WEBVTT 00:00:01.270 --> 00:00:04.366 In this unit, we're going to have a look at a way 00:00:04.366 --> 00:00:05.398 of combining two vectors. 00:00:06.460 --> 00:00:08.584 This method is called the vector 00:00:08.584 --> 00:00:12.341 product. And it's called the vector product because when we 00:00:12.341 --> 00:00:16.358 combine the two vectors in this way, the result that will get is 00:00:16.358 --> 00:00:20.495 another vector. OK, let's start with the two vectors. 00:00:23.410 --> 00:00:26.994 So suppose we have a vector A. 00:00:27.840 --> 00:00:29.130 And another vector. 00:00:29.860 --> 00:00:37.320 Be. And we have drawn these vectors A&B so that 00:00:37.320 --> 00:00:39.060 their tails coincide. 00:00:39.560 --> 00:00:44.070 And so that we can label the angle between A&B's 00:00:44.070 --> 00:00:45.423 theater like that. 00:00:46.610 --> 00:00:49.718 So we start with two vectors and we're going to do some 00:00:49.718 --> 00:00:51.790 calculations with these two vectors to generate what's 00:00:51.790 --> 00:00:52.826 called the vector product. 00:00:53.570 --> 00:00:58.370 And the vector product is defined to be the length of the 00:00:58.370 --> 00:01:00.370 first vector. That's the length 00:01:00.370 --> 00:01:05.070 of A. Multiplied by the length of the second vector, the length 00:01:05.070 --> 00:01:11.840 of be. Multiplied this time by the sign of the angle between a 00:01:11.840 --> 00:01:15.895 envy. Length of the first one, length of the second one, and 00:01:15.895 --> 00:01:17.785 the sign of the angle between 00:01:17.785 --> 00:01:22.080 the two vectors. Now that's all well and good, but this isn't a 00:01:22.080 --> 00:01:25.460 vector. This is a scalar. This is just a number times a number 00:01:25.460 --> 00:01:29.315 times a number. And if we're going to get a result which is a 00:01:29.315 --> 00:01:30.665 vector we've got to give some 00:01:30.665 --> 00:01:34.995 direction to this. If we look back at the diagram, will notice 00:01:34.995 --> 00:01:38.460 that if we choose any two vectors, these two vectors lie 00:01:38.460 --> 00:01:42.240 in a plane. They form a plane in which they both lie. 00:01:43.340 --> 00:01:47.608 When we calculate the vector product of these two vectors, we 00:01:47.608 --> 00:01:51.488 define the direction of the vector product to be the 00:01:51.488 --> 00:01:55.368 direction which is perpendicular to the plane containing A&B. So 00:01:55.368 --> 00:02:00.800 if we have a vector which is perpendicular to a man to be and 00:02:00.800 --> 00:02:05.456 to enter the plane containing an be, this gives us the direction 00:02:05.456 --> 00:02:07.008 of the vector product. 00:02:07.930 --> 00:02:11.360 And in fact, there are two possible directions which are 00:02:11.360 --> 00:02:14.790 perpendicular to this plane, because In addition to the one 00:02:14.790 --> 00:02:17.877 that I've drawn upwards like this, There's also one. 00:02:18.660 --> 00:02:22.970 That's downwards like that, so we have two possible directions 00:02:22.970 --> 00:02:26.418 which are perpendicular to the plane containing A&B. 00:02:27.170 --> 00:02:30.870 And we have a convention for deciding which direction we 00:02:30.870 --> 00:02:34.312 should choose. Now, this convention is variously called 00:02:34.312 --> 00:02:38.238 the right hand screw rule or the right hand thumb rule, so I'm 00:02:38.238 --> 00:02:40.352 going to show you both of these 00:02:40.352 --> 00:02:45.768 different conventions. First of all, the right hand screw rule. 00:02:46.770 --> 00:02:50.325 If you take a conventional screwdriver and right handed 00:02:50.325 --> 00:02:53.880 screw, and if you imagine turning the screwdriver handle 00:02:53.880 --> 00:02:58.620 so that we turn from the direction of a round to the 00:02:58.620 --> 00:03:02.965 direction of be. So I'm turning my screwdriver In this sense. 00:03:03.840 --> 00:03:05.230 That way around like that. 00:03:06.410 --> 00:03:10.029 Imagine the direction in which the screw would advance. So in 00:03:10.029 --> 00:03:12.990 turning from the sensor data, the sense of B. 00:03:13.880 --> 00:03:15.158 The scroll advance. 00:03:15.720 --> 00:03:21.330 Upwards. So it's this upwards direction here, which we take as 00:03:21.330 --> 00:03:24.957 the direction we require for evaluating this vector product. 00:03:25.850 --> 00:03:31.526 Let me denote a unit vector in this direction by N. 00:03:32.510 --> 00:03:35.513 So if we want to calculate the vector product of amb, this is 00:03:35.513 --> 00:03:36.668 going to be its magnitude. 00:03:37.260 --> 00:03:40.980 And the direction we want is one in the direction of this 00:03:40.980 --> 00:03:45.320 unit vector N. So I put a unit vector N there to give the 00:03:45.320 --> 00:03:46.870 direction to this quantity here. 00:03:48.120 --> 00:03:51.045 There's another way, which sometimes people use to define 00:03:51.045 --> 00:03:53.970 the direction of the vector product, which is equivalent, 00:03:53.970 --> 00:03:57.870 and it's called the right hand thumb rule. Let me explain this 00:03:57.870 --> 00:04:01.770 as well. Imagine you take your fingers of your right hand and 00:04:01.770 --> 00:04:05.995 you curl them in the sense going round from a round towards be. 00:04:05.995 --> 00:04:07.620 So you curl your fingers. 00:04:08.360 --> 00:04:13.028 In the direction from a round to be then the thumb points 00:04:13.028 --> 00:04:16.140 in the required direction of the vector product. 00:04:18.540 --> 00:04:19.810 Yet another way of thinking 00:04:19.810 --> 00:04:24.180 about this. Still with the right hand is to point the first 00:04:24.180 --> 00:04:26.046 finger in the direction of A. 00:04:26.800 --> 00:04:28.455 The middle finger in the 00:04:28.455 --> 00:04:33.684 direction of B. And then the thumb points in the required 00:04:33.684 --> 00:04:35.824 direction of the vector product 00:04:35.824 --> 00:04:40.710 of A&B. So using either the right hand screw rule or the 00:04:40.710 --> 00:04:45.013 right hand thumb rule, we can deduce that when we want to find 00:04:45.013 --> 00:04:48.654 the vector product of these two vectors A&B, the direction that 00:04:48.654 --> 00:04:52.295 we require is the one that's moving upwards on this diagram 00:04:52.295 --> 00:04:53.950 here, not the one moving 00:04:53.950 --> 00:04:58.200 downwards. So the vector product is defined as the length of the 00:04:58.200 --> 00:05:01.560 first times the length of the second times the sign of the 00:05:01.560 --> 00:05:03.240 angle between the two times this 00:05:03.240 --> 00:05:07.970 unit vector N. Which is defined in a sense defined by the right 00:05:07.970 --> 00:05:09.746 hand thumb rule or the right 00:05:09.746 --> 00:05:14.435 hand screw rule. Now we have a notation for the vector product 00:05:14.435 --> 00:05:19.910 and we write the vector product of A&B, like this A and we use a 00:05:19.910 --> 00:05:21.370 time sign or across. 00:05:21.890 --> 00:05:25.640 And so sometimes instead of calling this the vector product, 00:05:25.640 --> 00:05:29.765 we sometimes call this the cross product and throughout the rest 00:05:29.765 --> 00:05:34.265 of this unit, sometimes you'll hear me refer to this either as 00:05:34.265 --> 00:05:37.265 the cross product product, all the vector product. 00:05:37.520 --> 00:05:40.868 And I'll use these two words 00:05:40.868 --> 00:05:44.414 interchangeably. So that's the definition will need. I'll also 00:05:44.414 --> 00:05:48.144 mention that some authors and some lecturers will use a 00:05:48.144 --> 00:05:52.247 different notation again, and you might see a cross be written 00:05:52.247 --> 00:05:55.604 using this wedge symbol like this, and that's equally 00:05:55.604 --> 00:06:00.080 acceptable and often people will use the wedge as well to define 00:06:00.080 --> 00:06:01.199 the vector product. 00:06:02.150 --> 00:06:05.474 It's very important that you put this symbol in either the wedge 00:06:05.474 --> 00:06:08.798 or the cross because you don't want to just write down to 00:06:08.798 --> 00:06:12.122 vectors like that when you might mean the vector product or you 00:06:12.122 --> 00:06:15.169 in fact might mean a scalar product. Or you might mean 00:06:15.169 --> 00:06:18.216 something else, so don't use that sort of notation. Make sure 00:06:18.216 --> 00:06:21.540 you're very explicit about when you want to use a vector product 00:06:21.540 --> 00:06:23.479 by putting the time sign or the 00:06:23.479 --> 00:06:28.252 wedge sign in. Let's look at what happens now. When we 00:06:28.252 --> 00:06:33.699 calculate the vector product of A&B. But we do it in a different 00:06:33.699 --> 00:06:37.051 order, so we look at B cross a. 00:06:38.570 --> 00:06:42.530 Now, as before, we want the modulus of the first factor, 00:06:42.530 --> 00:06:44.690 which is the modulus of be. 00:06:45.280 --> 00:06:49.370 Multiplied by the modulus of the second vector, the modular 00:06:49.370 --> 00:06:54.876 survey. We want the sign of the angle between B&A, which is 00:06:54.876 --> 00:06:57.252 still the sign of angle theater. 00:06:58.020 --> 00:06:59.510 And we want to direction. 00:07:00.580 --> 00:07:04.612 Now again using our right hand screw or right hand thumb rule 00:07:04.612 --> 00:07:09.652 we can try to find the sense we require so that as we turn from 00:07:09.652 --> 00:07:14.020 B round to a, this time from the first round of the second 00:07:14.020 --> 00:07:17.716 vector, reoccuring my fingers in the sense from be round to 00:07:17.716 --> 00:07:20.740 a. You'll see now that the thumb points downwards. 00:07:22.210 --> 00:07:26.513 So this time we want a vector which is a unit vector pointing 00:07:26.513 --> 00:07:27.837 downwards instead of upwards. 00:07:28.920 --> 00:07:33.618 Now unit vector downwards must be minus N hot. 00:07:34.200 --> 00:07:36.167 If the output vector was an hat. 00:07:36.860 --> 00:07:42.514 So this time the required direction of be cross a is 00:07:42.514 --> 00:07:44.056 minus an hat. 00:07:45.570 --> 00:07:50.162 If we bring the minus sign out to the front, you'll see that we 00:07:50.162 --> 00:07:54.426 get modulus of be modulus of a sine theater. An hat with the 00:07:54.426 --> 00:07:58.690 minus sign at the front now, and if you examine this vector with 00:07:58.690 --> 00:08:02.626 the vector we had before, when we calculated a cross, B will 00:08:02.626 --> 00:08:06.234 see that the magnitudes of these two vectors are the same, 00:08:06.234 --> 00:08:10.498 because we've got a modulus of a modulus of being sign theater in 00:08:10.498 --> 00:08:14.720 each. But the directions are now different, because where is the 00:08:14.720 --> 00:08:18.428 direction of this one? Was an hat the direction now is minus 00:08:18.428 --> 00:08:22.146 and hat. And we see that this quantity in here. 00:08:23.580 --> 00:08:25.498 Is the same as we had here. 00:08:26.070 --> 00:08:31.978 So in fact, what we've found is that B Cross A is the negative 00:08:31.978 --> 00:08:34.510 of a crossbite. This is very 00:08:34.510 --> 00:08:39.274 important. When we interchange the order of the two vectors. 00:08:40.100 --> 00:08:44.377 We get a different answer be cross a is in fact the negative 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 00:08:49.980 --> 00:08:56.835 across A. Across be is not equal to be across A. 00:08:57.560 --> 00:09:03.538 And in fact, what is true is that B cross a is the negative 00:09:03.538 --> 00:09:05.246 of a cross be. 00:09:05.350 --> 00:09:11.600 So we say that the vector product is not commutative. 00:09:11.600 --> 00:09:16.501 So what that means in practice is that when you've got to find 00:09:16.501 --> 00:09:20.648 the vector product of two vectors, you must be very clear 00:09:20.648 --> 00:09:25.172 about the order in which you want to carry out the operation. 00:09:26.370 --> 00:09:32.070 Now there's a second property of the vector product. I want to 00:09:32.070 --> 00:09:37.295 explain to you, and it's called the distributivity of the vector 00:09:37.295 --> 00:09:42.995 product over addition. What does that mean? It means that if we 00:09:42.995 --> 00:09:49.645 have a vector A and we want to find the cross product with the 00:09:49.645 --> 00:09:52.970 sum of two vectors B Plus C. 00:09:53.250 --> 00:09:58.294 We can evaluate this or remove the brackets in the way that you 00:09:58.294 --> 00:10:01.786 would normally expect algebraic expressions to be evaluated. In 00:10:01.786 --> 00:10:05.278 other words, the cross product distributes itself over this 00:10:05.278 --> 00:10:09.546 edition. In other words, that means that we workout a crossed 00:10:09.546 --> 00:10:16.760 with B. And then because of this addition here, we add that to a 00:10:16.760 --> 00:10:18.290 crossed with C. 00:10:18.530 --> 00:10:22.996 So this is the distributivity rule which allows you to remove 00:10:22.996 --> 00:10:25.026 brackets in a natural way. 00:10:25.840 --> 00:10:30.096 It's also works through the way around, so if we had B Plus C 00:10:30.096 --> 00:10:33.886 first. And we want to cross that 00:10:33.886 --> 00:10:36.772 with a. We do this in a way you 00:10:36.772 --> 00:10:39.179 would expect. Be cross a. 00:10:39.860 --> 00:10:43.195 Plus 00:10:43.195 --> 00:10:50.078 secrecy. And together those rules 00:10:50.078 --> 00:10:53.626 are called the distributivity 00:10:53.626 --> 00:10:58.576 rules. I will use those in a little bit later, little bit 00:10:58.576 --> 00:11:00.356 later on in this unit. 00:11:02.170 --> 00:11:07.549 Another property I want to explain to you is what happens 00:11:07.549 --> 00:11:12.928 when the two vectors that we're interested in our parallel. So 00:11:12.928 --> 00:11:18.796 let's look at what happens when we want to find the vector 00:11:18.796 --> 00:11:20.752 product of parallel vectors. 00:11:23.040 --> 00:11:29.412 Suppose now then we have a Vector A and another vector B 00:11:29.412 --> 00:11:35.784 where A&B are parallel vectors. Is that parallel and we make the 00:11:35.784 --> 00:11:41.094 tales of these two vectors coincide? Then the angle between 00:11:41.094 --> 00:11:47.466 them will be 0. So when the vectors are parallel theater is 00:11:47.466 --> 00:11:54.175 0. So when we come to work out across VB, the modulus 00:11:54.175 --> 00:11:59.730 of a modulus of B, the sign, this time of 0. 00:12:00.040 --> 00:12:02.238 And the sign of 0 is 0. 00:12:02.900 --> 00:12:06.607 So when the two vectors are parallel, the magnitude of this 00:12:06.607 --> 00:12:11.662 vector turns out to be not. So in fact what we get is the zero 00:12:11.662 --> 00:12:14.592 vector. So for two parallel 00:12:14.592 --> 00:12:18.505 vectors. The vector product is the zero vector. 00:12:19.460 --> 00:12:25.368 We now want to start to look at how we can calculate the vector 00:12:25.368 --> 00:12:30.010 product when the two vectors are given in Cartesian form. Before 00:12:30.010 --> 00:12:35.074 we do that, I'd like to develop some results which will be 00:12:35.074 --> 00:12:40.551 particularly important. In this diagram I've shown a 3 00:12:40.551 --> 00:12:45.798 dimensional coordinate system, so we can see an X 00:12:45.798 --> 00:12:49.296 axis or Y axis and it 00:12:49.296 --> 00:12:54.543 said access. And superimposed on this system I've got a unit 00:12:54.543 --> 00:12:59.541 vector I in the X direction unit vector J in the Y direction and 00:12:59.541 --> 00:13:01.683 unit vector K and the zed 00:13:01.683 --> 00:13:07.046 direction. And what I want to do is explore what happens when we 00:13:07.046 --> 00:13:12.268 cross these unit vectors so we workout I cross J or J Cross K 00:13:12.268 --> 00:13:14.506 etc. And let's see what happens. 00:13:14.520 --> 00:13:17.995 Suppose we want I crossed 00:13:17.995 --> 00:13:23.570 with J. Well, by definition, the vector product of iron Jay will 00:13:23.570 --> 00:13:26.024 be the modulus of the first 00:13:26.024 --> 00:13:31.150 vector. And we want the modulus of this vector I now this is a 00:13:31.150 --> 00:13:32.700 unit vector remember, so its 00:13:32.700 --> 00:13:36.074 modulus is one. Times the modulus of the second vector, 00:13:36.074 --> 00:13:40.862 and again the modulus of J is the modulus of a unit vector. So 00:13:40.862 --> 00:13:41.888 again, it's one. 00:13:42.790 --> 00:13:45.274 Times the sign of the angle 00:13:45.274 --> 00:13:48.600 between. I&J, that's the sign of 00:13:48.600 --> 00:13:54.221 90 degrees. And then we've got to give it a direction, and the 00:13:54.221 --> 00:13:58.793 direction is that defined by the right hand screw rule or right 00:13:58.793 --> 00:14:03.365 hand thumb rule. So as we imagine, turning from Iran to J. 00:14:03.365 --> 00:14:07.556 Imagine curling the fingers around from I to J. The thumb 00:14:07.556 --> 00:14:11.747 points in the upward direction points in the K direction. So 00:14:11.747 --> 00:14:16.700 when we work out across J, the direction of the result will be 00:14:16.700 --> 00:14:19.367 Kay Kay being a unit vector in 00:14:19.367 --> 00:14:22.672 the direction. Which is perpendicular to the plane 00:14:22.672 --> 00:14:26.758 containing I&J. Now, this simplifies a great deal because 00:14:26.758 --> 00:14:32.518 the sign of 90 side of 90 degrees is one. So we've got 1 * 00:14:32.518 --> 00:14:34.822 1 * 1, which is 1K. 00:14:34.840 --> 00:14:41.200 So we've got this important result that I cross. Jay is K. 00:14:41.300 --> 00:14:47.618 Let's look at what happens now when we work out. Jake Ross I. 00:14:48.220 --> 00:14:51.892 When we work out Jake Ross, I were doing the vector product 00:14:51.892 --> 00:14:55.564 in the opposite order to which we did it when we calculated 00:14:55.564 --> 00:14:59.848 across J and we know from what we've just done that we get a 00:14:59.848 --> 00:15:02.602 vector of the same magnitude of the opposite sense, 00:15:02.602 --> 00:15:06.274 opposite direction. So when we work out Jake Ross, I in fact 00:15:06.274 --> 00:15:09.028 will get minus K, which is another important result. 00:15:10.430 --> 00:15:16.094 Similarly, if we work out, say, for example J Cross K, let's 00:15:16.094 --> 00:15:17.982 look at that one. 00:15:18.450 --> 00:15:23.299 Again, you want the modulus of the first one, which is one the 00:15:23.299 --> 00:15:28.148 modulus of the second one, which is one the sign of the angle 00:15:28.148 --> 00:15:33.270 between J&K. Which is the sign of 90 degrees. And again you 00:15:33.270 --> 00:15:37.362 want to direction defined by the right hand screw rule and J 00:15:37.362 --> 00:15:41.454 Cross K being where they are here. If you kill your fingers 00:15:41.454 --> 00:15:45.546 in the sense around from the direction of J round to the 00:15:45.546 --> 00:15:49.638 direction of K and imagine which way your thumb will move, your 00:15:49.638 --> 00:15:51.684 thumb will move in the eye 00:15:51.684 --> 00:15:55.410 direction. So Jake Ross K equals 00:15:55.410 --> 00:16:00.530 I. 1 * 1 * 1 gives you just the eye. 00:16:01.360 --> 00:16:06.530 So this is another important result. J Cross K is I. 00:16:06.530 --> 00:16:11.051 And equivalently, if we reverse, the order will get cake. Ross 00:16:11.051 --> 00:16:15.161 Jay is minus I from the result we had before. 00:16:15.720 --> 00:16:20.400 So we've dealt with I crossing with JJ, crossing with K. What 00:16:20.400 --> 00:16:22.350 about I crossing with K? 00:16:22.980 --> 00:16:27.236 If we workout I Cross K again. Modulus of the first one is one 00:16:27.236 --> 00:16:31.188 modulus of the second one is one and I encounter at 90 degrees. 00:16:31.188 --> 00:16:35.140 So with the sign of 90 degrees and we want a direction again 00:16:35.140 --> 00:16:39.092 and again with the right hand screw rule I cross K will give 00:16:39.092 --> 00:16:43.652 you a direction which is in this time in the sense of minus J we 00:16:43.652 --> 00:16:47.604 just look at that for a minute. If you imagine to curling your 00:16:47.604 --> 00:16:50.644 fingers in the sense of round from my towards K. 00:16:52.410 --> 00:16:58.434 Then the thumb will point in the direction of minus J, so 00:16:58.434 --> 00:17:04.960 we've got I Cross K is minus J, or equivalently K Cross I 00:17:04.960 --> 00:17:05.964 equals J. 00:17:08.240 --> 00:17:10.580 So again important results. 00:17:11.420 --> 00:17:16.110 Now, it's important that you can remember how you calculate 00:17:16.110 --> 00:17:22.207 vector product of the eyes in the Jays. In the case that I'm 00:17:22.207 --> 00:17:27.835 going to suggest a way that you might remember this if you 00:17:27.835 --> 00:17:32.994 write down the IJ&K in a cyclic order like that, and 00:17:32.994 --> 00:17:37.215 imagine moving around this cycle in a clockwise sense. 00:17:38.370 --> 00:17:43.323 And if you want to workout, I cross Jay. The result will be 00:17:43.323 --> 00:17:45.228 the next vector round K. 00:17:45.230 --> 00:17:51.854 If you want to workout J Cross K. 00:17:52.390 --> 00:17:57.370 The result will be the next vector round when we move around 00:17:57.370 --> 00:17:59.030 clockwise, which is I. 00:18:00.780 --> 00:18:04.510 And finally came across I. 00:18:04.620 --> 00:18:06.000 Same argument. 00:18:07.040 --> 00:18:11.192 Next vector round is Jay, so that's that's an easy way of 00:18:11.192 --> 00:18:14.998 remembering the vector products of the eyes and Jason case. And 00:18:14.998 --> 00:18:18.804 clearly when you reverse the order of any of these, you 00:18:18.804 --> 00:18:21.918 introduce a minus sign. Alternatively, if you wanted for 00:18:21.918 --> 00:18:26.416 example K Cross J, you'd realize that to move from kata ju moving 00:18:26.416 --> 00:18:30.914 anticlockwise, so K Cross Jay is minus I, that's the way that I 00:18:30.914 --> 00:18:32.644 find helpful to remember these 00:18:32.644 --> 00:18:38.960 results. We're now in a position to calculate the vector product 00:18:38.960 --> 00:18:41.915 of two vectors given in 00:18:41.915 --> 00:18:49.049 Cartesian form. Suppose we start with the Vector A and let's 00:18:49.049 --> 00:18:55.661 suppose this is a general 3 dimensional vector, a one I plus 00:18:55.661 --> 00:19:01.722 A2J plus a 3K where A1A2A3 or any numbers we choose. 00:19:02.580 --> 00:19:09.543 Suppose our second vector B, again arbitrary is be one. I 00:19:09.543 --> 00:19:12.708 be 2 J and B3K. 00:19:12.820 --> 00:19:17.930 And we set about trying to calculate the vector product 00:19:17.930 --> 00:19:25.190 across be. So we want the first one A1I 00:19:25.190 --> 00:19:28.990 Plus A2J plus a 3K. 00:19:28.990 --> 00:19:36.150 I'm going to cross it with B1IB2J and 00:19:36.150 --> 00:19:39.786 B3K. Now this is a little bit laborious and it's going to take 00:19:39.786 --> 00:19:43.206 a little bit of time to develop it at the end of the day, will 00:19:43.206 --> 00:19:45.486 end up with a formula which is relatively simple for 00:19:45.486 --> 00:19:49.144 calculating the vector product. So we start to remove these 00:19:49.144 --> 00:19:53.038 brackets in the way that we've learned about previously. We use 00:19:53.038 --> 00:19:57.286 the distributive law to say that the first component here a one 00:19:57.286 --> 00:20:02.810 I. And then the 2nd and then the third distributes themselves 00:20:02.810 --> 00:20:08.725 over the second vector here. So will get a one. I crossed with 00:20:08.725 --> 00:20:11.000 all of this second vector. 00:20:11.030 --> 00:20:17.820 The One I plus B2J plus B3K. 00:20:18.540 --> 00:20:21.864 Then added to the second vector 00:20:21.864 --> 00:20:25.390 here. A2J Crossed with. 00:20:26.040 --> 00:20:27.380 This whole vector here. 00:20:27.910 --> 00:20:35.260 The One I plus B2J plus 00:20:35.260 --> 00:20:41.904 B3K. And finally, the third one here, a 3K. 00:20:41.910 --> 00:20:48.438 Crossed with the whole of this vector B1 00:20:48.438 --> 00:20:51.702 I add B2J ad 00:20:51.702 --> 00:20:56.380 D3K. So at that stage we've used the distributive law wants to 00:20:56.380 --> 00:20:57.780 expand this first bracket. 00:20:58.830 --> 00:21:03.081 We can use it again because now we've got a vector here crossed 00:21:03.081 --> 00:21:06.351 with three vectors added together and we can use the 00:21:06.351 --> 00:21:09.621 distributive law to distribute this. A one I cross product 00:21:09.621 --> 00:21:11.583 across each of these three terms 00:21:11.583 --> 00:21:16.561 here. Then again, to distribute this across these three terms 00:21:16.561 --> 00:21:21.709 and to distribute this across these three terms. So if we do 00:21:21.709 --> 00:21:24.712 that, we'll get a one. I crossed 00:21:24.712 --> 00:21:31.397 with B1I. Added to a one I cross B2J. 00:21:32.630 --> 00:21:36.326 Do you want I cross B2J? 00:21:37.310 --> 00:21:40.490 Added to a one I. 00:21:40.490 --> 00:21:44.030 Crossed with B3K. 00:21:44.030 --> 00:21:47.306 So that's taken care of removing the brackets from 00:21:47.306 --> 00:21:48.762 this first term here. 00:21:50.010 --> 00:21:55.600 Let's move to the second term. We've got a 2 J crossed with 00:21:55.600 --> 00:22:02.380 B1I. A2 J crossed with 00:22:02.380 --> 00:22:09.638 B2J. And a two J 00:22:09.638 --> 00:22:13.364 crossed with B3K. 00:22:13.370 --> 00:22:19.859 And that's taking care of this term. 00:22:21.100 --> 00:22:24.250 And finally we use the distributive law once more to 00:22:24.250 --> 00:22:28.345 remove the brackets over. Here will get a 3K cross be one I. 00:22:29.350 --> 00:22:35.606 Plus a 3K cross 00:22:35.606 --> 00:22:38.734 be 2 00:22:38.734 --> 00:22:46.055 J. And finally, a 3K cross 00:22:46.055 --> 00:22:51.650 be 3K. So we get all these nine terms. In fact, here 00:22:51.650 --> 00:22:56.405 now it's not as bad as it looks, because some of this is going to 00:22:56.405 --> 00:22:59.892 cancel out, and in particular one of the things you may 00:22:59.892 --> 00:23:03.379 remember we said was that if you have two parallel vectors. 00:23:04.000 --> 00:23:05.860 Their vector product is 0. 00:23:06.650 --> 00:23:12.149 Now these two vectors A1I and B1 I a parallel because both of 00:23:12.149 --> 00:23:16.802 them are pointing in the direction of Vector I. So these 00:23:16.802 --> 00:23:22.301 are these are parallel and so the vector product is 0, so that 00:23:22.301 --> 00:23:23.570 will become zero. 00:23:24.550 --> 00:23:29.980 For the same reason, the A2 J Crosby to Jay is 0 because A2 J 00:23:29.980 --> 00:23:32.152 is parallel to be 2 J. 00:23:33.480 --> 00:23:40.080 And a 3K Crosby 3K is 0 because a 3K and B3K apparel vectors so 00:23:40.080 --> 00:23:43.786 they disappear. So that's reduced it to six terms. 00:23:44.550 --> 00:23:50.190 Now, what about this term here? Let's look at a one I cross be 2 00:23:50.190 --> 00:23:54.570 J. If you work out the vector product of these, we've got a 00:23:54.570 --> 00:23:57.990 vector in the direction of. I crossed with a vector in the 00:23:57.990 --> 00:24:01.695 direction of Jay, and we've seen already that if you have an I 00:24:01.695 --> 00:24:05.685 cross OJ the result is K. So when we workout a one across be 00:24:05.685 --> 00:24:09.675 2, J will write down the length of the first, the length of the 00:24:09.675 --> 00:24:14.892 2nd. And then the direction is going to be such that it's at 00:24:14.892 --> 00:24:19.585 right angles to I and TJ in a sense defined by the right 00:24:19.585 --> 00:24:23.556 hand screw rule. And that sense is K. So when we 00:24:23.556 --> 00:24:27.166 simplify this first term here, it'll just simplify to A1B2K. 00:24:29.460 --> 00:24:30.920 What about this one here? 00:24:31.850 --> 00:24:35.516 This direct this factor here a one is in the direction of I. 00:24:36.690 --> 00:24:41.149 This ones in the direction of K and we've already seen that if 00:24:41.149 --> 00:24:45.608 we workout I cross K let me remind you of our little diagram 00:24:45.608 --> 00:24:50.067 we had. I JK this cycle of vectors here. If we want a 00:24:50.067 --> 00:24:53.497 vector in the direction of across with vector in the 00:24:53.497 --> 00:24:56.241 direction of K were coming anticlockwise around this 00:24:56.241 --> 00:25:00.014 diagram and I Cross K is going to be minus J. 00:25:00.600 --> 00:25:05.528 So when we come to workout this term, we want the length of the 00:25:05.528 --> 00:25:10.104 first term A1 length of the second one which is B3. But I 00:25:10.104 --> 00:25:13.028 Cross K. Is going to be minus J. 00:25:13.620 --> 00:25:17.164 So that start with 00:25:17.164 --> 00:25:22.110 that. What about this one? Again, length of the first one 00:25:22.110 --> 00:25:27.080 is A2. Length of the second one is B1, and a Jake Ross I. 00:25:28.440 --> 00:25:32.160 For the diagram again Jake Ross I moving anticlockwise around 00:25:32.160 --> 00:25:36.252 this circle, J Cross Eye is going to be minus K. 00:25:36.270 --> 00:25:42.000 And also this one will have a 2 J Crosby 3K length of the first 00:25:42.000 --> 00:25:47.348 one is A2 length of the second one is B3 and Jake Ross K. 00:25:48.110 --> 00:25:51.750 Clockwise now Jake Ross K is I. 00:25:52.800 --> 00:25:59.100 We dealt with that one and the last two terms, a 3K cross be 00:25:59.100 --> 00:26:05.850 one. I will be a 3B One and K Cross. I came across. I will 00:26:05.850 --> 00:26:13.576 be J. And last of all, a 3K cross B2J will be a 00:26:13.576 --> 00:26:19.802 3B2K Cross JK cross. Jay going anticlockwise will be minus I. 00:26:21.110 --> 00:26:25.543 And these six times if we study them now, you realize that two 00:26:25.543 --> 00:26:29.976 of the terms of involved K2 of the terms involved Jay and two 00:26:29.976 --> 00:26:34.068 of the terms involved I. So we can collect those like terms 00:26:34.068 --> 00:26:37.819 together and in terms of eyes, there will be a 2B3I 00:26:37.860 --> 00:26:45.504 And and minus a 3B2I. So just the items will give you 00:26:45.504 --> 00:26:47.415 this term here. 00:26:49.430 --> 00:26:53.238 Just the Jay terms will be a 00:26:53.238 --> 00:26:56.655 3B one. Minus 00:26:56.655 --> 00:27:00.410 A1B3? There the Jay terms. 00:27:01.280 --> 00:27:04.635 And the Kay terms, there's 00:27:04.635 --> 00:27:08.290 an A1B2. Minus an 00:27:08.290 --> 00:27:13.006 A2B one. And those are the key terms. 00:27:13.600 --> 00:27:17.550 So All in all, with now reduced this complicated calculation 00:27:17.550 --> 00:27:22.290 down to one which just gives us a formula for calculating a 00:27:22.290 --> 00:27:26.240 cross be in terms of the components of the original 00:27:26.240 --> 00:27:30.190 vectors, and that's an important formula that will now illustrate 00:27:30.190 --> 00:27:31.770 in the following example. 00:27:32.440 --> 00:27:39.425 OK, so to illustrate this formula with an example, let me 00:27:39.425 --> 00:27:46.410 write down the formula again across be is given by a 00:27:46.410 --> 00:27:50.220 two B 3 - 8 three 00:27:50.220 --> 00:27:56.470 B2. Plus a 3B one minus 00:27:56.470 --> 00:27:59.830 a one V3J. 00:28:00.900 --> 00:28:07.985 Plus A1 B 2 - 8, two B1K. So that's the formula will 00:28:07.985 --> 00:28:11.255 use. Let's look at a specific 00:28:11.255 --> 00:28:18.556 example. And let's choose A to be the Vector 00:28:18.556 --> 00:28:26.176 4I Plus 3J plus 7K. And let's suppose that B 00:28:26.176 --> 00:28:33.034 is the vector to I plus 5J plus 4K. 00:28:33.720 --> 00:28:39.510 So to use the formula we need to identify A1A2A3B1B2B3 00:28:39.510 --> 00:28:42.984 the components, so a one will 00:28:42.984 --> 00:28:50.033 be 4. A2 will be 3 and a three will be 7B. One will be 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 00:28:56.850 --> 00:29:01.261 now is to take these numbers and substitute them in the 00:29:01.261 --> 00:29:05.672 appropriate place in the formula. So will do that and see 00:29:05.672 --> 00:29:08.880 what we get across be. We want a 00:29:08.880 --> 00:29:12.380 2B3. Which is 3 * 4. 00:29:13.060 --> 00:29:16.342 Which is 12 00:29:16.342 --> 00:29:23.520 subtract A3B2. Which is 7 * 5, which is 35 and that 00:29:23.520 --> 00:29:26.642 will give us the I component of 00:29:26.642 --> 00:29:33.640 the answer. Plus a 3B One which is 7 * 00:29:33.640 --> 00:29:40.590 2 which is 14. Subtract A1B3 which is 4 * 00:29:40.590 --> 00:29:43.370 4 which is 16. 00:29:44.510 --> 00:29:48.162 And that will give us the J component of the answer. 00:29:48.940 --> 00:29:52.650 And finally. A1B2? 00:29:53.470 --> 00:30:00.295 Which is 4 * 5, which is 20. Subtract a 2B one which is 3 00:30:00.295 --> 00:30:06.665 * 2, which is 6 and that will give us the key component of 00:30:06.665 --> 00:30:07.575 the answer. 00:30:08.750 --> 00:30:14.606 So just tidying this up 12. Subtract 35 is minus 23 I. 00:30:15.430 --> 00:30:18.800 14 subtract 16 is minus 00:30:18.800 --> 00:30:26.104 2 J. And 20 subtract 6 is plus 14K. 00:30:26.950 --> 00:30:31.592 That's the result of calculating the vector product of these two 00:30:31.592 --> 00:30:36.656 vectors A&B and you'll notice that the answer we get is indeed 00:30:36.656 --> 00:30:41.283 another vector. Now that example that we've just seen shows that 00:30:41.283 --> 00:30:45.100 it's a little bit cumbersome to try to workout of vector 00:30:45.100 --> 00:30:48.223 product, and for those of you familiar with mathematical 00:30:48.223 --> 00:30:51.346 objects, called determinants, there's another way which we can 00:30:51.346 --> 00:30:54.816 use to calculate the vector product and I'll illustrate it. 00:30:54.816 --> 00:30:58.286 Now, if you've never seen a determinant before, it doesn't 00:30:58.286 --> 00:31:02.103 really matter because you should still get the Gist of what's 00:31:02.103 --> 00:31:05.920 going on here. If we want to calculate a cross be. 00:31:06.660 --> 00:31:10.477 We can evaluate this as a determinant, which is an object 00:31:10.477 --> 00:31:14.641 with two straight lines down the size like this along the first 00:31:14.641 --> 00:31:18.111 line will write the three unit vectors I Jane K. 00:31:18.640 --> 00:31:22.710 On the second line, will write the components of the first 00:31:22.710 --> 00:31:26.780 vector, the first vector being a as components A1A2 and A3. 00:31:27.940 --> 00:31:32.698 And on the last line, the line below will write the cover 3 00:31:32.698 --> 00:31:35.992 components of the vector be which is B1B 2B3. 00:31:36.770 --> 00:31:39.604 Now, as I say, when you evaluate the determinant, you do it like 00:31:39.604 --> 00:31:42.002 this, and if you've never seen it before, it doesn't matter. 00:31:42.002 --> 00:31:44.618 Just watch what I do and we'll see how to do it. 00:31:45.930 --> 00:31:50.012 Imagine first of all, that we cover up the row and column with 00:31:50.012 --> 00:31:51.268 the I in it. 00:31:52.560 --> 00:31:55.376 This first entry here corrupt their own column and look at 00:31:55.376 --> 00:31:59.380 what's left. We've got four entries left and what we do is 00:31:59.380 --> 00:32:02.968 we calculate the product of the entries from the top left to the 00:32:02.968 --> 00:32:06.503 bottom right. And subtract the product of the entries from the 00:32:06.503 --> 00:32:08.045 top right to the bottom left. 00:32:08.780 --> 00:32:11.419 In other words, we workout a 2B3. 00:32:12.460 --> 00:32:19.441 Subtract A3B2. That's 82B3, subtract 00:32:19.441 --> 00:32:26.828 A3B2. So that Operation A2 B 3 - 8 three B2 is what will 00:32:26.828 --> 00:32:30.572 give us the eye components of the vector product. 00:32:32.270 --> 00:32:35.469 Then we come to the J component. 00:32:36.080 --> 00:32:41.405 And again, we cover up the row and the column with a J in it, 00:32:41.405 --> 00:32:45.310 and we do the same thing. We calculate the product A1B3. 00:32:46.500 --> 00:32:49.269 Subtract A3B, One. 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 00:32:56.664 --> 00:33:01.086 work a determinant out, the convention is that we change the 00:33:01.086 --> 00:33:03.498 sign of this middle term here. 00:33:05.600 --> 00:33:09.296 Finally, we moved to the last entry here on the 1st Row. 00:33:09.850 --> 00:33:14.166 And we cover up their own column with a K in and do 00:33:14.166 --> 00:33:17.154 the same thing again, A1B2 subtract a 2B one. 00:33:18.760 --> 00:33:23.258 A1B2 subtract 8 two B1 and that will give us the key component 00:33:23.258 --> 00:33:27.064 and this formula is equivalent to the formula that we just 00:33:27.064 --> 00:33:30.524 developed earlier on. The only difference is there's a minus 00:33:30.524 --> 00:33:35.022 sign here, but when you apply the minus sign to these terms in 00:33:35.022 --> 00:33:39.174 here, this will swap them around and you'll get the same formula 00:33:39.174 --> 00:33:40.558 as we have before. 00:33:41.470 --> 00:33:46.690 So let's repeat the previous example, doing it using these 00:33:46.690 --> 00:33:53.476 determinants we had that a was the vector 4I Plus 3J plus 7K, 00:33:53.476 --> 00:34:00.784 and B was the vector to I plus 5J Plus 4K, so will find 00:34:00.784 --> 00:34:06.004 the vector product. But this time will use the determinant. 00:34:06.560 --> 00:34:13.526 Always first row right down the unit vectors IJ&K. 00:34:14.490 --> 00:34:18.726 Always in the 2nd row, the three components of the first vector, 00:34:18.726 --> 00:34:22.609 which is A the three components being 4, three and Seven. 00:34:22.630 --> 00:34:28.119 And the last line, the three components of the second vector. 00:34:28.119 --> 00:34:29.616 25 and four. 00:34:29.620 --> 00:34:34.760 And then we evaluate this. As I said before, imagine 00:34:34.760 --> 00:34:39.900 crossing out the row and the column with the IN. 00:34:41.930 --> 00:34:47.780 And calculating the product 3 * 4 - 7 * 5, which is 3, four 00:34:47.780 --> 00:34:53.240 12 - 7, five 35 and that will give you the I component of 00:34:53.240 --> 00:34:54.020 the answer. 00:34:55.350 --> 00:34:57.654 Cross out their own column with the Jays in. 00:34:58.990 --> 00:35:06.080 4 * 4 - 7 * 2, which is four 00:35:06.080 --> 00:35:09.639 416-7214. That will give you the J component and as always with 00:35:09.639 --> 00:35:11.970 determinants, we change the sign of this middle term. 00:35:12.720 --> 00:35:16.200 And finally cross out the row and column with a K in. 00:35:16.940 --> 00:35:18.540 We want 4 * 5. 00:35:19.040 --> 00:35:23.618 Which is 20. Subtract 3 * 2 which is 6. So we've got 20 00:35:23.618 --> 00:35:27.542 subtract 6 and that will give you the cake component of the 00:35:27.542 --> 00:35:32.120 answer and just to tidy it all up, 12 subtract 35 will give you 00:35:32.120 --> 00:35:38.140 minus 23 I. 16 subtract 14 is 2 with the minus sign. There is 00:35:38.140 --> 00:35:39.310 minus 2 J. 00:35:39.820 --> 00:35:44.045 And 20 subtract 6 is 14K and that's the answer we got before 00:35:44.045 --> 00:35:47.620 this method that we've used to expand the determinant is called 00:35:47.620 --> 00:35:51.195 expansion along the first row and those of you that know 00:35:51.195 --> 00:35:54.445 determinants will find this very straightforward and those of you 00:35:54.445 --> 00:35:58.020 that haven't. All you need to know about determinants is what 00:35:58.020 --> 00:35:59.320 we've just done here. 00:36:00.510 --> 00:36:06.395 We now want to look at some applications of the vector 00:36:06.395 --> 00:36:09.825 product. And the first application I want to 00:36:09.825 --> 00:36:13.400 introduce you to is how to find a vector which is 00:36:13.400 --> 00:36:15.025 perpendicular to two given vectors. 00:36:35.040 --> 00:36:40.584 And the easiest way to illustrate this is by using an 00:36:40.584 --> 00:36:45.624 example. So let's suppose are two given vectors. Are these 00:36:45.624 --> 00:36:50.664 supposing the vector A is I plus 3J minus 2K? 00:36:50.690 --> 00:36:57.350 The second given vector B is 5. I minus 3K. 00:36:59.110 --> 00:37:04.193 Now you remember when we defined the vector product of the two of 00:37:04.193 --> 00:37:08.103 two vectors A&B, the direction of the result was perpendicular 00:37:08.103 --> 00:37:09.276 both to a. 00:37:09.780 --> 00:37:13.882 And to be, and indeed the plane which contains A and be. So if 00:37:13.882 --> 00:37:17.398 we want to find a vector which is perpendicular to these two 00:37:17.398 --> 00:37:20.914 given vectors or we have to do is find the vector product 00:37:20.914 --> 00:37:28.098 across be. So we do that a cross B and again will use the 00:37:28.098 --> 00:37:31.521 determinants, firstrow being the unit vectors IJ&K. 00:37:32.520 --> 00:37:37.850 The 2nd row being the components of A the first vector, which are 00:37:37.850 --> 00:37:39.900 1, three and minus 2. 00:37:39.910 --> 00:37:44.145 And the last drove being the components of the second vector, 00:37:44.145 --> 00:37:49.150 which is B and those components are 5 zero because there are no 00:37:49.150 --> 00:37:50.305 JS in here. 00:37:50.840 --> 00:37:53.120 And minus three from the case. 00:37:54.130 --> 00:37:57.730 So let's workout this determinant. So when we work it 00:37:57.730 --> 00:38:01.690 out, we cross out the row and a column containing I. 00:38:02.410 --> 00:38:06.062 And we calculate the products of these entries that are left 00:38:06.062 --> 00:38:09.050 three times, minus three subtract minus two times. Not. 00:38:09.050 --> 00:38:13.034 So we want three times minus three, which is minus 9 subtract 00:38:13.034 --> 00:38:16.686 minus two times. Not. So we're subtracting zero, and that will 00:38:16.686 --> 00:38:19.010 give you the I component of the 00:38:19.010 --> 00:38:24.716 result. Then we want to move to the J component, cross out the 00:38:24.716 --> 00:38:29.132 role in the column with the Jays in, and again evaluate the 00:38:29.132 --> 00:38:33.548 product's one times. Minus three is minus 3, subtract minus 2 * 00:38:33.548 --> 00:38:37.964 5, so we're subtracting minus 10, which is the same as adding 00:38:37.964 --> 00:38:41.554 10. And you remember we change the sign of the middle term. 00:38:43.090 --> 00:38:46.480 Finally, last of all the 00:38:46.480 --> 00:38:51.828 K component. Cross out the row in the column with a K in. 00:38:52.760 --> 00:38:57.828 And the products that are left are 1 * 0, which is 0. Subtract 00:38:57.828 --> 00:39:02.896 3 five 15, so it's 0 subtract 15. And if we tidy it, what 00:39:02.896 --> 00:39:05.068 we've got, they'll be minus nine 00:39:05.068 --> 00:39:12.428 I. This 10 subtract 3 is 7 so it will be minus Seven 00:39:12.428 --> 00:39:16.250 J. Minus 15. 00:39:16.250 --> 00:39:23.450 And this vector that we've found here is perpendicular to both A 00:39:23.450 --> 00:39:25.250 and to be. 00:39:25.800 --> 00:39:27.480 And we've solved the problem. 00:39:28.280 --> 00:39:31.540 Sometimes you asked to find a unit vector which is 00:39:31.540 --> 00:39:34.800 perpendicular to two given vectors. So if this problem had 00:39:34.800 --> 00:39:38.712 been find a unit vector perpendicular to a into B, or we 00:39:38.712 --> 00:39:43.276 have to do is calculate a Crosby and then find a unit vector in 00:39:43.276 --> 00:39:48.380 this direction. Now to find a unit vector in the direction of 00:39:48.380 --> 00:39:53.392 any given vector or we have to do is divide the vector by its 00:39:53.392 --> 00:39:57.330 modulus. It's a general result, the unit vector in the direction 00:39:57.330 --> 00:40:01.984 of any vector is found by taking the vector and dividing it by 00:40:01.984 --> 00:40:06.582 its modulus. So if we want a unit vector in the direction of 00:40:06.582 --> 00:40:07.638 a cross be. 00:40:07.850 --> 00:40:13.362 All we have to do is divide minus nine. I minus Seven J 00:40:13.362 --> 00:40:18.874 minus 15 K by the modulus of that vector and the modulus of 00:40:18.874 --> 00:40:23.962 this vector is found by finding the square root of minus 9 00:40:23.962 --> 00:40:27.700 squared. Add 2 - 7 squared. 00:40:28.230 --> 00:40:33.261 Added 2 - 15 squared and if you do that calculation you'll find 00:40:33.261 --> 00:40:37.905 out that this number at the bottom is the square root of 00:40:37.905 --> 00:40:43.323 355, so I can write my unit vector in this form one over the 00:40:43.323 --> 00:40:44.871 square root of 355. 00:40:45.550 --> 00:40:52.690 Minus 9 - 7 J minus 15 K, so that's now a unit vector. 00:40:53.480 --> 00:40:57.040 Which is perpendicular to the two given factors. 00:40:59.370 --> 00:41:05.970 A second application that I want to introduce you to is a 00:41:05.970 --> 00:41:12.020 geometrical one. We can use the vector product to calculate the 00:41:12.020 --> 00:41:14.220 area of a parallelogram. 00:41:14.880 --> 00:41:22.560 Let's suppose we have 00:41:22.560 --> 00:41:26.400 a parallelogram. 00:41:29.740 --> 00:41:36.520 And let me denote. 00:41:37.570 --> 00:41:40.860 A vector along one of the sides 00:41:40.860 --> 00:41:44.825 as be. And along this side here 00:41:44.825 --> 00:41:49.959 as see. And this angle in the parallelogram here is theater. 00:41:50.590 --> 00:41:54.922 Now this is a parallelogram so that sides parallel to this side 00:41:54.922 --> 00:41:58.893 and this sides parallel to that side. The area of a 00:41:58.893 --> 00:42:00.698 parallelogram is the length of 00:42:00.698 --> 00:42:03.735 the base. Times the perpendicular height. Let 00:42:03.735 --> 00:42:06.080 me put this perpendicular height in here. 00:42:08.990 --> 00:42:11.982 So there's a perpendicular and let's call this 00:42:11.982 --> 00:42:13.104 perpendicular height age. 00:42:14.380 --> 00:42:18.916 Now if we focus our attention on this right angle triangle in 00:42:18.916 --> 00:42:23.074 here, we can do a bit of trigonometry in here to 00:42:23.074 --> 00:42:24.208 calculate this perpendicular 00:42:24.208 --> 00:42:29.620 height H. In particular, if we find the sign of Theta, remember 00:42:29.620 --> 00:42:33.899 that the sign is the opposite over the hypotenuse. We can 00:42:33.899 --> 00:42:35.844 write down that sign Theta. 00:42:35.920 --> 00:42:39.450 Is the opposite side, which is H, the perpendicular height 00:42:39.450 --> 00:42:42.980 divided by the hypotenuse. That's the length of this side. 00:42:43.490 --> 00:42:48.586 And the length of this side is just the length of this vector, 00:42:48.586 --> 00:42:51.330 see, so that's the modulus of C. 00:42:52.080 --> 00:42:57.408 If we rearrange this formula, we can get a formula for the 00:42:57.408 --> 00:43:02.736 perpendicular height age and we can write it as modulus of C 00:43:02.736 --> 00:43:07.470 sign theater. We're now in a position to write down the area 00:43:07.470 --> 00:43:10.530 of the parallelogram. The area of the parallelogram is the 00:43:10.530 --> 00:43:13.590 length of the base times the perpendicular height and the 00:43:13.590 --> 00:43:17.568 length of the base is just the modulus of this vector. Be now 00:43:17.568 --> 00:43:19.098 it's just modulus of be. 00:43:20.350 --> 00:43:24.490 What's the length of the base when we multiply it by the 00:43:24.490 --> 00:43:26.905 perpendicular height, the perpendicular height is the 00:43:26.905 --> 00:43:28.630 modulus of C sign theater. 00:43:29.350 --> 00:43:30.918 If you look at what we've got 00:43:30.918 --> 00:43:35.330 here now. We've got the modulus of a vector, the modulus of 00:43:35.330 --> 00:43:39.482 another vector times the sign of the angle in between the two 00:43:39.482 --> 00:43:43.634 vectors, and this is just the definition of the modulus of the 00:43:43.634 --> 00:43:47.786 vector product be crossy, so that is just the modulus of the 00:43:47.786 --> 00:43:52.012 cross, see. So this is an important result. If we ever 00:43:52.012 --> 00:43:55.468 want to find the area of a parallelogram and we know that 00:43:55.468 --> 00:43:59.212 two of the sides are represented by vector being vector C or we 00:43:59.212 --> 00:44:02.956 have to do to find the area of the parallelogram is find the 00:44:02.956 --> 00:44:06.124 vector product be crossy and take the modulus of the answer 00:44:06.124 --> 00:44:09.004 that we get? That's a very straightforward way of finding 00:44:09.004 --> 00:44:10.444 the area of a parallelogram. 00:44:11.460 --> 00:44:17.999 The final application I want to look at is to finding the volume 00:44:17.999 --> 00:44:22.023 of the parallelepiped now parallelepiped such as that 00:44:22.023 --> 00:44:28.562 shown here is A6 faced solid where each of the faces is a 00:44:28.562 --> 00:44:32.582 parallelogram. And the opposite faces are identical 00:44:32.582 --> 00:44:36.346 parallelograms. Now, if we want the volume of this 00:44:36.346 --> 00:44:39.613 parallelepiped, we want to find the area of the base and 00:44:39.613 --> 00:44:42.880 multiply it by the perpendicular height. So let's put that in. 00:44:43.960 --> 00:44:46.640 Extend the top here. 00:44:47.180 --> 00:44:50.210 So that we can see what the perpendicular height is. 00:44:53.840 --> 00:44:57.656 And let's call that perpendicular height H. Now in 00:44:57.656 --> 00:45:02.320 fact, this perpendicular height is the component of a which is 00:45:02.320 --> 00:45:06.136 in the direction which is normal to the base. 00:45:06.980 --> 00:45:12.102 Now we've seen that a direction which is normal to the base can 00:45:12.102 --> 00:45:15.648 be obtained by finding the vector product be crossy. 00:45:15.648 --> 00:45:21.164 Remember when we find be cross C we get a vector which is normal, 00:45:21.164 --> 00:45:26.286 so the component of a in the direction which is normal to the 00:45:26.286 --> 00:45:31.014 base is given by a dotted with a unit vector in this 00:45:31.014 --> 00:45:34.560 perpendicular direction, which is a unit vector in the 00:45:34.560 --> 00:45:36.924 direction be across see so this 00:45:36.924 --> 00:45:39.210 formula. Will give us this perpendicular height. 00:45:40.770 --> 00:45:45.148 Now we want to multiply this perpendicular height by the area 00:45:45.148 --> 00:45:49.924 of the base. We've already seen that the base area in the 00:45:49.924 --> 00:45:53.904 previous application was just the modulus of be cross, see. 00:45:54.800 --> 00:45:59.398 So In other words, the volume of the parallelepiped, let's call 00:45:59.398 --> 00:46:05.250 that V is going to be a dot B Cross SI unit vector multiplied 00:46:05.250 --> 00:46:08.176 by the modulus of be cross, see. 00:46:10.020 --> 00:46:15.259 Now remember when you find the unit vector, one way of doing it 00:46:15.259 --> 00:46:19.692 is to find the cross product and divide by the length. 00:46:19.700 --> 00:46:23.756 So you'll see when we recognize it in that form, you'll see 00:46:23.756 --> 00:46:26.798 there's a bee crossy modulus here, Annabi, Crossy modulus 00:46:26.798 --> 00:46:28.488 here, and those will cancel. 00:46:29.840 --> 00:46:34.415 And what we're left with is that if we want to find the volume of 00:46:34.415 --> 00:46:37.770 the parallelepiped, all we have to do is evaluate the dot 00:46:37.770 --> 00:46:42.040 product of A with the vector be cross. See now that may or may 00:46:42.040 --> 00:46:45.700 not give you a positive or negative answer. So what we do 00:46:45.700 --> 00:46:49.665 normally is say that if we want the volume we want the modulus 00:46:49.665 --> 00:46:53.935 of a dotted with be Cross C and that is the formula for the 00:46:53.935 --> 00:46:55.155 volume of the parallelepiped. 00:46:55.170 --> 00:46:59.620 We look at one final example which illustrates the previous 00:46:59.620 --> 00:47:04.515 formula that the volume of the parallelepiped is given by the 00:47:04.515 --> 00:47:08.965 modulus of a dotted with the vector product of BNC. 00:47:08.970 --> 00:47:16.310 Let's suppose that a is the vector 3I Plus 2J 00:47:16.310 --> 00:47:23.524 Plus K. B is the vector, two I plus J 00:47:23.524 --> 00:47:30.653 Plus K. And see is the vector I plus 00:47:30.653 --> 00:47:32.780 2J plus 4K. 00:47:32.780 --> 00:47:36.320 So those three vectors represent the three edges of the 00:47:36.320 --> 00:47:40.787 parallelepiped. OK, first of all, to apply this formula we 00:47:40.787 --> 00:47:43.771 want to workout the vector product of B&C. 00:47:43.820 --> 00:47:50.255 And I'll use the determinants again, be crossy first row. As 00:47:50.255 --> 00:47:52.010 always I JK. 00:47:52.550 --> 00:47:57.410 The 2nd row we want the three components of the first vector, 00:47:57.410 --> 00:47:58.625 which are 211. 00:47:58.630 --> 00:48:02.338 And then the last row we want the three components of the 00:48:02.338 --> 00:48:04.192 second vector, which are 1, two 00:48:04.192 --> 00:48:07.404 and four. And then we proceed to evaluate the determinant 00:48:07.404 --> 00:48:11.603 crossing out the row and column with the eyes in will get ones, 00:48:11.603 --> 00:48:16.125 four is 4. Subtract ones too is 2, four subtract 2 is 2, and 00:48:16.125 --> 00:48:20.324 that will give you the number of eyes in the solution to I. 00:48:20.990 --> 00:48:25.792 Then we come to the Jays. We want 248 subtract 1 is one is 00:48:25.792 --> 00:48:30.937 one, so it's 8. Subtract 1 is 7 and then we change the signs of 00:48:30.937 --> 00:48:35.053 the middle term. So we'll get minus Seven J and finally for 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 00:48:41.570 --> 00:48:43.285 end up with plus 3. 00:48:44.360 --> 00:48:46.262 So that's the vector product of 00:48:46.262 --> 00:48:50.508 BNC. And we now need to find the scalar product the dot product 00:48:50.508 --> 00:48:52.461 of A with this result that we've 00:48:52.461 --> 00:48:57.270 just obtained. And I'll use the column vector notation to do 00:48:57.270 --> 00:49:01.780 this because it's easy to identify the terms we need to 00:49:01.780 --> 00:49:06.290 multiply together the vector a as a column vector is 321. 00:49:06.310 --> 00:49:11.014 We're going to dot it with be cross, see which we've just 00:49:11.014 --> 00:49:13.758 found is 2 - 7 and three. 00:49:14.470 --> 00:49:17.680 And you'll remember that to calculate this dot product we 00:49:17.680 --> 00:49:19.927 multiply corresponding components together and then add 00:49:19.927 --> 00:49:22.174 up the results. So we want 3 00:49:22.174 --> 00:49:29.470 twos at 6. Two times minus 7 - 14 and 1. Three is 00:49:29.470 --> 00:49:35.590 3. So we've got 9 subtract 14, which is minus 5. 00:49:35.590 --> 00:49:40.067 Finally, to find the volume of the parallelepiped, we find the 00:49:40.067 --> 00:49:45.765 modulus of this answer, so V will be simply 5, so five is its 00:49:45.765 --> 00:49:49.175 volume. And that's an application of both scalar 00:49:49.175 --> 00:49:50.950 product and the vector product.