00:14:19,344 --> 00:14:23,392 I've results that you should remember that if you took the 198 00:14:23,392 --> 00:14:27,808 vector I and dotted it with itself, that would be the same 199 00:14:27,808 --> 00:14:29,648 as dotting Jay with itself. 200 00:14:31,060 --> 00:14:35,785 And it's the same as Dot Inc. A with itself, and in any case we 201 00:14:35,785 --> 00:14:37,045 get the answer 1. 202 00:14:38,850 --> 00:14:46,007 If we don't, I with J we get zero if we got I with K if we 203 00:14:46,007 --> 00:14:51,480 get zero and if we got J with K we also get 0. 204 00:14:52,480 --> 00:14:55,312 And all those results are particularly important, and 205 00:14:55,312 --> 00:14:58,852 once that you should become familiar with as you work 206 00:14:58,852 --> 00:15:00,976 through the exercises accompanying this unit. 207 00:15:05,890 --> 00:15:10,674 Now I want to use those results about the dot products of those 208 00:15:10,674 --> 00:15:15,458 unit vectors to find a way to calculate the dot product of two 209 00:15:15,458 --> 00:15:19,138 arbitrary vectors that are given in Cartesian form. So let's 210 00:15:19,138 --> 00:15:22,450 choose two arbitrary vectors. Let's suppose the first one. 211 00:15:23,750 --> 00:15:26,060 Is a one. 212 00:15:27,430 --> 00:15:28,010 I. 213 00:15:29,830 --> 00:15:35,120 A2, J and a 3K. 214 00:15:36,520 --> 00:15:38,062 So this is now an arbitrary 215 00:15:38,062 --> 00:15:44,094 vector. A where a one A2 and a three are the three cartesian 216 00:15:44,094 --> 00:15:46,154 components of the vector A. 217 00:15:47,270 --> 00:15:52,926 And let's suppose vector B similarly is B1I. 218 00:15:54,500 --> 00:16:00,593 Plus B2J Plus V3K and again B1B2B3 arbitrary numbers, 219 00:16:00,593 --> 00:16:06,686 so this is an arbitrary vector in three dimensions. 220 00:16:08,260 --> 00:16:12,280 What I want to do is workout the dot product of A&B. 221 00:16:13,510 --> 00:16:17,434 So let's work it through a dot B equals. 222 00:16:18,900 --> 00:16:21,268 It's the first vector which is all this. 223 00:16:31,180 --> 00:16:36,060 Dotted with the second vector be, which is all this. 224 00:16:46,100 --> 00:16:51,170 OK. Now what we want to do now is use our 225 00:16:51,170 --> 00:16:54,329 distributivity rule to be able to workout to expand 226 00:16:54,329 --> 00:16:57,137 these brackets and workout the individual dot products. 227 00:16:58,310 --> 00:17:02,524 And we can do that using the rules that we know that in turn 228 00:17:02,524 --> 00:17:04,029 each one of these vectors. 229 00:17:04,850 --> 00:17:07,775 In the first bracket, multiplies each of these in 230 00:17:07,775 --> 00:17:11,025 the SEC bracket and just the normal algebraic way. So 231 00:17:11,025 --> 00:17:15,575 first of all we want a one I dotted with B1 I those terms. 232 00:17:24,480 --> 00:17:26,196 Then we want a one I. 233 00:17:26,880 --> 00:17:28,800 Dotted with B2J 234 00:17:37,020 --> 00:17:41,388 and finally a one I dotted with B3K. 235 00:17:47,490 --> 00:17:50,290 So that takes care of this first 236 00:17:50,290 --> 00:17:54,730 vector here. And we moved to the second vector. So now we 237 00:17:54,730 --> 00:17:57,470 want plus A2, J, dotted with each of these three. 238 00:18:01,200 --> 00:18:04,024 So it's a 2 J dotted with B1I. 239 00:18:09,120 --> 00:18:11,450 A2 J dotted with B2J. 240 00:18:17,180 --> 00:18:19,686 And a two J dotted with B3K. 241 00:18:27,000 --> 00:18:30,443 And finally, this loss vector in the first bracket dotted with 242 00:18:30,443 --> 00:18:34,199 each of these vectors in the SEC bracket will be a 3K. 243 00:18:37,350 --> 00:18:38,940 Dotted with B1I 244 00:18:40,860 --> 00:18:43,500 plus a 3K dotted with B2J. 245 00:18:48,500 --> 00:18:51,524 And finally, a 3K dotted with B3K. 246 00:18:56,540 --> 00:19:00,016 Now this looks horrendous. With all these nine terms in here, 247 00:19:00,016 --> 00:19:04,124 but a lot of this is going to cancel out now. You'll remember 248 00:19:04,124 --> 00:19:06,968 that any vectors which are perpendicular to each other. 249 00:19:08,000 --> 00:19:10,065 Have a dot product which is 0. 250 00:19:11,590 --> 00:19:12,640 Now this vector. 251 00:19:14,320 --> 00:19:16,742 Is in the eye direction and this 252 00:19:16,742 --> 00:19:21,792 vector. Is in the J direction, so a one I is perpendicular to 253 00:19:21,792 --> 00:19:22,830 be 2 J. 254 00:19:23,890 --> 00:19:28,420 So where we have the item dotted with the J term, this must be 0. 255 00:19:31,840 --> 00:19:36,076 Similarly, any vector in the direction of I any multiple of I 256 00:19:36,076 --> 00:19:38,900 must be perpendicular to any multiple of K. 257 00:19:39,900 --> 00:19:42,147 So this dot product is also zero. 258 00:19:43,890 --> 00:19:47,850 And similarly for the A2JB1I, those vectors are perpendicular. 259 00:19:47,850 --> 00:19:50,050 Their dot product is 0. 260 00:19:52,620 --> 00:19:55,955 A2J dot B3K. A perpendicular 261 00:19:55,955 --> 00:20:01,200 that's zero. Similarly, that will be 0 and that will be 0. 262 00:20:01,940 --> 00:20:05,117 And we'll be left with three non zero terms. 263 00:20:07,310 --> 00:20:08,610 What about this term here? 264 00:20:10,170 --> 00:20:15,386 A1 I be one I now this is a multiple of I and this is a 265 00:20:15,386 --> 00:20:17,342 multiple of I. So these two 266 00:20:17,342 --> 00:20:22,251 vectors are parallel. The angle between them is 0, so when we 267 00:20:22,251 --> 00:20:26,319 work out the dot product will want the length of the first 268 00:20:26,319 --> 00:20:28,014 one, which is a one. 269 00:20:28,810 --> 00:20:30,310 The length of the second one, 270 00:20:30,310 --> 00:20:31,510 which is. He wants. 271 00:20:33,170 --> 00:20:36,052 And the cosine of the angle between the two. The angle 272 00:20:36,052 --> 00:20:41,649 between the two. Is zero and the cosine of 0 is one, so we get a 273 00:20:41,649 --> 00:20:43,264 one B 1 * 1. 274 00:20:44,340 --> 00:20:47,740 So this first term, whilst it looked quite complicated, just 275 00:20:47,740 --> 00:20:49,440 simplifies to a 1B one. 276 00:20:52,090 --> 00:20:54,884 Similarly, these two vectors here. This is a vector in the 277 00:20:54,884 --> 00:20:58,186 direction of J and this is a vector in the direction of Jay. 278 00:20:58,830 --> 00:21:03,000 So the angle between the two of them is 0, the cosine of 0 is 279 00:21:03,000 --> 00:21:04,946 one, so when we workout this dot 280 00:21:04,946 --> 00:21:09,453 product we want. The length of the first one, which is a 2. The 281 00:21:09,453 --> 00:21:10,911 length of the second one which 282 00:21:10,911 --> 00:21:16,373 is B2. Multiplied by the cosine of 0. Cosine of 0 is one, so we 283 00:21:16,373 --> 00:21:17,657 just get a 2B2. 284 00:21:19,150 --> 00:21:22,102 And finally, same argument applies here. This is a 285 00:21:22,102 --> 00:21:26,038 multiple of K. This is a multiple of K, so the vectors 286 00:21:26,038 --> 00:21:29,974 are parallel. The angle between them is zero and the cosine of 287 00:21:29,974 --> 00:21:34,894 0 is one. So we get the length of the first being A3 length of 288 00:21:34,894 --> 00:21:39,486 the second being B3 and the cosine of 0 cosine of 0 is one, 289 00:21:39,486 --> 00:21:42,766 so the whole lot. All of this complicated stuff just 290 00:21:42,766 --> 00:21:46,702 simplifies down at the end of the day to this very important 291 00:21:46,702 --> 00:21:47,358 result here. 292 00:21:49,810 --> 00:21:55,036 And what this result says is if you want to find the scalar 293 00:21:55,036 --> 00:21:59,056 product of two vectors A&B which are given in this 294 00:21:59,056 --> 00:22:03,880 cartesian form, like that, all we need to do is multiply the 295 00:22:03,880 --> 00:22:08,302 I components together. You see the A1B one is the eye 296 00:22:08,302 --> 00:22:09,508 components multiplied together. 297 00:22:11,410 --> 00:22:13,374 Multiply the J components 298 00:22:13,374 --> 00:22:18,458 together A2B2. Multiply the K components together A3B3. 299 00:22:19,370 --> 00:22:21,860 And then finally add up these 300 00:22:21,860 --> 00:22:24,400 results. So the dot product. 301 00:22:25,420 --> 00:22:26,488 Is the sum. 302 00:22:27,170 --> 00:22:30,579 Of the products of the corresponding components. 303 00:22:32,220 --> 00:22:35,751 And you'll notice that the result we get here on the 304 00:22:35,751 --> 00:22:39,603 right hand side doesn't have any eyes or Jays orkez in the 305 00:22:39,603 --> 00:22:43,134 answer anymore. This is purely a number. This is purely a 306 00:22:43,134 --> 00:22:46,344 scalar, as we'd expect when we calculate a scalar product. 307 00:22:48,460 --> 00:22:52,660 Point out also that this expression on the right, which 308 00:22:52,660 --> 00:22:57,280 is the sum of the products of the corresponding components is 309 00:22:57,280 --> 00:23:00,640 also sometimes referred to as the inner product. 310 00:23:02,330 --> 00:23:07,530 Of A&B, so on occasions you may hear some staff or reading some 311 00:23:07,530 --> 00:23:11,530 textbooks that this is called the inner product of Bambi. 312 00:23:19,520 --> 00:23:22,424 OK, so we have a formula now for the dot product. 313 00:23:24,800 --> 00:23:31,807 It's a one B 1 + 8, two B2 plus A3B3. 314 00:23:35,210 --> 00:23:38,928 Can very important results and this will allow us to calculate 315 00:23:38,928 --> 00:23:42,308 the scalar product of two vectors given in cartesian form. 316 00:23:42,308 --> 00:23:44,336 Let me give you an example. 317 00:23:49,030 --> 00:23:56,950 Suppose the first vector a is this one 4I Plus 3J minus. 318 00:23:56,950 --> 00:23:58,930 Sorry plus 7K. 319 00:24:00,620 --> 00:24:03,952 And Suppose B is the vector two 320 00:24:03,952 --> 00:24:06,480 I. +5 J. 321 00:24:07,980 --> 00:24:08,910 Plus 4K. 322 00:24:11,300 --> 00:24:16,459 Suppose we want to calculate the scalar product A dot B. 323 00:24:18,390 --> 00:24:22,636 Well, this formula tells us that we multiply the corresponding I 324 00:24:22,636 --> 00:24:27,268 components together A1B one. This is a one for this is B1 325 00:24:27,268 --> 00:24:31,514 which is 2. So we multiply the four by the two. 326 00:24:35,290 --> 00:24:38,530 We multiply the corresponding J components together. A2B2 well, 327 00:24:38,530 --> 00:24:43,570 this is a two and this is B2, so we want 3 * 5. 328 00:24:47,080 --> 00:24:50,280 And then we multiply the corresponding K components 329 00:24:50,280 --> 00:24:53,080 together. A3B3, which is 7 * 4. 330 00:24:56,050 --> 00:25:03,415 When we do that and we add up all the results and will get 8 331 00:25:03,415 --> 00:25:10,289 + 3, five, 15, Seven, 428. And if we work those out we shall 332 00:25:10,289 --> 00:25:17,163 get 2815 and eight to add up to 51. So the dot product of 333 00:25:17,163 --> 00:25:19,618 A&B is the scalar 51. 334 00:25:22,420 --> 00:25:25,940 Now, sometimes, instead of writing it all out like this, we 335 00:25:25,940 --> 00:25:29,780 can use column vector notation and some people find it a bit 336 00:25:29,780 --> 00:25:33,620 easier to work with column vectors. So just let me show you 337 00:25:33,620 --> 00:25:37,460 how we do the same calculation. If we had these vectors given 338 00:25:37,460 --> 00:25:41,300 not in this form, but as column vectors, suppose we had a 339 00:25:41,300 --> 00:25:47,258 written as 437. And be written AS254. 340 00:25:49,020 --> 00:25:51,306 Then the a dotted with B. 341 00:25:52,170 --> 00:25:57,786 Would be 437 dotted with 254. 342 00:26:00,350 --> 00:26:02,886 Now to multiply corresponding components together is quite 343 00:26:02,886 --> 00:26:07,641 easy to see what they are now because we want 4 * 2, which is 344 00:26:07,641 --> 00:26:13,820 8. 3 * 5 which is 15 and 7 * 4, which is 28. And if we had 345 00:26:13,820 --> 00:26:17,850 those that will get 51 like we did over here. So some people 346 00:26:17,850 --> 00:26:21,880 might find it a bit easier just to write them as column vectors 347 00:26:21,880 --> 00:26:24,360 and then just read off the corresponding components, 348 00:26:24,360 --> 00:26:27,770 multiply them together, and add them up to get the result. 349 00:26:32,200 --> 00:26:36,088 Now it's so important that you know how to calculate a scalar 350 00:26:36,088 --> 00:26:39,976 product that I'm going to give you one more example, just to 351 00:26:39,976 --> 00:26:42,244 make sure that you understand the process. 352 00:26:43,420 --> 00:26:48,060 And will go straight into column vector notation. Suppose the 353 00:26:48,060 --> 00:26:51,308 vector A is the vector minus 6. 354 00:26:51,870 --> 00:26:54,420 3 - 11. 355 00:26:55,710 --> 00:26:59,014 That's minus six I plus 3J minus 356 00:26:59,014 --> 00:27:06,532 11 K. And let's suppose the vector B is the 357 00:27:06,532 --> 00:27:11,600 vector 12:04 that's twelve IOJ plus 4K. 358 00:27:14,670 --> 00:27:20,354 So the dot product A dot B will be minus six 3 - 11. 359 00:27:21,520 --> 00:27:24,348 Dotted with twelve 04. 360 00:27:26,050 --> 00:27:29,504 And in the column vector notation, it's so easy to work 361 00:27:29,504 --> 00:27:33,272 this out. We want minus 6 * 12, which is minus 72. 362 00:27:35,410 --> 00:27:37,680 3 * 0 is 0. 363 00:27:40,100 --> 00:27:43,604 And minus 11 * 4 is minus 44. 364 00:27:44,980 --> 00:27:46,764 And if we add these up, we get. 365 00:27:48,180 --> 00:27:51,210 Minus 116 366 00:27:53,080 --> 00:27:57,030 so the dot product of A&B is just the sum. 367 00:27:57,850 --> 00:28:00,986 Of the products of the corresponding components. 368 00:28:06,430 --> 00:28:11,022 I want to move on now to look at some applications of the scalar 369 00:28:11,022 --> 00:28:13,974 product and the first application is one that I 370 00:28:13,974 --> 00:28:17,910 mentioned very early on and it's to do with testing whether or 371 00:28:17,910 --> 00:28:19,550 not two given vectors are 372 00:28:19,550 --> 00:28:23,799 perpendicular or not. Now you remember from the definition of 373 00:28:23,799 --> 00:28:27,506 the scalar product of two vectors, 8 be that is the 374 00:28:27,506 --> 00:28:31,550 modulus of the first one times the modulus of the second one 375 00:28:31,550 --> 00:28:35,257 times the cosine of the angle in between the two vectors. 376 00:28:36,280 --> 00:28:39,460 Now, if we find that when we work this out 377 00:28:39,460 --> 00:28:41,050 that the result is 0. 378 00:28:42,330 --> 00:28:47,048 Then either the modulus of a must be 0, or the modulus of be 379 00:28:47,048 --> 00:28:50,418 must be 0. Or Alternatively, cosine theater must be 0. 380 00:28:51,540 --> 00:28:55,740 Now, if we know that the two given vectors do not have a zero 381 00:28:55,740 --> 00:28:59,640 length, In other words, the modulus of a can't be 0, and the 382 00:28:59,640 --> 00:29:01,440 modulus of B can't be 0. 383 00:29:02,070 --> 00:29:06,714 Then the only conclusion we can draw is that cosine theater must 384 00:29:06,714 --> 00:29:12,132 be 0, and from that we did use that theater must be 90 degrees. 385 00:29:13,290 --> 00:29:16,865 In other words, if we take the dot product of two vectors and 386 00:29:16,865 --> 00:29:18,790 we find we get the answer 0. 387 00:29:19,800 --> 00:29:24,510 Then, provided the two factors were nonzero vectors, we can 388 00:29:24,510 --> 00:29:28,278 deduce the amb must be perpendicular vectors, so. 389 00:29:30,360 --> 00:29:32,110 If A&B. 390 00:29:34,090 --> 00:29:35,839 A nonzero vectors. 391 00:29:41,660 --> 00:29:42,550 Search that. 392 00:29:46,000 --> 00:29:49,575 Hey, don't be when we work it out. Turns out to be 0. 393 00:29:50,420 --> 00:29:53,640 Then A&B. 394 00:29:55,780 --> 00:29:56,710 Are perpendicular. 395 00:30:02,530 --> 00:30:03,630 There at right angles. 396 00:30:04,300 --> 00:30:08,216 Another word we sometimes use is orthogonal. You may hear that, 397 00:30:08,216 --> 00:30:11,420 particularly in more advanced work, two vectors at right 398 00:30:11,420 --> 00:30:14,624 angles to vectors which are perpendicular are sometimes said 399 00:30:14,624 --> 00:30:15,692 to be orthogonal. 400 00:30:16,660 --> 00:30:20,566 So we have a test here we take the two vectors, we find their 401 00:30:20,566 --> 00:30:24,193 dot product and if the answer that we get zero then we deduce 402 00:30:24,193 --> 00:30:27,262 that the two vectors must be perpendicular. So that's a very 403 00:30:27,262 --> 00:30:30,889 easy test we can apply to see if two vectors are Purple ****. 404 00:30:33,560 --> 00:30:34,946 Let me give you an example. 405 00:30:39,840 --> 00:30:46,522 Suppose we have the Vector A, which is the vector three 2 - 406 00:30:46,522 --> 00:30:52,690 1 and the Vector B which is 1 - 2 - 1. 407 00:30:54,580 --> 00:30:58,701 Now I think you would agree that just by looking at these vectors 408 00:30:58,701 --> 00:31:00,603 three, I +2 J Minus K. 409 00:31:01,530 --> 00:31:05,618 And one I minus two J minus K. You'd have no idea just from 410 00:31:05,618 --> 00:31:08,830 looking at them written down like that. Whether or not these 411 00:31:08,830 --> 00:31:10,582 vectors where at right angles to 412 00:31:10,582 --> 00:31:15,800 each other. We can apply this dot product test and workout a 413 00:31:15,800 --> 00:31:19,870 baby to see what happens. So let's workout a dot B. 414 00:31:21,390 --> 00:31:27,166 I want the dot product of three 2 - 1 with 1 - 2 - 1. 415 00:31:28,550 --> 00:31:30,020 And we'll get 3 ones or three. 416 00:31:30,950 --> 00:31:33,380 Two times minus 2 - 4. 417 00:31:33,940 --> 00:31:38,168 And minus one times minus one is plus one. So we've got three and 418 00:31:38,168 --> 00:31:43,000 one which is 4 - 4 which is 0. So you'll see that when we work 419 00:31:43,000 --> 00:31:46,624 out the dot product of these two nonzero vectors, the answer that 420 00:31:46,624 --> 00:31:47,832 we get is 0. 421 00:31:48,660 --> 00:31:52,296 So we can deduce from this result that this vector a must 422 00:31:52,296 --> 00:31:55,629 be at right angles to this vector B and that's something 423 00:31:55,629 --> 00:31:59,265 that's not obvious just from looking at it. So we've got a 424 00:31:59,265 --> 00:32:02,598 very useful test there to test for. The two vectors are 425 00:32:02,598 --> 00:32:03,507 perpendicular or not. 426 00:32:09,320 --> 00:32:11,980 Now another important application of the scalar 427 00:32:11,980 --> 00:32:16,160 product is to finding the angle between two given vectors. Let 428 00:32:16,160 --> 00:32:20,340 me remind you of the definition of the scalar product again 429 00:32:20,340 --> 00:32:24,900 because we will need that the scalar product of A&B is found 430 00:32:24,900 --> 00:32:29,460 by taking the length of the first vector times the length of 431 00:32:29,460 --> 00:32:33,640 the second vector times the cosine of the angle in between 432 00:32:33,640 --> 00:32:35,160 the two vectors so. 433 00:32:35,270 --> 00:32:38,286 If we get a vector, if we get two vectors in Cartesian form. 434 00:32:39,190 --> 00:32:40,665 And we calculate their dot 435 00:32:40,665 --> 00:32:45,238 product. And if we know how to calculate the modulus of each of 436 00:32:45,238 --> 00:32:48,826 those two vectors, then the only thing in this expression that we 437 00:32:48,826 --> 00:32:52,414 don't know is cosine Theta, so will be able to use this 438 00:32:52,414 --> 00:32:55,404 definition to calculate cosine theater the cosine of the angle 439 00:32:55,404 --> 00:32:58,693 between the two vectors, from which we can deduce the angle 440 00:32:58,693 --> 00:33:01,476 itself. OK, let's rearrange 441 00:33:01,476 --> 00:33:06,540 this. If we divide both sides by the modular surveying, the 442 00:33:06,540 --> 00:33:11,860 modulus of B will be able to get cosine theater is a dot B 443 00:33:11,860 --> 00:33:16,420 divided by the modulus of a modulus of B and that's the 444 00:33:16,420 --> 00:33:20,980 formula that we're going to use in a minute to find cosine, 445 00:33:20,980 --> 00:33:22,500 Theta and hence theater. 446 00:33:23,940 --> 00:33:27,252 Now we've done a lot of work already in this unit on 447 00:33:27,252 --> 00:33:30,564 calculating the dot product. Let me just remind you a little bit 448 00:33:30,564 --> 00:33:34,152 about how you find the modulus of a vector when it's given in 449 00:33:34,152 --> 00:33:41,640 cartesian form. If we have a Vector A, which is a one 450 00:33:41,640 --> 00:33:45,270 I plus A2J plus a 3K. 451 00:33:46,470 --> 00:33:51,370 Then the modulus of this vector is found by squaring 452 00:33:51,370 --> 00:33:53,820 the individual components, adding them. 453 00:33:55,890 --> 00:33:59,894 And finally, taking the square root of the result. So this is a 454 00:33:59,894 --> 00:34:03,590 formula that we're going to use for finding the modulus of a 455 00:34:03,590 --> 00:34:06,978 vector in Cartesian form, and this has been covered in an 456 00:34:06,978 --> 00:34:10,366 early you unit. If you need to look back at that. 457 00:34:14,830 --> 00:34:17,701 Let's look at a specific example where we want to find 458 00:34:17,701 --> 00:34:19,006 the angle between two vectors. 459 00:34:23,760 --> 00:34:26,140 So the problem is find the angle. 460 00:34:34,840 --> 00:34:41,128 Between the two vectors, I'm going to choose are A which is 461 00:34:41,128 --> 00:34:43,748 4I Plus 3J plus 7K. 462 00:34:45,320 --> 00:34:46,410 And B. 463 00:34:47,740 --> 00:34:54,149 Which is 2 I plus 5J Plus 4K, so two vectors given in 464 00:34:54,149 --> 00:34:59,079 cartesian form and the problem is to find the angle 465 00:34:59,079 --> 00:35:00,065 between them. 466 00:35:01,300 --> 00:35:06,370 And we have this result that we just reduced the cosine of the 467 00:35:06,370 --> 00:35:10,660 required angle. Is the dot product 8B divided by the length 468 00:35:10,660 --> 00:35:15,730 of a times the length of be? So that's the formula that we're 469 00:35:15,730 --> 00:35:16,900 going to use. 470 00:35:18,470 --> 00:35:23,926 OK, a dot B now. In fact a dot B has already been evaluated in an 471 00:35:23,926 --> 00:35:28,359 earlier example, but I'll just remind you of that a dot B. If 472 00:35:28,359 --> 00:35:31,428 we use the column vector notation will be 437. 473 00:35:32,460 --> 00:35:34,359 Dotted with 254 474 00:35:37,210 --> 00:35:44,347 which is 428-3515 and Seven 428. And if you work that out, you 475 00:35:44,347 --> 00:35:46,543 find you get 51. 476 00:35:48,130 --> 00:35:50,715 So all we need to do is calculate now the modular 477 00:35:50,715 --> 00:35:53,535 survey in the modulus of B and then we've got everything we 478 00:35:53,535 --> 00:35:56,355 need to know to put in the right hand side of this 479 00:35:56,355 --> 00:35:56,590 formula. 480 00:35:58,010 --> 00:35:59,518 OK, the modular survey. 481 00:36:02,030 --> 00:36:06,650 Now the modulus of a is found by taking the square root of the 482 00:36:06,650 --> 00:36:10,280 squares of these components added up. So we want 4 squared. 483 00:36:11,470 --> 00:36:12,570 3 squared 484 00:36:13,620 --> 00:36:14,700 7 squared. 485 00:36:16,700 --> 00:36:19,730 Which is going to be the square root of 16. 486 00:36:21,030 --> 00:36:26,670 330977's of 49 and if you work that out, you'll find that 487 00:36:26,670 --> 00:36:29,490 that's the square root of 74. 488 00:36:31,860 --> 00:36:37,827 Similarly, the modulus of B will be the square root of 2 squared. 489 00:36:39,350 --> 00:36:42,298 +5 squared, +4 squared. 490 00:36:44,400 --> 00:36:51,862 Which is the square root of 4 + 25 + 16 and 25 + 491 00:36:51,862 --> 00:36:54,527 4 + 16 is 45. 492 00:36:55,890 --> 00:36:58,740 So we've got all the ingredients we need now to 493 00:36:58,740 --> 00:36:59,880 pop into this formula. 494 00:37:01,190 --> 00:37:06,572 Cosine theater will be a dotted with B, which we found was 51 495 00:37:06,572 --> 00:37:11,540 divided by the modular survey, which is the square root of 74. 496 00:37:12,810 --> 00:37:16,982 The modulus of B, which is the square root of 45 now will need 497 00:37:16,982 --> 00:37:18,770 a Calculator to work this out. 498 00:37:21,650 --> 00:37:22,949 We want 51. 499 00:37:24,980 --> 00:37:27,745 Divided by the square root of 74. 500 00:37:29,090 --> 00:37:30,390 Divided by the square root. 501 00:37:31,370 --> 00:37:38,560 45 Which is not .8838 to 4 decimal places. That's the 502 00:37:38,560 --> 00:37:43,609 cosine of the angle between the two vectors A&B. 503 00:37:44,440 --> 00:37:46,306 If we find the inverse cosine. 504 00:37:49,340 --> 00:37:51,610 Of .8838. 505 00:37:54,160 --> 00:37:58,228 Which is 27.90 degrees. 506 00:38:00,980 --> 00:38:04,566 So there we've seen how we can use the scalar product. 507 00:38:06,040 --> 00:38:09,883 To find the angle between two vectors when they're 508 00:38:09,883 --> 00:38:11,591 given in Cartesian form. 509 00:38:17,270 --> 00:38:21,314 Now there's one more application I'd like to tell you about, and 510 00:38:21,314 --> 00:38:25,021 it's concerned with finding the components of one vector in the 511 00:38:25,021 --> 00:38:28,728 direction of a second vector. Let me give you an example. 512 00:38:30,860 --> 00:38:33,308 Suppose we have a vector A. 513 00:38:35,170 --> 00:38:36,556 Let me have a second vector. 514 00:38:38,300 --> 00:38:38,990 B. 515 00:38:42,280 --> 00:38:46,988 And I've drawn these vectors so that their tails coincide and. 516 00:38:47,580 --> 00:38:51,280 As usual, the angle between the two vectors is theater. 517 00:38:52,440 --> 00:38:54,260 Let's call this .0. 518 00:38:54,850 --> 00:38:58,546 And what I'm going to do is I'm going to drop a 519 00:38:58,546 --> 00:39:01,010 perpendicular from this point, which I'll call B 520 00:39:01,010 --> 00:39:01,318 down. 521 00:39:04,330 --> 00:39:06,100 To meet the vector A. 522 00:39:09,510 --> 00:39:11,640 Let's call this point here a. 523 00:39:13,100 --> 00:39:15,347 And we can regard the vector OB. 524 00:39:16,810 --> 00:39:21,046 As the sum of the vector from O2 Capital A and then 525 00:39:21,046 --> 00:39:25,282 the vector from A to B. In other words, the vector OB. 526 00:39:27,190 --> 00:39:28,940 Is the sum of OA. 527 00:39:31,680 --> 00:39:32,988 Plus a bee. 528 00:39:36,550 --> 00:39:40,598 Now this vector here, which we've called OK, that vector is 529 00:39:40,598 --> 00:39:45,750 said to be the component of B in the direction of a. You'll see 530 00:39:45,750 --> 00:39:50,902 we can regard be as being made up of two components, one in the 531 00:39:50,902 --> 00:39:56,054 direction of A and one which is at right angles to that. So this 532 00:39:56,054 --> 00:39:57,158 component, oh A. 533 00:39:57,960 --> 00:40:00,910 Is the component of B in the direction of a. 534 00:40:18,340 --> 00:40:21,980 And what we're going to do is we're going to see how we can 535 00:40:21,980 --> 00:40:24,060 use the scalar product to find this component. 536 00:40:26,070 --> 00:40:30,282 Let's call this length here the length from O to a. Let's call 537 00:40:30,282 --> 00:40:34,930 that little. And then let's focus our attention on this 538 00:40:34,930 --> 00:40:36,070 right angled triangle. 539 00:40:36,830 --> 00:40:38,300 Oba. 540 00:40:39,470 --> 00:40:41,514 So there's a right angle in there. 541 00:40:43,320 --> 00:40:47,390 Now, using our knowledge of trigonometry, we can write down 542 00:40:47,390 --> 00:40:51,867 the cosine of this angle is adjacent over hypotenuse, so the 543 00:40:51,867 --> 00:40:53,088 cosine of Theta. 544 00:40:54,640 --> 00:40:57,874 Is this length L the adjacent side? 545 00:40:59,350 --> 00:41:01,878 Over the hypotenuse, now the hypotenuse is the 546 00:41:01,878 --> 00:41:03,142 length of this side. 547 00:41:04,650 --> 00:41:07,410 And this is the length of this side is the modulus of 548 00:41:07,410 --> 00:41:07,870 vector be. 549 00:41:12,020 --> 00:41:16,299 Now, In other words, if we rearrange this L, the component 550 00:41:16,299 --> 00:41:22,134 of B in the direction of a week and right as the modulus of be 551 00:41:22,134 --> 00:41:23,690 times, the cosine Theta. 552 00:41:27,750 --> 00:41:31,290 Now we've already got an expression for the cosine of 553 00:41:31,290 --> 00:41:35,538 this angle in the previous work that we've done, we already know 554 00:41:35,538 --> 00:41:40,140 that the cosine of feta can be written as the dot product for 555 00:41:40,140 --> 00:41:45,070 amb. Divided by the modular survey times the modulus of be, 556 00:41:45,070 --> 00:41:47,344 that's the result we just had. 557 00:41:48,470 --> 00:41:51,200 So if you look at this expression now, you'll see 558 00:41:51,200 --> 00:41:54,476 with the modulus B in the numerator and a modulus of be 559 00:41:54,476 --> 00:41:55,841 in the denominator, they cancel. 560 00:41:56,900 --> 00:41:58,420 So we can write this. 561 00:41:59,170 --> 00:42:03,535 As a not be over the modulus of A. 562 00:42:06,760 --> 00:42:08,730 Let me write that down again on the next page. 563 00:42:20,970 --> 00:42:24,546 So this remember L is the component of be in the direction 564 00:42:24,546 --> 00:42:28,718 of a. We can write it like this. I'm going to Rearrange this a 565 00:42:28,718 --> 00:42:32,592 little bit to write it as a divided by the modulus of A. 566 00:42:34,000 --> 00:42:35,908 Dotted with effective be. 567 00:42:37,240 --> 00:42:40,771 And because of the result we got very early on the 568 00:42:40,771 --> 00:42:43,981 commutativity of the dot product. We can write this as 569 00:42:43,981 --> 00:42:46,870 B dot A divided by the modulus of A. 570 00:42:49,400 --> 00:42:53,313 Now when you take any vector and you divide it by its modulus, 571 00:42:53,313 --> 00:42:57,226 what you get is a unit vector in the direction of that vector, 572 00:42:57,226 --> 00:43:01,440 and we write a unit vector as the vector with a little hat on 573 00:43:01,440 --> 00:43:05,353 like that. So this is a unit vector in the direction of A. 574 00:43:14,770 --> 00:43:18,502 So this is an important result we've got. We've got L, which 575 00:43:18,502 --> 00:43:22,856 is the component of be in the direction of a can be found by 576 00:43:22,856 --> 00:43:26,588 taking the dot product of be with a unit vector in the 577 00:43:26,588 --> 00:43:27,521 direction of A. 578 00:43:31,340 --> 00:43:37,794 So suppose we wish to find the component of Vector B which is 3 579 00:43:37,794 --> 00:43:43,787 I plus J Plus 4K in the direction of the Vector A, which 580 00:43:43,787 --> 00:43:46,553 is I minus J Plus K. 581 00:43:48,180 --> 00:43:53,306 What we want to do is evaluate B dotted with a 582 00:43:53,306 --> 00:43:55,636 divided by the modular survey. 583 00:43:59,190 --> 00:44:03,026 Now we can do that as follows. We can workout the dot product B 584 00:44:03,026 --> 00:44:07,136 dot A and then divide by the modulus of a. So B Dot A is 585 00:44:07,136 --> 00:44:09,602 going to be 3 * 1, which is 3. 586 00:44:10,190 --> 00:44:15,970 Plus one times minus one which is minus 1 + 4 * 1 which is 4 or 587 00:44:15,970 --> 00:44:20,390 divided by the modulus of a which is the square root of 1 588 00:44:20,390 --> 00:44:24,130 squared plus minus one squared plus one squared, which is the 589 00:44:24,130 --> 00:44:25,490 square root of 3. 590 00:44:27,570 --> 00:44:32,988 If we work that out, will get 3 - 1 is 2 + 4 six six over Route 591 00:44:32,988 --> 00:44:37,503 3. This means that the component of B in the direction of a is 6. 592 00:44:37,503 --> 00:44:41,115 Over Route 3. We might want to write that in an alternative 593 00:44:41,115 --> 00:44:45,329 form just to tidy it up so we don't have the square root in 594 00:44:45,329 --> 00:44:48,640 the denominator and we can do that by multiplying top and 595 00:44:48,640 --> 00:44:52,553 bottom by the square root of 3, which will produce Route 3 times 596 00:44:52,553 --> 00:44:56,165 route 3 in the denominator, which is 3 and three into six 597 00:44:56,165 --> 00:44:58,884 goes twice. So that would just simplify to 598 00:44:58,884 --> 00:45:00,404 two square root of 3. 599 00:45:02,560 --> 00:45:06,169 So in this unit we've introduced the scalar product. 600 00:45:06,760 --> 00:45:10,225 Learn how to calculate it and looked at some of its 601 00:45:10,225 --> 00:45:11,170 properties and applications.