1 00:00:01,270 --> 00:00:04,366 In this unit, we're going to have a look at a way 2 00:00:04,366 --> 00:00:05,398 of combining two vectors. 3 00:00:06,460 --> 00:00:08,584 This method is called the vector 4 00:00:08,584 --> 00:00:12,341 product. And it's called the vector product because when we 5 00:00:12,341 --> 00:00:16,358 combine the two vectors in this way, the result that will get is 6 00:00:16,358 --> 00:00:20,495 another vector. OK, let's start with the two vectors. 7 00:00:23,410 --> 00:00:26,994 So suppose we have a vector A. 8 00:00:27,840 --> 00:00:29,130 And another vector. 9 00:00:29,860 --> 00:00:37,320 Be. And we have drawn these vectors A&B so that 10 00:00:37,320 --> 00:00:39,060 their tails coincide. 11 00:00:39,560 --> 00:00:44,070 And so that we can label the angle between A&B's 12 00:00:44,070 --> 00:00:45,423 theater like that. 13 00:00:46,610 --> 00:00:49,718 So we start with two vectors and we're going to do some 14 00:00:49,718 --> 00:00:51,790 calculations with these two vectors to generate what's 15 00:00:51,790 --> 00:00:52,826 called the vector product. 16 00:00:53,570 --> 00:00:58,370 And the vector product is defined to be the length of the 17 00:00:58,370 --> 00:01:00,370 first vector. That's the length 18 00:01:00,370 --> 00:01:05,070 of A. Multiplied by the length of the second vector, the length 19 00:01:05,070 --> 00:01:11,840 of be. Multiplied this time by the sign of the angle between a 20 00:01:11,840 --> 00:01:15,895 envy. Length of the first one, length of the second one, and 21 00:01:15,895 --> 00:01:17,785 the sign of the angle between 22 00:01:17,785 --> 00:01:22,080 the two vectors. Now that's all well and good, but this isn't a 23 00:01:22,080 --> 00:01:25,460 vector. This is a scalar. This is just a number times a number 24 00:01:25,460 --> 00:01:29,315 times a number. And if we're going to get a result which is a 25 00:01:29,315 --> 00:01:30,665 vector we've got to give some 26 00:01:30,665 --> 00:01:34,995 direction to this. If we look back at the diagram, will notice 27 00:01:34,995 --> 00:01:38,460 that if we choose any two vectors, these two vectors lie 28 00:01:38,460 --> 00:01:42,240 in a plane. They form a plane in which they both lie. 29 00:01:43,340 --> 00:01:47,608 When we calculate the vector product of these two vectors, we 30 00:01:47,608 --> 00:01:51,488 define the direction of the vector product to be the 31 00:01:51,488 --> 00:01:55,368 direction which is perpendicular to the plane containing A&B. So 32 00:01:55,368 --> 00:02:00,800 if we have a vector which is perpendicular to a man to be and 33 00:02:00,800 --> 00:02:05,456 to enter the plane containing an be, this gives us the direction 34 00:02:05,456 --> 00:02:07,008 of the vector product. 35 00:02:07,930 --> 00:02:11,360 And in fact, there are two possible directions which are 36 00:02:11,360 --> 00:02:14,790 perpendicular to this plane, because In addition to the one 37 00:02:14,790 --> 00:02:17,877 that I've drawn upwards like this, There's also one. 38 00:02:18,660 --> 00:02:22,970 That's downwards like that, so we have two possible directions 39 00:02:22,970 --> 00:02:26,418 which are perpendicular to the plane containing A&B. 40 00:02:27,170 --> 00:02:30,870 And we have a convention for deciding which direction we 41 00:02:30,870 --> 00:02:34,312 should choose. Now, this convention is variously called 42 00:02:34,312 --> 00:02:38,238 the right hand screw rule or the right hand thumb rule, so I'm 43 00:02:38,238 --> 00:02:40,352 going to show you both of these 44 00:02:40,352 --> 00:02:45,768 different conventions. First of all, the right hand screw rule. 45 00:02:46,770 --> 00:02:50,325 If you take a conventional screwdriver and right handed 46 00:02:50,325 --> 00:02:53,880 screw, and if you imagine turning the screwdriver handle 47 00:02:53,880 --> 00:02:58,620 so that we turn from the direction of a round to the 48 00:02:58,620 --> 00:03:02,965 direction of be. So I'm turning my screwdriver In this sense. 49 00:03:03,840 --> 00:03:05,230 That way around like that. 50 00:03:06,410 --> 00:03:10,029 Imagine the direction in which the screw would advance. So in 51 00:03:10,029 --> 00:03:12,990 turning from the sensor data, the sense of B. 52 00:03:13,880 --> 00:03:15,158 The scroll advance. 53 00:03:15,720 --> 00:03:21,330 Upwards. So it's this upwards direction here, which we take as 54 00:03:21,330 --> 00:03:24,957 the direction we require for evaluating this vector product. 55 00:03:25,850 --> 00:03:31,526 Let me denote a unit vector in this direction by N. 56 00:03:32,510 --> 00:03:35,513 So if we want to calculate the vector product of amb, this is 57 00:03:35,513 --> 00:03:36,668 going to be its magnitude. 58 00:03:37,260 --> 00:03:40,980 And the direction we want is one in the direction of this 59 00:03:40,980 --> 00:03:45,320 unit vector N. So I put a unit vector N there to give the 60 00:03:45,320 --> 00:03:46,870 direction to this quantity here. 61 00:03:48,120 --> 00:03:51,045 There's another way, which sometimes people use to define 62 00:03:51,045 --> 00:03:53,970 the direction of the vector product, which is equivalent, 63 00:03:53,970 --> 00:03:57,870 and it's called the right hand thumb rule. Let me explain this 64 00:03:57,870 --> 00:04:01,770 as well. Imagine you take your fingers of your right hand and 65 00:04:01,770 --> 00:04:05,995 you curl them in the sense going round from a round towards be. 66 00:04:05,995 --> 00:04:07,620 So you curl your fingers. 67 00:04:08,360 --> 00:04:13,028 In the direction from a round to be then the thumb points 68 00:04:13,028 --> 00:04:16,140 in the required direction of the vector product. 69 00:04:18,540 --> 00:04:19,810 Yet another way of thinking 70 00:04:19,810 --> 00:04:24,180 about this. Still with the right hand is to point the first 71 00:04:24,180 --> 00:04:26,046 finger in the direction of A. 72 00:04:26,800 --> 00:04:28,455 The middle finger in the 73 00:04:28,455 --> 00:04:33,684 direction of B. And then the thumb points in the required 74 00:04:33,684 --> 00:04:35,824 direction of the vector product 75 00:04:35,824 --> 00:04:40,710 of A&B. So using either the right hand screw rule or the 76 00:04:40,710 --> 00:04:45,013 right hand thumb rule, we can deduce that when we want to find 77 00:04:45,013 --> 00:04:48,654 the vector product of these two vectors A&B, the direction that 78 00:04:48,654 --> 00:04:52,295 we require is the one that's moving upwards on this diagram 79 00:04:52,295 --> 00:04:53,950 here, not the one moving 80 00:04:53,950 --> 00:04:58,200 downwards. So the vector product is defined as the length of the 81 00:04:58,200 --> 00:05:01,560 first times the length of the second times the sign of the 82 00:05:01,560 --> 00:05:03,240 angle between the two times this 83 00:05:03,240 --> 00:05:07,970 unit vector N. Which is defined in a sense defined by the right 84 00:05:07,970 --> 00:05:09,746 hand thumb rule or the right 85 00:05:09,746 --> 00:05:14,435 hand screw rule. Now we have a notation for the vector product 86 00:05:14,435 --> 00:05:19,910 and we write the vector product of A&B, like this A and we use a 87 00:05:19,910 --> 00:05:21,370 time sign or across. 88 00:05:21,890 --> 00:05:25,640 And so sometimes instead of calling this the vector product, 89 00:05:25,640 --> 00:05:29,765 we sometimes call this the cross product and throughout the rest 90 00:05:29,765 --> 00:05:34,265 of this unit, sometimes you'll hear me refer to this either as 91 00:05:34,265 --> 00:05:37,265 the cross product product, all the vector product. 92 00:05:37,520 --> 00:05:40,868 And I'll use these two words 93 00:05:40,868 --> 00:05:44,414 interchangeably. So that's the definition will need. I'll also 94 00:05:44,414 --> 00:05:48,144 mention that some authors and some lecturers will use a 95 00:05:48,144 --> 00:05:52,247 different notation again, and you might see a cross be written 96 00:05:52,247 --> 00:05:55,604 using this wedge symbol like this, and that's equally 97 00:05:55,604 --> 00:06:00,080 acceptable and often people will use the wedge as well to define 98 00:06:00,080 --> 00:06:01,199 the vector product. 99 00:06:02,150 --> 00:06:05,474 It's very important that you put this symbol in either the wedge 100 00:06:05,474 --> 00:06:08,798 or the cross because you don't want to just write down to 101 00:06:08,798 --> 00:06:12,122 vectors like that when you might mean the vector product or you 102 00:06:12,122 --> 00:06:15,169 in fact might mean a scalar product. Or you might mean 103 00:06:15,169 --> 00:06:18,216 something else, so don't use that sort of notation. Make sure 104 00:06:18,216 --> 00:06:21,540 you're very explicit about when you want to use a vector product 105 00:06:21,540 --> 00:06:23,479 by putting the time sign or the 106 00:06:23,479 --> 00:06:28,252 wedge sign in. Let's look at what happens now. When we 107 00:06:28,252 --> 00:06:33,699 calculate the vector product of A&B. But we do it in a different 108 00:06:33,699 --> 00:06:37,051 order, so we look at B cross a. 109 00:06:38,570 --> 00:06:42,530 Now, as before, we want the modulus of the first factor, 110 00:06:42,530 --> 00:06:44,690 which is the modulus of be. 111 00:06:45,280 --> 00:06:49,370 Multiplied by the modulus of the second vector, the modular 112 00:06:49,370 --> 00:06:54,876 survey. We want the sign of the angle between B&A, which is 113 00:06:54,876 --> 00:06:57,252 still the sign of angle theater. 114 00:06:58,020 --> 00:06:59,510 And we want to direction. 115 00:07:00,580 --> 00:07:04,612 Now again using our right hand screw or right hand thumb rule 116 00:07:04,612 --> 00:07:09,652 we can try to find the sense we require so that as we turn from 117 00:07:09,652 --> 00:07:14,020 B round to a, this time from the first round of the second 118 00:07:14,020 --> 00:07:17,716 vector, reoccuring my fingers in the sense from be round to 119 00:07:17,716 --> 00:07:20,740 a. You'll see now that the thumb points downwards. 120 00:07:22,210 --> 00:07:26,513 So this time we want a vector which is a unit vector pointing 121 00:07:26,513 --> 00:07:27,837 downwards instead of upwards. 122 00:07:28,920 --> 00:07:33,618 Now unit vector downwards must be minus N hot. 123 00:07:34,200 --> 00:07:36,167 If the output vector was an hat. 124 00:07:36,860 --> 00:07:42,514 So this time the required direction of be cross a is 125 00:07:42,514 --> 00:07:44,056 minus an hat. 126 00:07:45,570 --> 00:07:50,162 If we bring the minus sign out to the front, you'll see that we 127 00:07:50,162 --> 00:07:54,426 get modulus of be modulus of a sine theater. An hat with the 128 00:07:54,426 --> 00:07:58,690 minus sign at the front now, and if you examine this vector with 129 00:07:58,690 --> 00:08:02,626 the vector we had before, when we calculated a cross, B will 130 00:08:02,626 --> 00:08:06,234 see that the magnitudes of these two vectors are the same, 131 00:08:06,234 --> 00:08:10,498 because we've got a modulus of a modulus of being sign theater in 132 00:08:10,498 --> 00:08:14,720 each. But the directions are now different, because where is the 133 00:08:14,720 --> 00:08:18,428 direction of this one? Was an hat the direction now is minus 134 00:08:18,428 --> 00:08:22,146 and hat. And we see that this quantity in here. 135 00:08:23,580 --> 00:08:25,498 Is the same as we had here. 136 00:08:26,070 --> 00:08:31,978 So in fact, what we've found is that B Cross A is the negative 137 00:08:31,978 --> 00:08:34,510 of a crossbite. This is very 138 00:08:34,510 --> 00:08:39,274 important. When we interchange the order of the two vectors. 139 00:08:40,100 --> 00:08:44,377 We get a different answer be cross a is in fact the negative 140 00:08:44,377 --> 00:08:49,980 of across be. So it's not true that a cross B is the same as be 141 00:08:49,980 --> 00:08:56,835 across A. Across be is not equal to be across A. 142 00:08:57,560 --> 00:09:03,538 And in fact, what is true is that B cross a is the negative 143 00:09:03,538 --> 00:09:05,246 of a cross be. 144 00:09:05,350 --> 00:09:11,600 So we say that the vector product is not commutative. 145 00:09:11,600 --> 00:09:16,501 So what that means in practice is that when you've got to find 146 00:09:16,501 --> 00:09:20,648 the vector product of two vectors, you must be very clear 147 00:09:20,648 --> 00:09:25,172 about the order in which you want to carry out the operation. 148 00:09:26,370 --> 00:09:32,070 Now there's a second property of the vector product. I want to 149 00:09:32,070 --> 00:09:37,295 explain to you, and it's called the distributivity of the vector 150 00:09:37,295 --> 00:09:42,995 product over addition. What does that mean? It means that if we 151 00:09:42,995 --> 00:09:49,645 have a vector A and we want to find the cross product with the 152 00:09:49,645 --> 00:09:52,970 sum of two vectors B Plus C. 153 00:09:53,250 --> 00:09:58,294 We can evaluate this or remove the brackets in the way that you 154 00:09:58,294 --> 00:10:01,786 would normally expect algebraic expressions to be evaluated. In 155 00:10:01,786 --> 00:10:05,278 other words, the cross product distributes itself over this 156 00:10:05,278 --> 00:10:09,546 edition. In other words, that means that we workout a crossed 157 00:10:09,546 --> 00:10:16,760 with B. And then because of this addition here, we add that to a 158 00:10:16,760 --> 00:10:18,290 crossed with C. 159 00:10:18,530 --> 00:10:22,996 So this is the distributivity rule which allows you to remove 160 00:10:22,996 --> 00:10:25,026 brackets in a natural way. 161 00:10:25,840 --> 00:10:30,096 It's also works through the way around, so if we had B Plus C 162 00:10:30,096 --> 00:10:33,886 first. And we want to cross that 163 00:10:33,886 --> 00:10:36,772 with a. We do this in a way you 164 00:10:36,772 --> 00:10:39,179 would expect. Be cross a. 165 00:10:39,860 --> 00:10:43,195 Plus 166 00:10:43,195 --> 00:10:50,078 secrecy. And together those rules 167 00:10:50,078 --> 00:10:53,626 are called the distributivity 168 00:10:53,626 --> 00:10:58,576 rules. I will use those in a little bit later, little bit 169 00:10:58,576 --> 00:11:00,356 later on in this unit. 170 00:11:02,170 --> 00:11:07,549 Another property I want to explain to you is what happens 171 00:11:07,549 --> 00:11:12,928 when the two vectors that we're interested in our parallel. So 172 00:11:12,928 --> 00:11:18,796 let's look at what happens when we want to find the vector 173 00:11:18,796 --> 00:11:20,752 product of parallel vectors. 174 00:11:23,040 --> 00:11:29,412 Suppose now then we have a Vector A and another vector B 175 00:11:29,412 --> 00:11:35,784 where A&B are parallel vectors. Is that parallel and we make the 176 00:11:35,784 --> 00:11:41,094 tales of these two vectors coincide? Then the angle between 177 00:11:41,094 --> 00:11:47,466 them will be 0. So when the vectors are parallel theater is 178 00:11:47,466 --> 00:11:54,175 0. So when we come to work out across VB, the modulus 179 00:11:54,175 --> 00:11:59,730 of a modulus of B, the sign, this time of 0. 180 00:12:00,040 --> 00:12:02,238 And the sign of 0 is 0. 181 00:12:02,900 --> 00:12:06,607 So when the two vectors are parallel, the magnitude of this 182 00:12:06,607 --> 00:12:11,662 vector turns out to be not. So in fact what we get is the zero 183 00:12:11,662 --> 00:12:14,592 vector. So for two parallel 184 00:12:14,592 --> 00:12:18,505 vectors. The vector product is the zero vector. 185 00:12:19,460 --> 00:12:25,368 We now want to start to look at how we can calculate the vector 186 00:12:25,368 --> 00:12:30,010 product when the two vectors are given in Cartesian form. Before 187 00:12:30,010 --> 00:12:35,074 we do that, I'd like to develop some results which will be 188 00:12:35,074 --> 00:12:40,551 particularly important. In this diagram I've shown a 3 189 00:12:40,551 --> 00:12:45,798 dimensional coordinate system, so we can see an X 190 00:12:45,798 --> 00:12:49,296 axis or Y axis and it 191 00:12:49,296 --> 00:12:54,543 said access. And superimposed on this system I've got a unit 192 00:12:54,543 --> 00:12:59,541 vector I in the X direction unit vector J in the Y direction and 193 00:12:59,541 --> 00:13:01,683 unit vector K and the zed 194 00:13:01,683 --> 00:13:07,046 direction. And what I want to do is explore what happens when we 195 00:13:07,046 --> 00:13:12,268 cross these unit vectors so we workout I cross J or J Cross K 196 00:13:12,268 --> 00:13:14,506 etc. And let's see what happens. 197 00:13:14,520 --> 00:13:17,995 Suppose we want I crossed 198 00:13:17,995 --> 00:13:23,570 with J. Well, by definition, the vector product of iron Jay will 199 00:13:23,570 --> 00:13:26,024 be the modulus of the first 200 00:13:26,024 --> 00:13:31,150 vector. And we want the modulus of this vector I now this is a 201 00:13:31,150 --> 00:13:32,700 unit vector remember, so its 202 00:13:32,700 --> 00:13:36,074 modulus is one. Times the modulus of the second vector, 203 00:13:36,074 --> 00:13:40,862 and again the modulus of J is the modulus of a unit vector. So 204 00:13:40,862 --> 00:13:41,888 again, it's one. 205 00:13:42,790 --> 00:13:45,274 Times the sign of the angle 206 00:13:45,274 --> 00:13:48,600 between. I&J, that's the sign of 207 00:13:48,600 --> 00:13:54,221 90 degrees. And then we've got to give it a direction, and the 208 00:13:54,221 --> 00:13:58,793 direction is that defined by the right hand screw rule or right 209 00:13:58,793 --> 00:14:03,365 hand thumb rule. So as we imagine, turning from Iran to J. 210 00:14:03,365 --> 00:14:07,556 Imagine curling the fingers around from I to J. The thumb 211 00:14:07,556 --> 00:14:11,747 points in the upward direction points in the K direction. So 212 00:14:11,747 --> 00:14:16,700 when we work out across J, the direction of the result will be 213 00:14:16,700 --> 00:14:19,367 Kay Kay being a unit vector in 214 00:14:19,367 --> 00:14:22,672 the direction. Which is perpendicular to the plane 215 00:14:22,672 --> 00:14:26,758 containing I&J. Now, this simplifies a great deal because 216 00:14:26,758 --> 00:14:32,518 the sign of 90 side of 90 degrees is one. So we've got 1 * 217 00:14:32,518 --> 00:14:34,822 1 * 1, which is 1K. 218 00:14:34,840 --> 00:14:41,200 So we've got this important result that I cross. Jay is K. 219 00:14:41,300 --> 00:14:47,618 Let's look at what happens now when we work out. Jake Ross I. 220 00:14:48,220 --> 00:14:51,892 When we work out Jake Ross, I were doing the vector product 221 00:14:51,892 --> 00:14:55,564 in the opposite order to which we did it when we calculated 222 00:14:55,564 --> 00:14:59,848 across J and we know from what we've just done that we get a 223 00:14:59,848 --> 00:15:02,602 vector of the same magnitude of the opposite sense, 224 00:15:02,602 --> 00:15:06,274 opposite direction. So when we work out Jake Ross, I in fact 225 00:15:06,274 --> 00:15:09,028 will get minus K, which is another important result. 226 00:15:10,430 --> 00:15:16,094 Similarly, if we work out, say, for example J Cross K, let's 227 00:15:16,094 --> 00:15:17,982 look at that one. 228 00:15:18,450 --> 00:15:23,299 Again, you want the modulus of the first one, which is one the 229 00:15:23,299 --> 00:15:28,148 modulus of the second one, which is one the sign of the angle 230 00:15:28,148 --> 00:15:33,270 between J&K. Which is the sign of 90 degrees. And again you 231 00:15:33,270 --> 00:15:37,362 want to direction defined by the right hand screw rule and J 232 00:15:37,362 --> 00:15:41,454 Cross K being where they are here. If you kill your fingers 233 00:15:41,454 --> 00:15:45,546 in the sense around from the direction of J round to the 234 00:15:45,546 --> 00:15:49,638 direction of K and imagine which way your thumb will move, your 235 00:15:49,638 --> 00:15:51,684 thumb will move in the eye 236 00:15:51,684 --> 00:15:55,410 direction. So Jake Ross K equals 237 00:15:55,410 --> 00:16:00,530 I. 1 * 1 * 1 gives you just the eye. 238 00:16:01,360 --> 00:16:06,530 So this is another important result. J Cross K is I. 239 00:16:06,530 --> 00:16:11,051 And equivalently, if we reverse, the order will get cake. Ross 240 00:16:11,051 --> 00:16:15,161 Jay is minus I from the result we had before. 241 00:16:15,720 --> 00:16:20,400 So we've dealt with I crossing with JJ, crossing with K. What 242 00:16:20,400 --> 00:16:22,350 about I crossing with K? 243 00:16:22,980 --> 00:16:27,236 If we workout I Cross K again. Modulus of the first one is one 244 00:16:27,236 --> 00:16:31,188 modulus of the second one is one and I encounter at 90 degrees. 245 00:16:31,188 --> 00:16:35,140 So with the sign of 90 degrees and we want a direction again 246 00:16:35,140 --> 00:16:39,092 and again with the right hand screw rule I cross K will give 247 00:16:39,092 --> 00:16:43,652 you a direction which is in this time in the sense of minus J we 248 00:16:43,652 --> 00:16:47,604 just look at that for a minute. If you imagine to curling your 249 00:16:47,604 --> 00:16:50,644 fingers in the sense of round from my towards K. 250 00:16:52,410 --> 00:16:58,434 Then the thumb will point in the direction of minus J, so 251 00:16:58,434 --> 00:17:04,960 we've got I Cross K is minus J, or equivalently K Cross I 252 00:17:04,960 --> 00:17:05,964 equals J. 253 00:17:08,240 --> 00:17:10,580 So again important results. 254 00:17:11,420 --> 00:17:16,110 Now, it's important that you can remember how you calculate 255 00:17:16,110 --> 00:17:22,207 vector product of the eyes in the Jays. In the case that I'm 256 00:17:22,207 --> 00:17:27,835 going to suggest a way that you might remember this if you 257 00:17:27,835 --> 00:17:32,994 write down the IJ&K in a cyclic order like that, and 258 00:17:32,994 --> 00:17:37,215 imagine moving around this cycle in a clockwise sense. 259 00:17:38,370 --> 00:17:43,323 And if you want to workout, I cross Jay. The result will be 260 00:17:43,323 --> 00:17:45,228 the next vector round K. 261 00:17:45,230 --> 00:17:51,854 If you want to workout J Cross K. 262 00:17:52,390 --> 00:17:57,370 The result will be the next vector round when we move around 263 00:17:57,370 --> 00:17:59,030 clockwise, which is I. 264 00:18:00,780 --> 00:18:04,510 And finally came across I. 265 00:18:04,620 --> 00:18:06,000 Same argument. 266 00:18:07,040 --> 00:18:11,192 Next vector round is Jay, so that's that's an easy way of 267 00:18:11,192 --> 00:18:14,998 remembering the vector products of the eyes and Jason case. And 268 00:18:14,998 --> 00:18:18,804 clearly when you reverse the order of any of these, you 269 00:18:18,804 --> 00:18:21,918 introduce a minus sign. Alternatively, if you wanted for 270 00:18:21,918 --> 00:18:26,416 example K Cross J, you'd realize that to move from kata ju moving 271 00:18:26,416 --> 00:18:30,914 anticlockwise, so K Cross Jay is minus I, that's the way that I 272 00:18:30,914 --> 00:18:32,644 find helpful to remember these 273 00:18:32,644 --> 00:18:38,960 results. We're now in a position to calculate the vector product 274 00:18:38,960 --> 00:18:41,915 of two vectors given in 275 00:18:41,915 --> 00:18:49,049 Cartesian form. Suppose we start with the Vector A and let's 276 00:18:49,049 --> 00:18:55,661 suppose this is a general 3 dimensional vector, a one I plus 277 00:18:55,661 --> 00:19:01,722 A2J plus a 3K where A1A2A3 or any numbers we choose. 278 00:19:02,580 --> 00:19:09,543 Suppose our second vector B, again arbitrary is be one. I 279 00:19:09,543 --> 00:19:12,708 be 2 J and B3K. 280 00:19:12,820 --> 00:19:17,930 And we set about trying to calculate the vector product 281 00:19:17,930 --> 00:19:25,190 across be. So we want the first one A1I 282 00:19:25,190 --> 00:19:28,990 Plus A2J plus a 3K. 283 00:19:28,990 --> 00:19:36,150 I'm going to cross it with B1IB2J and 284 00:19:36,150 --> 00:19:39,786 B3K. Now this is a little bit laborious and it's going to take 285 00:19:39,786 --> 00:19:43,206 a little bit of time to develop it at the end of the day, will 286 00:19:43,206 --> 00:19:45,486 end up with a formula which is relatively simple for 287 00:19:45,486 --> 00:19:49,144 calculating the vector product. So we start to remove these 288 00:19:49,144 --> 00:19:53,038 brackets in the way that we've learned about previously. We use 289 00:19:53,038 --> 00:19:57,286 the distributive law to say that the first component here a one 290 00:19:57,286 --> 00:20:02,810 I. And then the 2nd and then the third distributes themselves 291 00:20:02,810 --> 00:20:08,725 over the second vector here. So will get a one. I crossed with 292 00:20:08,725 --> 00:20:11,000 all of this second vector. 293 00:20:11,030 --> 00:20:17,820 The One I plus B2J plus B3K. 294 00:20:18,540 --> 00:20:21,864 Then added to the second vector 295 00:20:21,864 --> 00:20:25,390 here. A2J Crossed with. 296 00:20:26,040 --> 00:20:27,380 This whole vector here. 297 00:20:27,910 --> 00:20:35,260 The One I plus B2J plus 298 00:20:35,260 --> 00:20:41,904 B3K. And finally, the third one here, a 3K. 299 00:20:41,910 --> 00:20:48,438 Crossed with the whole of this vector B1 300 00:20:48,438 --> 00:20:51,702 I add B2J ad 301 00:20:51,702 --> 00:20:56,380 D3K. So at that stage we've used the distributive law wants to 302 00:20:56,380 --> 00:20:57,780 expand this first bracket. 303 00:20:58,830 --> 00:21:03,081 We can use it again because now we've got a vector here crossed 304 00:21:03,081 --> 00:21:06,351 with three vectors added together and we can use the 305 00:21:06,351 --> 00:21:09,621 distributive law to distribute this. A one I cross product 306 00:21:09,621 --> 00:21:11,583 across each of these three terms 307 00:21:11,583 --> 00:21:16,561 here. Then again, to distribute this across these three terms 308 00:21:16,561 --> 00:21:21,709 and to distribute this across these three terms. So if we do 309 00:21:21,709 --> 00:21:24,712 that, we'll get a one. I crossed 310 00:21:24,712 --> 00:21:31,397 with B1I. Added to a one I cross B2J. 311 00:21:32,630 --> 00:21:36,326 Do you want I cross B2J? 312 00:21:37,310 --> 00:21:40,490 Added to a one I. 313 00:21:40,490 --> 00:21:44,030 Crossed with B3K. 314 00:21:44,030 --> 00:21:47,306 So that's taken care of removing the brackets from 315 00:21:47,306 --> 00:21:48,762 this first term here. 316 00:21:50,010 --> 00:21:55,600 Let's move to the second term. We've got a 2 J crossed with 317 00:21:55,600 --> 00:22:02,380 B1I. A2 J crossed with 318 00:22:02,380 --> 00:22:09,638 B2J. And a two J 319 00:22:09,638 --> 00:22:13,364 crossed with B3K. 320 00:22:13,370 --> 00:22:19,859 And that's taking care of this term. 321 00:22:21,100 --> 00:22:24,250 And finally we use the distributive law once more to 322 00:22:24,250 --> 00:22:28,345 remove the brackets over. Here will get a 3K cross be one I. 323 00:22:29,350 --> 00:22:35,606 Plus a 3K cross 324 00:22:35,606 --> 00:22:38,734 be 2 325 00:22:38,734 --> 00:22:46,055 J. And finally, a 3K cross 326 00:22:46,055 --> 00:22:51,650 be 3K. So we get all these nine terms. In fact, here 327 00:22:51,650 --> 00:22:56,405 now it's not as bad as it looks, because some of this is going to 328 00:22:56,405 --> 00:22:59,892 cancel out, and in particular one of the things you may 329 00:22:59,892 --> 00:23:03,379 remember we said was that if you have two parallel vectors. 330 00:23:04,000 --> 00:23:05,860 Their vector product is 0. 331 00:23:06,650 --> 00:23:12,149 Now these two vectors A1I and B1 I a parallel because both of 332 00:23:12,149 --> 00:23:16,802 them are pointing in the direction of Vector I. So these 333 00:23:16,802 --> 00:23:22,301 are these are parallel and so the vector product is 0, so that 334 00:23:22,301 --> 00:23:23,570 will become zero. 335 00:23:24,550 --> 00:23:29,980 For the same reason, the A2 J Crosby to Jay is 0 because A2 J 336 00:23:29,980 --> 00:23:32,152 is parallel to be 2 J. 337 00:23:33,480 --> 00:23:40,080 And a 3K Crosby 3K is 0 because a 3K and B3K apparel vectors so 338 00:23:40,080 --> 00:23:43,786 they disappear. So that's reduced it to six terms. 339 00:23:44,550 --> 00:23:50,190 Now, what about this term here? Let's look at a one I cross be 2 340 00:23:50,190 --> 00:23:54,570 J. If you work out the vector product of these, we've got a 341 00:23:54,570 --> 00:23:57,990 vector in the direction of. I crossed with a vector in the 342 00:23:57,990 --> 00:24:01,695 direction of Jay, and we've seen already that if you have an I 343 00:24:01,695 --> 00:24:05,685 cross OJ the result is K. So when we workout a one across be 344 00:24:05,685 --> 00:24:09,675 2, J will write down the length of the first, the length of the 345 00:24:09,675 --> 00:24:14,892 2nd. And then the direction is going to be such that it's at 346 00:24:14,892 --> 00:24:19,585 right angles to I and TJ in a sense defined by the right 347 00:24:19,585 --> 00:24:23,556 hand screw rule. And that sense is K. So when we 348 00:24:23,556 --> 00:24:27,166 simplify this first term here, it'll just simplify to A1B2K. 349 00:24:29,460 --> 00:24:30,920 What about this one here? 350 00:24:31,850 --> 00:24:35,516 This direct this factor here a one is in the direction of I. 351 00:24:36,690 --> 00:24:41,149 This ones in the direction of K and we've already seen that if 352 00:24:41,149 --> 00:24:45,608 we workout I cross K let me remind you of our little diagram 353 00:24:45,608 --> 00:24:50,067 we had. I JK this cycle of vectors here. If we want a 354 00:24:50,067 --> 00:24:53,497 vector in the direction of across with vector in the 355 00:24:53,497 --> 00:24:56,241 direction of K were coming anticlockwise around this 356 00:24:56,241 --> 00:25:00,014 diagram and I Cross K is going to be minus J. 357 00:25:00,600 --> 00:25:05,528 So when we come to workout this term, we want the length of the 358 00:25:05,528 --> 00:25:10,104 first term A1 length of the second one which is B3. But I 359 00:25:10,104 --> 00:25:13,028 Cross K. Is going to be minus J. 360 00:25:13,620 --> 00:25:17,164 So that start with 361 00:25:17,164 --> 00:25:22,110 that. What about this one? Again, length of the first one 362 00:25:22,110 --> 00:25:27,080 is A2. Length of the second one is B1, and a Jake Ross I. 363 00:25:28,440 --> 00:25:32,160 For the diagram again Jake Ross I moving anticlockwise around 364 00:25:32,160 --> 00:25:36,252 this circle, J Cross Eye is going to be minus K. 365 00:25:36,270 --> 00:25:42,000 And also this one will have a 2 J Crosby 3K length of the first 366 00:25:42,000 --> 00:25:47,348 one is A2 length of the second one is B3 and Jake Ross K. 367 00:25:48,110 --> 00:25:51,750 Clockwise now Jake Ross K is I. 368 00:25:52,800 --> 00:25:59,100 We dealt with that one and the last two terms, a 3K cross be 369 00:25:59,100 --> 00:26:05,850 one. I will be a 3B One and K Cross. I came across. I will 370 00:26:05,850 --> 00:26:13,576 be J. And last of all, a 3K cross B2J will be a 371 00:26:13,576 --> 00:26:19,802 3B2K Cross JK cross. Jay going anticlockwise will be minus I. 372 00:26:21,110 --> 00:26:25,543 And these six times if we study them now, you realize that two 373 00:26:25,543 --> 00:26:29,976 of the terms of involved K2 of the terms involved Jay and two 374 00:26:29,976 --> 00:26:34,068 of the terms involved I. So we can collect those like terms 375 00:26:34,068 --> 00:26:37,819 together and in terms of eyes, there will be a 2B3I 376 00:26:37,860 --> 00:26:45,504 And and minus a 3B2I. So just the items will give you 377 00:26:45,504 --> 00:26:47,415 this term here. 378 00:26:49,430 --> 00:26:53,238 Just the Jay terms will be a 379 00:26:53,238 --> 00:26:56,655 3B one. Minus 380 00:26:56,655 --> 00:27:00,410 A1B3? There the Jay terms. 381 00:27:01,280 --> 00:27:04,635 And the Kay terms, there's 382 00:27:04,635 --> 00:27:08,290 an A1B2. Minus an 383 00:27:08,290 --> 00:27:13,006 A2B one. And those are the key terms. 384 00:27:13,600 --> 00:27:17,550 So All in all, with now reduced this complicated calculation 385 00:27:17,550 --> 00:27:22,290 down to one which just gives us a formula for calculating a 386 00:27:22,290 --> 00:27:26,240 cross be in terms of the components of the original 387 00:27:26,240 --> 00:27:30,190 vectors, and that's an important formula that will now illustrate 388 00:27:30,190 --> 00:27:31,770 in the following example. 389 00:27:32,440 --> 00:27:39,425 OK, so to illustrate this formula with an example, let me 390 00:27:39,425 --> 00:27:46,410 write down the formula again across be is given by a 391 00:27:46,410 --> 00:27:50,220 two B 3 - 8 three 392 00:27:50,220 --> 00:27:56,470 B2. Plus a 3B one minus 393 00:27:56,470 --> 00:27:59,830 a one V3J. 394 00:28:00,900 --> 00:28:07,985 Plus A1 B 2 - 8, two B1K. So that's the formula will 395 00:28:07,985 --> 00:28:11,255 use. Let's look at a specific 396 00:28:11,255 --> 00:28:18,556 example. And let's choose A to be the Vector 397 00:28:18,556 --> 00:28:26,176 4I Plus 3J plus 7K. And let's suppose that B 398 00:28:26,176 --> 00:28:33,034 is the vector to I plus 5J plus 4K. 399 00:28:33,720 --> 00:28:39,510 So to use the formula we need to identify A1A2A3B1B2B3 400 00:28:39,510 --> 00:28:42,984 the components, so a one will 401 00:28:42,984 --> 00:28:50,033 be 4. A2 will be 3 and a three will be 7B. One will be 402 00:28:50,033 --> 00:28:56,850 two, B2 will be 5 and be three, will be 4 and all we need to do 403 00:28:56,850 --> 00:29:01,261 now is to take these numbers and substitute them in the 404 00:29:01,261 --> 00:29:05,672 appropriate place in the formula. So will do that and see 405 00:29:05,672 --> 00:29:08,880 what we get across be. We want a 406 00:29:08,880 --> 00:29:12,380 2B3. Which is 3 * 4. 407 00:29:13,060 --> 00:29:16,342 Which is 12 408 00:29:16,342 --> 00:29:23,520 subtract A3B2. Which is 7 * 5, which is 35 and that 409 00:29:23,520 --> 00:29:26,642 will give us the I component of 410 00:29:26,642 --> 00:29:33,640 the answer. Plus a 3B One which is 7 * 411 00:29:33,640 --> 00:29:40,590 2 which is 14. Subtract A1B3 which is 4 * 412 00:29:40,590 --> 00:29:43,370 4 which is 16. 413 00:29:44,510 --> 00:29:48,162 And that will give us the J component of the answer. 414 00:29:48,940 --> 00:29:52,650 And finally. A1B2? 415 00:29:53,470 --> 00:30:00,295 Which is 4 * 5, which is 20. Subtract a 2B one which is 3 416 00:30:00,295 --> 00:30:06,665 * 2, which is 6 and that will give us the key component of 417 00:30:06,665 --> 00:30:07,575 the answer. 418 00:30:08,750 --> 00:30:14,606 So just tidying this up 12. Subtract 35 is minus 23 I. 419 00:30:15,430 --> 00:30:18,800 14 subtract 16 is minus 420 00:30:18,800 --> 00:30:26,104 2 J. And 20 subtract 6 is plus 14K. 421 00:30:26,950 --> 00:30:31,592 That's the result of calculating the vector product of these two 422 00:30:31,592 --> 00:30:36,656 vectors A&B and you'll notice that the answer we get is indeed 423 00:30:36,656 --> 00:30:41,283 another vector. Now that example that we've just seen shows that 424 00:30:41,283 --> 00:30:45,100 it's a little bit cumbersome to try to workout of vector 425 00:30:45,100 --> 00:30:48,223 product, and for those of you familiar with mathematical 426 00:30:48,223 --> 00:30:51,346 objects, called determinants, there's another way which we can 427 00:30:51,346 --> 00:30:54,816 use to calculate the vector product and I'll illustrate it. 428 00:30:54,816 --> 00:30:58,286 Now, if you've never seen a determinant before, it doesn't 429 00:30:58,286 --> 00:31:02,103 really matter because you should still get the Gist of what's 430 00:31:02,103 --> 00:31:05,920 going on here. If we want to calculate a cross be. 431 00:31:06,660 --> 00:31:10,477 We can evaluate this as a determinant, which is an object 432 00:31:10,477 --> 00:31:14,641 with two straight lines down the size like this along the first 433 00:31:14,641 --> 00:31:18,111 line will write the three unit vectors I Jane K. 434 00:31:18,640 --> 00:31:22,710 On the second line, will write the components of the first 435 00:31:22,710 --> 00:31:26,780 vector, the first vector being a as components A1A2 and A3. 436 00:31:27,940 --> 00:31:32,698 And on the last line, the line below will write the cover 3 437 00:31:32,698 --> 00:31:35,992 components of the vector be which is B1B 2B3. 438 00:31:36,770 --> 00:31:39,604 Now, as I say, when you evaluate the determinant, you do it like 439 00:31:39,604 --> 00:31:42,002 this, and if you've never seen it before, it doesn't matter. 440 00:31:42,002 --> 00:31:44,618 Just watch what I do and we'll see how to do it. 441 00:31:45,930 --> 00:31:50,012 Imagine first of all, that we cover up the row and column with 442 00:31:50,012 --> 00:31:51,268 the I in it. 443 00:31:52,560 --> 00:31:55,376 This first entry here corrupt their own column and look at 444 00:31:55,376 --> 00:31:59,380 what's left. We've got four entries left and what we do is 445 00:31:59,380 --> 00:32:02,968 we calculate the product of the entries from the top left to the 446 00:32:02,968 --> 00:32:06,503 bottom right. And subtract the product of the entries from the 447 00:32:06,503 --> 00:32:08,045 top right to the bottom left. 448 00:32:08,780 --> 00:32:11,419 In other words, we workout a 2B3. 449 00:32:12,460 --> 00:32:19,441 Subtract A3B2. That's 82B3, subtract 450 00:32:19,441 --> 00:32:26,828 A3B2. So that Operation A2 B 3 - 8 three B2 is what will 451 00:32:26,828 --> 00:32:30,572 give us the eye components of the vector product. 452 00:32:32,270 --> 00:32:35,469 Then we come to the J component. 453 00:32:36,080 --> 00:32:41,405 And again, we cover up the row and the column with a J in it, 454 00:32:41,405 --> 00:32:45,310 and we do the same thing. We calculate the product A1B3. 455 00:32:46,500 --> 00:32:49,269 Subtract A3B, One. 456 00:32:49,830 --> 00:32:56,664 And that will give us a one B 3 minus a 3B One J. But when we 457 00:32:56,664 --> 00:33:01,086 work a determinant out, the convention is that we change the 458 00:33:01,086 --> 00:33:03,498 sign of this middle term here. 459 00:33:05,600 --> 00:33:09,296 Finally, we moved to the last entry here on the 1st Row. 460 00:33:09,850 --> 00:33:14,166 And we cover up their own column with a K in and do 461 00:33:14,166 --> 00:33:17,154 the same thing again, A1B2 subtract a 2B one. 462 00:33:18,760 --> 00:33:23,258 A1B2 subtract 8 two B1 and that will give us the key component 463 00:33:23,258 --> 00:33:27,064 and this formula is equivalent to the formula that we just 464 00:33:27,064 --> 00:33:30,524 developed earlier on. The only difference is there's a minus 465 00:33:30,524 --> 00:33:35,022 sign here, but when you apply the minus sign to these terms in 466 00:33:35,022 --> 00:33:39,174 here, this will swap them around and you'll get the same formula 467 00:33:39,174 --> 00:33:40,558 as we have before. 468 00:33:41,470 --> 00:33:46,690 So let's repeat the previous example, doing it using these 469 00:33:46,690 --> 00:33:53,476 determinants we had that a was the vector 4I Plus 3J plus 7K, 470 00:33:53,476 --> 00:34:00,784 and B was the vector to I plus 5J Plus 4K, so will find 471 00:34:00,784 --> 00:34:06,004 the vector product. But this time will use the determinant. 472 00:34:06,560 --> 00:34:13,526 Always first row right down the unit vectors IJ&K. 473 00:34:14,490 --> 00:34:18,726 Always in the 2nd row, the three components of the first vector, 474 00:34:18,726 --> 00:34:22,609 which is A the three components being 4, three and Seven. 475 00:34:22,630 --> 00:34:28,119 And the last line, the three components of the second vector. 476 00:34:28,119 --> 00:34:29,616 25 and four. 477 00:34:29,620 --> 00:34:34,760 And then we evaluate this. As I said before, imagine 478 00:34:34,760 --> 00:34:39,900 crossing out the row and the column with the IN. 479 00:34:41,930 --> 00:34:47,780 And calculating the product 3 * 4 - 7 * 5, which is 3, four 480 00:34:47,780 --> 00:34:53,240 12 - 7, five 35 and that will give you the I component of 481 00:34:53,240 --> 00:34:54,020 the answer. 482 00:34:55,350 --> 00:34:57,654 Cross out their own column with the Jays in. 483 00:34:58,990 --> 00:35:06,080 4 * 4 - 7 * 2, which is four 484 00:35:06,080 --> 00:35:09,639 416-7214. That will give you the J component and as always with 485 00:35:09,639 --> 00:35:11,970 determinants, we change the sign of this middle term. 486 00:35:12,720 --> 00:35:16,200 And finally cross out the row and column with a K in. 487 00:35:16,940 --> 00:35:18,540 We want 4 * 5. 488 00:35:19,040 --> 00:35:23,618 Which is 20. Subtract 3 * 2 which is 6. So we've got 20 489 00:35:23,618 --> 00:35:27,542 subtract 6 and that will give you the cake component of the 490 00:35:27,542 --> 00:35:32,120 answer and just to tidy it all up, 12 subtract 35 will give you 491 00:35:32,120 --> 00:35:38,140 minus 23 I. 16 subtract 14 is 2 with the minus sign. There is 492 00:35:38,140 --> 00:35:39,310 minus 2 J. 493 00:35:39,820 --> 00:35:44,045 And 20 subtract 6 is 14K and that's the answer we got before 494 00:35:44,045 --> 00:35:47,620 this method that we've used to expand the determinant is called 495 00:35:47,620 --> 00:35:51,195 expansion along the first row and those of you that know 496 00:35:51,195 --> 00:35:54,445 determinants will find this very straightforward and those of you 497 00:35:54,445 --> 00:35:58,020 that haven't. All you need to know about determinants is what 498 00:35:58,020 --> 00:35:59,320 we've just done here. 499 00:36:00,510 --> 00:36:06,395 We now want to look at some applications of the vector 500 00:36:06,395 --> 00:36:09,825 product. And the first application I want to 501 00:36:09,825 --> 00:36:13,400 introduce you to is how to find a vector which is 502 00:36:13,400 --> 00:36:15,025 perpendicular to two given vectors. 503 00:36:35,040 --> 00:36:40,584 And the easiest way to illustrate this is by using an 504 00:36:40,584 --> 00:36:45,624 example. So let's suppose are two given vectors. Are these 505 00:36:45,624 --> 00:36:50,664 supposing the vector A is I plus 3J minus 2K? 506 00:36:50,690 --> 00:36:57,350 The second given vector B is 5. I minus 3K. 507 00:36:59,110 --> 00:37:04,193 Now you remember when we defined the vector product of the two of 508 00:37:04,193 --> 00:37:08,103 two vectors A&B, the direction of the result was perpendicular 509 00:37:08,103 --> 00:37:09,276 both to a. 510 00:37:09,780 --> 00:37:13,882 And to be, and indeed the plane which contains A and be. So if 511 00:37:13,882 --> 00:37:17,398 we want to find a vector which is perpendicular to these two 512 00:37:17,398 --> 00:37:20,914 given vectors or we have to do is find the vector product 513 00:37:20,914 --> 00:37:28,098 across be. So we do that a cross B and again will use the 514 00:37:28,098 --> 00:37:31,521 determinants, firstrow being the unit vectors IJ&K. 515 00:37:32,520 --> 00:37:37,850 The 2nd row being the components of A the first vector, which are 516 00:37:37,850 --> 00:37:39,900 1, three and minus 2. 517 00:37:39,910 --> 00:37:44,145 And the last drove being the components of the second vector, 518 00:37:44,145 --> 00:37:49,150 which is B and those components are 5 zero because there are no 519 00:37:49,150 --> 00:37:50,305 JS in here. 520 00:37:50,840 --> 00:37:53,120 And minus three from the case. 521 00:37:54,130 --> 00:37:57,730 So let's workout this determinant. So when we work it 522 00:37:57,730 --> 00:38:01,690 out, we cross out the row and a column containing I. 523 00:38:02,410 --> 00:38:06,062 And we calculate the products of these entries that are left 524 00:38:06,062 --> 00:38:09,050 three times, minus three subtract minus two times. Not. 525 00:38:09,050 --> 00:38:13,034 So we want three times minus three, which is minus 9 subtract 526 00:38:13,034 --> 00:38:16,686 minus two times. Not. So we're subtracting zero, and that will 527 00:38:16,686 --> 00:38:19,010 give you the I component of the 528 00:38:19,010 --> 00:38:24,716 result. Then we want to move to the J component, cross out the 529 00:38:24,716 --> 00:38:29,132 role in the column with the Jays in, and again evaluate the 530 00:38:29,132 --> 00:38:33,548 product's one times. Minus three is minus 3, subtract minus 2 * 531 00:38:33,548 --> 00:38:37,964 5, so we're subtracting minus 10, which is the same as adding 532 00:38:37,964 --> 00:38:41,554 10. And you remember we change the sign of the middle term. 533 00:38:43,090 --> 00:38:46,480 Finally, last of all the 534 00:38:46,480 --> 00:38:51,828 K component. Cross out the row in the column with a K in. 535 00:38:52,760 --> 00:38:57,828 And the products that are left are 1 * 0, which is 0. Subtract 536 00:38:57,828 --> 00:39:02,896 3 five 15, so it's 0 subtract 15. And if we tidy it, what 537 00:39:02,896 --> 00:39:05,068 we've got, they'll be minus nine 538 00:39:05,068 --> 00:39:12,428 I. This 10 subtract 3 is 7 so it will be minus Seven 539 00:39:12,428 --> 00:39:16,250 J. Minus 15. 540 00:39:16,250 --> 00:39:23,450 And this vector that we've found here is perpendicular to both A 541 00:39:23,450 --> 00:39:25,250 and to be. 542 00:39:25,800 --> 00:39:27,480 And we've solved the problem. 543 00:39:28,280 --> 00:39:31,540 Sometimes you asked to find a unit vector which is 544 00:39:31,540 --> 00:39:34,800 perpendicular to two given vectors. So if this problem had 545 00:39:34,800 --> 00:39:38,712 been find a unit vector perpendicular to a into B, or we 546 00:39:38,712 --> 00:39:43,276 have to do is calculate a Crosby and then find a unit vector in 547 00:39:43,276 --> 00:39:48,380 this direction. Now to find a unit vector in the direction of 548 00:39:48,380 --> 00:39:53,392 any given vector or we have to do is divide the vector by its 549 00:39:53,392 --> 00:39:57,330 modulus. It's a general result, the unit vector in the direction 550 00:39:57,330 --> 00:40:01,984 of any vector is found by taking the vector and dividing it by 551 00:40:01,984 --> 00:40:06,582 its modulus. So if we want a unit vector in the direction of 552 00:40:06,582 --> 00:40:07,638 a cross be. 553 00:40:07,850 --> 00:40:13,362 All we have to do is divide minus nine. I minus Seven J 554 00:40:13,362 --> 00:40:18,874 minus 15 K by the modulus of that vector and the modulus of 555 00:40:18,874 --> 00:40:23,962 this vector is found by finding the square root of minus 9 556 00:40:23,962 --> 00:40:27,700 squared. Add 2 - 7 squared. 557 00:40:28,230 --> 00:40:33,261 Added 2 - 15 squared and if you do that calculation you'll find 558 00:40:33,261 --> 00:40:37,905 out that this number at the bottom is the square root of 559 00:40:37,905 --> 00:40:43,323 355, so I can write my unit vector in this form one over the 560 00:40:43,323 --> 00:40:44,871 square root of 355. 561 00:40:45,550 --> 00:40:52,690 Minus 9 - 7 J minus 15 K, so that's now a unit vector. 562 00:40:53,480 --> 00:40:57,040 Which is perpendicular to the two given factors. 563 00:40:59,370 --> 00:41:05,970 A second application that I want to introduce you to is a 564 00:41:05,970 --> 00:41:12,020 geometrical one. We can use the vector product to calculate the 565 00:41:12,020 --> 00:41:14,220 area of a parallelogram. 566 00:41:14,880 --> 00:41:22,560 Let's suppose we have 567 00:41:22,560 --> 00:41:26,400 a parallelogram. 568 00:41:29,740 --> 00:41:36,520 And let me denote. 569 00:41:37,570 --> 00:41:40,860 A vector along one of the sides 570 00:41:40,860 --> 00:41:44,825 as be. And along this side here 571 00:41:44,825 --> 00:41:49,959 as see. And this angle in the parallelogram here is theater. 572 00:41:50,590 --> 00:41:54,922 Now this is a parallelogram so that sides parallel to this side 573 00:41:54,922 --> 00:41:58,893 and this sides parallel to that side. The area of a 574 00:41:58,893 --> 00:42:00,698 parallelogram is the length of 575 00:42:00,698 --> 00:42:03,735 the base. Times the perpendicular height. Let 576 00:42:03,735 --> 00:42:06,080 me put this perpendicular height in here. 577 00:42:08,990 --> 00:42:11,982 So there's a perpendicular and let's call this 578 00:42:11,982 --> 00:42:13,104 perpendicular height age. 579 00:42:14,380 --> 00:42:18,916 Now if we focus our attention on this right angle triangle in 580 00:42:18,916 --> 00:42:23,074 here, we can do a bit of trigonometry in here to 581 00:42:23,074 --> 00:42:24,208 calculate this perpendicular 582 00:42:24,208 --> 00:42:29,620 height H. In particular, if we find the sign of Theta, remember 583 00:42:29,620 --> 00:42:33,899 that the sign is the opposite over the hypotenuse. We can 584 00:42:33,899 --> 00:42:35,844 write down that sign Theta. 585 00:42:35,920 --> 00:42:39,450 Is the opposite side, which is H, the perpendicular height 586 00:42:39,450 --> 00:42:42,980 divided by the hypotenuse. That's the length of this side. 587 00:42:43,490 --> 00:42:48,586 And the length of this side is just the length of this vector, 588 00:42:48,586 --> 00:42:51,330 see, so that's the modulus of C. 589 00:42:52,080 --> 00:42:57,408 If we rearrange this formula, we can get a formula for the 590 00:42:57,408 --> 00:43:02,736 perpendicular height age and we can write it as modulus of C 591 00:43:02,736 --> 00:43:07,470 sign theater. We're now in a position to write down the area 592 00:43:07,470 --> 00:43:10,530 of the parallelogram. The area of the parallelogram is the 593 00:43:10,530 --> 00:43:13,590 length of the base times the perpendicular height and the 594 00:43:13,590 --> 00:43:17,568 length of the base is just the modulus of this vector. Be now 595 00:43:17,568 --> 00:43:19,098 it's just modulus of be. 596 00:43:20,350 --> 00:43:24,490 What's the length of the base when we multiply it by the 597 00:43:24,490 --> 00:43:26,905 perpendicular height, the perpendicular height is the 598 00:43:26,905 --> 00:43:28,630 modulus of C sign theater. 599 00:43:29,350 --> 00:43:30,918 If you look at what we've got 600 00:43:30,918 --> 00:43:35,330 here now. We've got the modulus of a vector, the modulus of 601 00:43:35,330 --> 00:43:39,482 another vector times the sign of the angle in between the two 602 00:43:39,482 --> 00:43:43,634 vectors, and this is just the definition of the modulus of the 603 00:43:43,634 --> 00:43:47,786 vector product be crossy, so that is just the modulus of the 604 00:43:47,786 --> 00:43:52,012 cross, see. So this is an important result. If we ever 605 00:43:52,012 --> 00:43:55,468 want to find the area of a parallelogram and we know that 606 00:43:55,468 --> 00:43:59,212 two of the sides are represented by vector being vector C or we 607 00:43:59,212 --> 00:44:02,956 have to do to find the area of the parallelogram is find the 608 00:44:02,956 --> 00:44:06,124 vector product be crossy and take the modulus of the answer 609 00:44:06,124 --> 00:44:09,004 that we get? That's a very straightforward way of finding 610 00:44:09,004 --> 00:44:10,444 the area of a parallelogram. 611 00:44:11,460 --> 00:44:17,999 The final application I want to look at is to finding the volume 612 00:44:17,999 --> 00:44:22,023 of the parallelepiped now parallelepiped such as that 613 00:44:22,023 --> 00:44:28,562 shown here is A6 faced solid where each of the faces is a 614 00:44:28,562 --> 00:44:32,582 parallelogram. And the opposite faces are identical 615 00:44:32,582 --> 00:44:36,346 parallelograms. Now, if we want the volume of this 616 00:44:36,346 --> 00:44:39,613 parallelepiped, we want to find the area of the base and 617 00:44:39,613 --> 00:44:42,880 multiply it by the perpendicular height. So let's put that in. 618 00:44:43,960 --> 00:44:46,640 Extend the top here. 619 00:44:47,180 --> 00:44:50,210 So that we can see what the perpendicular height is. 620 00:44:53,840 --> 00:44:57,656 And let's call that perpendicular height H. Now in 621 00:44:57,656 --> 00:45:02,320 fact, this perpendicular height is the component of a which is 622 00:45:02,320 --> 00:45:06,136 in the direction which is normal to the base. 623 00:45:06,980 --> 00:45:12,102 Now we've seen that a direction which is normal to the base can 624 00:45:12,102 --> 00:45:15,648 be obtained by finding the vector product be crossy. 625 00:45:15,648 --> 00:45:21,164 Remember when we find be cross C we get a vector which is normal, 626 00:45:21,164 --> 00:45:26,286 so the component of a in the direction which is normal to the 627 00:45:26,286 --> 00:45:31,014 base is given by a dotted with a unit vector in this 628 00:45:31,014 --> 00:45:34,560 perpendicular direction, which is a unit vector in the 629 00:45:34,560 --> 00:45:36,924 direction be across see so this 630 00:45:36,924 --> 00:45:39,210 formula. Will give us this perpendicular height. 631 00:45:40,770 --> 00:45:45,148 Now we want to multiply this perpendicular height by the area 632 00:45:45,148 --> 00:45:49,924 of the base. We've already seen that the base area in the 633 00:45:49,924 --> 00:45:53,904 previous application was just the modulus of be cross, see. 634 00:45:54,800 --> 00:45:59,398 So In other words, the volume of the parallelepiped, let's call 635 00:45:59,398 --> 00:46:05,250 that V is going to be a dot B Cross SI unit vector multiplied 636 00:46:05,250 --> 00:46:08,176 by the modulus of be cross, see. 637 00:46:10,020 --> 00:46:15,259 Now remember when you find the unit vector, one way of doing it 638 00:46:15,259 --> 00:46:19,692 is to find the cross product and divide by the length. 639 00:46:19,700 --> 00:46:23,756 So you'll see when we recognize it in that form, you'll see 640 00:46:23,756 --> 00:46:26,798 there's a bee crossy modulus here, Annabi, Crossy modulus 641 00:46:26,798 --> 00:46:28,488 here, and those will cancel. 642 00:46:29,840 --> 00:46:34,415 And what we're left with is that if we want to find the volume of 643 00:46:34,415 --> 00:46:37,770 the parallelepiped, all we have to do is evaluate the dot 644 00:46:37,770 --> 00:46:42,040 product of A with the vector be cross. See now that may or may 645 00:46:42,040 --> 00:46:45,700 not give you a positive or negative answer. So what we do 646 00:46:45,700 --> 00:46:49,665 normally is say that if we want the volume we want the modulus 647 00:46:49,665 --> 00:46:53,935 of a dotted with be Cross C and that is the formula for the 648 00:46:53,935 --> 00:46:55,155 volume of the parallelepiped. 649 00:46:55,170 --> 00:46:59,620 We look at one final example which illustrates the previous 650 00:46:59,620 --> 00:47:04,515 formula that the volume of the parallelepiped is given by the 651 00:47:04,515 --> 00:47:08,965 modulus of a dotted with the vector product of BNC. 652 00:47:08,970 --> 00:47:16,310 Let's suppose that a is the vector 3I Plus 2J 653 00:47:16,310 --> 00:47:23,524 Plus K. B is the vector, two I plus J 654 00:47:23,524 --> 00:47:30,653 Plus K. And see is the vector I plus 655 00:47:30,653 --> 00:47:32,780 2J plus 4K. 656 00:47:32,780 --> 00:47:36,320 So those three vectors represent the three edges of the 657 00:47:36,320 --> 00:47:40,787 parallelepiped. OK, first of all, to apply this formula we 658 00:47:40,787 --> 00:47:43,771 want to workout the vector product of B&C. 659 00:47:43,820 --> 00:47:50,255 And I'll use the determinants again, be crossy first row. As 660 00:47:50,255 --> 00:47:52,010 always I JK. 661 00:47:52,550 --> 00:47:57,410 The 2nd row we want the three components of the first vector, 662 00:47:57,410 --> 00:47:58,625 which are 211. 663 00:47:58,630 --> 00:48:02,338 And then the last row we want the three components of the 664 00:48:02,338 --> 00:48:04,192 second vector, which are 1, two 665 00:48:04,192 --> 00:48:07,404 and four. And then we proceed to evaluate the determinant 666 00:48:07,404 --> 00:48:11,603 crossing out the row and column with the eyes in will get ones, 667 00:48:11,603 --> 00:48:16,125 four is 4. Subtract ones too is 2, four subtract 2 is 2, and 668 00:48:16,125 --> 00:48:20,324 that will give you the number of eyes in the solution to I. 669 00:48:20,990 --> 00:48:25,792 Then we come to the Jays. We want 248 subtract 1 is one is 670 00:48:25,792 --> 00:48:30,937 one, so it's 8. Subtract 1 is 7 and then we change the signs of 671 00:48:30,937 --> 00:48:35,053 the middle term. So we'll get minus Seven J and finally for 672 00:48:35,053 --> 00:48:41,570 the case two tubes of 4 - 1 is one is one 4 - 1 is 3. So will 673 00:48:41,570 --> 00:48:43,285 end up with plus 3. 674 00:48:44,360 --> 00:48:46,262 So that's the vector product of 675 00:48:46,262 --> 00:48:50,508 BNC. And we now need to find the scalar product the dot product 676 00:48:50,508 --> 00:48:52,461 of A with this result that we've 677 00:48:52,461 --> 00:48:57,270 just obtained. And I'll use the column vector notation to do 678 00:48:57,270 --> 00:49:01,780 this because it's easy to identify the terms we need to 679 00:49:01,780 --> 00:49:06,290 multiply together the vector a as a column vector is 321. 680 00:49:06,310 --> 00:49:11,014 We're going to dot it with be cross, see which we've just 681 00:49:11,014 --> 00:49:13,758 found is 2 - 7 and three. 682 00:49:14,470 --> 00:49:17,680 And you'll remember that to calculate this dot product we 683 00:49:17,680 --> 00:49:19,927 multiply corresponding components together and then add 684 00:49:19,927 --> 00:49:22,174 up the results. So we want 3 685 00:49:22,174 --> 00:49:29,470 twos at 6. Two times minus 7 - 14 and 1. Three is 686 00:49:29,470 --> 00:49:35,590 3. So we've got 9 subtract 14, which is minus 5. 687 00:49:35,590 --> 00:49:40,067 Finally, to find the volume of the parallelepiped, we find the 688 00:49:40,067 --> 00:49:45,765 modulus of this answer, so V will be simply 5, so five is its 689 00:49:45,765 --> 00:49:49,175 volume. And that's an application of both scalar 690 00:49:49,175 --> 00:49:50,950 product and the vector product.