0:00:01.270,0:00:04.366
In this unit, we're going[br]to have a look at a way

0:00:04.366,0:00:05.398
of combining two vectors.

0:00:06.460,0:00:08.584
This method is called the vector

0:00:08.584,0:00:12.341
product. And it's called the[br]vector product because when we

0:00:12.341,0:00:16.358
combine the two vectors in this[br]way, the result that will get is

0:00:16.358,0:00:20.495
another vector. OK, let's start[br]with the two vectors.

0:00:23.410,0:00:26.994
So suppose we have a vector A.

0:00:27.840,0:00:29.130
And another vector.

0:00:29.860,0:00:37.320
Be. And we have[br]drawn these vectors A&B so that

0:00:37.320,0:00:39.060
their tails coincide.

0:00:39.560,0:00:44.070
And so that we can label[br]the angle between A&B's

0:00:44.070,0:00:45.423
theater like that.

0:00:46.610,0:00:49.718
So we start with two vectors and[br]we're going to do some

0:00:49.718,0:00:51.790
calculations with these two[br]vectors to generate what's

0:00:51.790,0:00:52.826
called the vector product.

0:00:53.570,0:00:58.370
And the vector product is[br]defined to be the length of the

0:00:58.370,0:01:00.370
first vector. That's the length

0:01:00.370,0:01:05.070
of A. Multiplied by the length[br]of the second vector, the length

0:01:05.070,0:01:11.840
of be. Multiplied this time by[br]the sign of the angle between a

0:01:11.840,0:01:15.895
envy. Length of the first one,[br]length of the second one, and

0:01:15.895,0:01:17.785
the sign of the angle between

0:01:17.785,0:01:22.080
the two vectors. Now that's all[br]well and good, but this isn't a

0:01:22.080,0:01:25.460
vector. This is a scalar. This[br]is just a number times a number

0:01:25.460,0:01:29.315
times a number. And if we're[br]going to get a result which is a

0:01:29.315,0:01:30.665
vector we've got to give some

0:01:30.665,0:01:34.995
direction to this. If we look[br]back at the diagram, will notice

0:01:34.995,0:01:38.460
that if we choose any two[br]vectors, these two vectors lie

0:01:38.460,0:01:42.240
in a plane. They form a plane in[br]which they both lie.

0:01:43.340,0:01:47.608
When we calculate the vector[br]product of these two vectors, we

0:01:47.608,0:01:51.488
define the direction of the[br]vector product to be the

0:01:51.488,0:01:55.368
direction which is perpendicular[br]to the plane containing A&B. So

0:01:55.368,0:02:00.800
if we have a vector which is[br]perpendicular to a man to be and

0:02:00.800,0:02:05.456
to enter the plane containing an[br]be, this gives us the direction

0:02:05.456,0:02:07.008
of the vector product.

0:02:07.930,0:02:11.360
And in fact, there are two[br]possible directions which are

0:02:11.360,0:02:14.790
perpendicular to this plane,[br]because In addition to the one

0:02:14.790,0:02:17.877
that I've drawn upwards like[br]this, There's also one.

0:02:18.660,0:02:22.970
That's downwards like that, so[br]we have two possible directions

0:02:22.970,0:02:26.418
which are perpendicular to the[br]plane containing A&B.

0:02:27.170,0:02:30.870
And we have a convention for[br]deciding which direction we

0:02:30.870,0:02:34.312
should choose. Now, this[br]convention is variously called

0:02:34.312,0:02:38.238
the right hand screw rule or the[br]right hand thumb rule, so I'm

0:02:38.238,0:02:40.352
going to show you both of these

0:02:40.352,0:02:45.768
different conventions. First of[br]all, the right hand screw rule.

0:02:46.770,0:02:50.325
If you take a conventional[br]screwdriver and right handed

0:02:50.325,0:02:53.880
screw, and if you imagine[br]turning the screwdriver handle

0:02:53.880,0:02:58.620
so that we turn from the[br]direction of a round to the

0:02:58.620,0:03:02.965
direction of be. So I'm turning[br]my screwdriver In this sense.

0:03:03.840,0:03:05.230
That way around like that.

0:03:06.410,0:03:10.029
Imagine the direction in which[br]the screw would advance. So in

0:03:10.029,0:03:12.990
turning from the sensor data,[br]the sense of B.

0:03:13.880,0:03:15.158
The scroll advance.

0:03:15.720,0:03:21.330
Upwards. So it's this upwards[br]direction here, which we take as

0:03:21.330,0:03:24.957
the direction we require for[br]evaluating this vector product.

0:03:25.850,0:03:31.526
Let me denote a unit vector in[br]this direction by N.

0:03:32.510,0:03:35.513
So if we want to calculate the[br]vector product of amb, this is

0:03:35.513,0:03:36.668
going to be its magnitude.

0:03:37.260,0:03:40.980
And the direction we want is[br]one in the direction of this

0:03:40.980,0:03:45.320
unit vector N. So I put a unit[br]vector N there to give the

0:03:45.320,0:03:46.870
direction to this quantity[br]here.

0:03:48.120,0:03:51.045
There's another way, which[br]sometimes people use to define

0:03:51.045,0:03:53.970
the direction of the vector[br]product, which is equivalent,

0:03:53.970,0:03:57.870
and it's called the right hand[br]thumb rule. Let me explain this

0:03:57.870,0:04:01.770
as well. Imagine you take your[br]fingers of your right hand and

0:04:01.770,0:04:05.995
you curl them in the sense going[br]round from a round towards be.

0:04:05.995,0:04:07.620
So you curl your fingers.

0:04:08.360,0:04:13.028
In the direction from a round[br]to be then the thumb points

0:04:13.028,0:04:16.140
in the required direction of[br]the vector product.

0:04:18.540,0:04:19.810
Yet another way of thinking

0:04:19.810,0:04:24.180
about this. Still with the right[br]hand is to point the first

0:04:24.180,0:04:26.046
finger in the direction of A.

0:04:26.800,0:04:28.455
The middle finger in the

0:04:28.455,0:04:33.684
direction of B. And then the[br]thumb points in the required

0:04:33.684,0:04:35.824
direction of the vector product

0:04:35.824,0:04:40.710
of A&B. So using either the[br]right hand screw rule or the

0:04:40.710,0:04:45.013
right hand thumb rule, we can[br]deduce that when we want to find

0:04:45.013,0:04:48.654
the vector product of these two[br]vectors A&B, the direction that

0:04:48.654,0:04:52.295
we require is the one that's[br]moving upwards on this diagram

0:04:52.295,0:04:53.950
here, not the one moving

0:04:53.950,0:04:58.200
downwards. So the vector product[br]is defined as the length of the

0:04:58.200,0:05:01.560
first times the length of the[br]second times the sign of the

0:05:01.560,0:05:03.240
angle between the two times this

0:05:03.240,0:05:07.970
unit vector N. Which is defined[br]in a sense defined by the right

0:05:07.970,0:05:09.746
hand thumb rule or the right

0:05:09.746,0:05:14.435
hand screw rule. Now we have a[br]notation for the vector product

0:05:14.435,0:05:19.910
and we write the vector product[br]of A&B, like this A and we use a

0:05:19.910,0:05:21.370
time sign or across.

0:05:21.890,0:05:25.640
And so sometimes instead of[br]calling this the vector product,

0:05:25.640,0:05:29.765
we sometimes call this the cross[br]product and throughout the rest

0:05:29.765,0:05:34.265
of this unit, sometimes you'll[br]hear me refer to this either as

0:05:34.265,0:05:37.265
the cross product product, all[br]the vector product.

0:05:37.520,0:05:40.868
And I'll use these two words

0:05:40.868,0:05:44.414
interchangeably. So that's the[br]definition will need. I'll also

0:05:44.414,0:05:48.144
mention that some authors and[br]some lecturers will use a

0:05:48.144,0:05:52.247
different notation again, and[br]you might see a cross be written

0:05:52.247,0:05:55.604
using this wedge symbol like[br]this, and that's equally

0:05:55.604,0:06:00.080
acceptable and often people will[br]use the wedge as well to define

0:06:00.080,0:06:01.199
the vector product.

0:06:02.150,0:06:05.474
It's very important that you put[br]this symbol in either the wedge

0:06:05.474,0:06:08.798
or the cross because you don't[br]want to just write down to

0:06:08.798,0:06:12.122
vectors like that when you might[br]mean the vector product or you

0:06:12.122,0:06:15.169
in fact might mean a scalar[br]product. Or you might mean

0:06:15.169,0:06:18.216
something else, so don't use[br]that sort of notation. Make sure

0:06:18.216,0:06:21.540
you're very explicit about when[br]you want to use a vector product

0:06:21.540,0:06:23.479
by putting the time sign or the

0:06:23.479,0:06:28.252
wedge sign in. Let's look at[br]what happens now. When we

0:06:28.252,0:06:33.699
calculate the vector product of[br]A&B. But we do it in a different

0:06:33.699,0:06:37.051
order, so we look at B cross a.

0:06:38.570,0:06:42.530
Now, as before, we want the[br]modulus of the first factor,

0:06:42.530,0:06:44.690
which is the modulus of be.

0:06:45.280,0:06:49.370
Multiplied by the modulus of the[br]second vector, the modular

0:06:49.370,0:06:54.876
survey. We want the sign of the[br]angle between B&A, which is

0:06:54.876,0:06:57.252
still the sign of angle theater.

0:06:58.020,0:06:59.510
And we want to direction.

0:07:00.580,0:07:04.612
Now again using our right hand[br]screw or right hand thumb rule

0:07:04.612,0:07:09.652
we can try to find the sense we[br]require so that as we turn from

0:07:09.652,0:07:14.020
B round to a, this time from[br]the first round of the second

0:07:14.020,0:07:17.716
vector, reoccuring my fingers[br]in the sense from be round to

0:07:17.716,0:07:20.740
a. You'll see now that the[br]thumb points downwards.

0:07:22.210,0:07:26.513
So this time we want a vector[br]which is a unit vector pointing

0:07:26.513,0:07:27.837
downwards instead of upwards.

0:07:28.920,0:07:33.618
Now unit vector downwards must[br]be minus N hot.

0:07:34.200,0:07:36.167
If the output vector was an hat.

0:07:36.860,0:07:42.514
So this time the required[br]direction of be cross a is

0:07:42.514,0:07:44.056
minus an hat.

0:07:45.570,0:07:50.162
If we bring the minus sign out[br]to the front, you'll see that we

0:07:50.162,0:07:54.426
get modulus of be modulus of a[br]sine theater. An hat with the

0:07:54.426,0:07:58.690
minus sign at the front now, and[br]if you examine this vector with

0:07:58.690,0:08:02.626
the vector we had before, when[br]we calculated a cross, B will

0:08:02.626,0:08:06.234
see that the magnitudes of these[br]two vectors are the same,

0:08:06.234,0:08:10.498
because we've got a modulus of a[br]modulus of being sign theater in

0:08:10.498,0:08:14.720
each. But the directions are now[br]different, because where is the

0:08:14.720,0:08:18.428
direction of this one? Was an[br]hat the direction now is minus

0:08:18.428,0:08:22.146
and hat. And we see that[br]this quantity in here.

0:08:23.580,0:08:25.498
Is the same as we had here.

0:08:26.070,0:08:31.978
So in fact, what we've found is[br]that B Cross A is the negative

0:08:31.978,0:08:34.510
of a crossbite. This is very

0:08:34.510,0:08:39.274
important. When we interchange[br]the order of the two vectors.

0:08:40.100,0:08:44.377
We get a different answer be[br]cross a is in fact the negative

0:08:44.377,0:08:49.980
of across be. So it's not true[br]that a cross B is the same as be

0:08:49.980,0:08:56.835
across A. Across be is not[br]equal to be across A.

0:08:57.560,0:09:03.538
And in fact, what is true is[br]that B cross a is the negative

0:09:03.538,0:09:05.246
of a cross be.

0:09:05.350,0:09:11.600
So we say that the vector[br]product is not commutative.

0:09:11.600,0:09:16.501
So what that means in practice[br]is that when you've got to find

0:09:16.501,0:09:20.648
the vector product of two[br]vectors, you must be very clear

0:09:20.648,0:09:25.172
about the order in which you[br]want to carry out the operation.

0:09:26.370,0:09:32.070
Now there's a second property of[br]the vector product. I want to

0:09:32.070,0:09:37.295
explain to you, and it's called[br]the distributivity of the vector

0:09:37.295,0:09:42.995
product over addition. What does[br]that mean? It means that if we

0:09:42.995,0:09:49.645
have a vector A and we want to[br]find the cross product with the

0:09:49.645,0:09:52.970
sum of two vectors B Plus C.

0:09:53.250,0:09:58.294
We can evaluate this or remove[br]the brackets in the way that you

0:09:58.294,0:10:01.786
would normally expect algebraic[br]expressions to be evaluated. In

0:10:01.786,0:10:05.278
other words, the cross product[br]distributes itself over this

0:10:05.278,0:10:09.546
edition. In other words, that[br]means that we workout a crossed

0:10:09.546,0:10:16.760
with B. And then because of this[br]addition here, we add that to a

0:10:16.760,0:10:18.290
crossed with C.

0:10:18.530,0:10:22.996
So this is the distributivity[br]rule which allows you to remove

0:10:22.996,0:10:25.026
brackets in a natural way.

0:10:25.840,0:10:30.096
It's also works through the way[br]around, so if we had B Plus C

0:10:30.096,0:10:33.886
first. And we want to cross that

0:10:33.886,0:10:36.772
with a. We do this in a way you

0:10:36.772,0:10:39.179
would expect. Be cross a.

0:10:39.860,0:10:43.195
Plus

0:10:43.195,0:10:50.078
secrecy.[br]And together those rules

0:10:50.078,0:10:53.626
are called the distributivity

0:10:53.626,0:10:58.576
rules. I will use those in a[br]little bit later, little bit

0:10:58.576,0:11:00.356
later on in this unit.

0:11:02.170,0:11:07.549
Another property I want to[br]explain to you is what happens

0:11:07.549,0:11:12.928
when the two vectors that we're[br]interested in our parallel. So

0:11:12.928,0:11:18.796
let's look at what happens when[br]we want to find the vector

0:11:18.796,0:11:20.752
product of parallel vectors.

0:11:23.040,0:11:29.412
Suppose now then we have a[br]Vector A and another vector B

0:11:29.412,0:11:35.784
where A&B are parallel vectors.[br]Is that parallel and we make the

0:11:35.784,0:11:41.094
tales of these two vectors[br]coincide? Then the angle between

0:11:41.094,0:11:47.466
them will be 0. So when the[br]vectors are parallel theater is

0:11:47.466,0:11:54.175
0. So when we come to[br]work out across VB, the modulus

0:11:54.175,0:11:59.730
of a modulus of B, the sign,[br]this time of 0.

0:12:00.040,0:12:02.238
And the sign of 0 is 0.

0:12:02.900,0:12:06.607
So when the two vectors are[br]parallel, the magnitude of this

0:12:06.607,0:12:11.662
vector turns out to be not. So[br]in fact what we get is the zero

0:12:11.662,0:12:14.592
vector. So for two parallel

0:12:14.592,0:12:18.505
vectors. The vector product is[br]the zero vector.

0:12:19.460,0:12:25.368
We now want to start to look at[br]how we can calculate the vector

0:12:25.368,0:12:30.010
product when the two vectors are[br]given in Cartesian form. Before

0:12:30.010,0:12:35.074
we do that, I'd like to develop[br]some results which will be

0:12:35.074,0:12:40.551
particularly important. In this[br]diagram I've shown a 3

0:12:40.551,0:12:45.798
dimensional coordinate system,[br]so we can see an X

0:12:45.798,0:12:49.296
axis or Y axis and it

0:12:49.296,0:12:54.543
said access. And superimposed on[br]this system I've got a unit

0:12:54.543,0:12:59.541
vector I in the X direction unit[br]vector J in the Y direction and

0:12:59.541,0:13:01.683
unit vector K and the zed

0:13:01.683,0:13:07.046
direction. And what I want to do[br]is explore what happens when we

0:13:07.046,0:13:12.268
cross these unit vectors so we[br]workout I cross J or J Cross K

0:13:12.268,0:13:14.506
etc. And let's see what happens.

0:13:14.520,0:13:17.995
Suppose we want I crossed

0:13:17.995,0:13:23.570
with J. Well, by definition, the[br]vector product of iron Jay will

0:13:23.570,0:13:26.024
be the modulus of the first

0:13:26.024,0:13:31.150
vector. And we want the modulus[br]of this vector I now this is a

0:13:31.150,0:13:32.700
unit vector remember, so its

0:13:32.700,0:13:36.074
modulus is one. Times the[br]modulus of the second vector,

0:13:36.074,0:13:40.862
and again the modulus of J is[br]the modulus of a unit vector. So

0:13:40.862,0:13:41.888
again, it's one.

0:13:42.790,0:13:45.274
Times the sign of the angle

0:13:45.274,0:13:48.600
between. I&J, that's the sign of

0:13:48.600,0:13:54.221
90 degrees. And then we've got[br]to give it a direction, and the

0:13:54.221,0:13:58.793
direction is that defined by the[br]right hand screw rule or right

0:13:58.793,0:14:03.365
hand thumb rule. So as we[br]imagine, turning from Iran to J.

0:14:03.365,0:14:07.556
Imagine curling the fingers[br]around from I to J. The thumb

0:14:07.556,0:14:11.747
points in the upward direction[br]points in the K direction. So

0:14:11.747,0:14:16.700
when we work out across J, the[br]direction of the result will be

0:14:16.700,0:14:19.367
Kay Kay being a unit vector in

0:14:19.367,0:14:22.672
the direction. Which is[br]perpendicular to the plane

0:14:22.672,0:14:26.758
containing I&J. Now, this[br]simplifies a great deal because

0:14:26.758,0:14:32.518
the sign of 90 side of 90[br]degrees is one. So we've got 1 *

0:14:32.518,0:14:34.822
1 * 1, which is 1K.

0:14:34.840,0:14:41.200
So we've got this important[br]result that I cross. Jay is K.

0:14:41.300,0:14:47.618
Let's look at what happens now[br]when we work out. Jake Ross I.

0:14:48.220,0:14:51.892
When we work out Jake Ross, I[br]were doing the vector product

0:14:51.892,0:14:55.564
in the opposite order to which[br]we did it when we calculated

0:14:55.564,0:14:59.848
across J and we know from what[br]we've just done that we get a

0:14:59.848,0:15:02.602
vector of the same magnitude[br]of the opposite sense,

0:15:02.602,0:15:06.274
opposite direction. So when we[br]work out Jake Ross, I in fact

0:15:06.274,0:15:09.028
will get minus K, which is[br]another important result.

0:15:10.430,0:15:16.094
Similarly, if we work out, say,[br]for example J Cross K, let's

0:15:16.094,0:15:17.982
look at that one.

0:15:18.450,0:15:23.299
Again, you want the modulus of[br]the first one, which is one the

0:15:23.299,0:15:28.148
modulus of the second one, which[br]is one the sign of the angle

0:15:28.148,0:15:33.270
between J&K. Which is the sign[br]of 90 degrees. And again you

0:15:33.270,0:15:37.362
want to direction defined by the[br]right hand screw rule and J

0:15:37.362,0:15:41.454
Cross K being where they are[br]here. If you kill your fingers

0:15:41.454,0:15:45.546
in the sense around from the[br]direction of J round to the

0:15:45.546,0:15:49.638
direction of K and imagine which[br]way your thumb will move, your

0:15:49.638,0:15:51.684
thumb will move in the eye

0:15:51.684,0:15:55.410
direction. So Jake Ross K equals

0:15:55.410,0:16:00.530
I. 1 * 1 * 1 gives you[br]just the eye.

0:16:01.360,0:16:06.530
So this is another important[br]result. J Cross K is I.

0:16:06.530,0:16:11.051
And equivalently, if we reverse,[br]the order will get cake. Ross

0:16:11.051,0:16:15.161
Jay is minus I from the result[br]we had before.

0:16:15.720,0:16:20.400
So we've dealt with I crossing[br]with JJ, crossing with K. What

0:16:20.400,0:16:22.350
about I crossing with K?

0:16:22.980,0:16:27.236
If we workout I Cross K again.[br]Modulus of the first one is one

0:16:27.236,0:16:31.188
modulus of the second one is one[br]and I encounter at 90 degrees.

0:16:31.188,0:16:35.140
So with the sign of 90 degrees[br]and we want a direction again

0:16:35.140,0:16:39.092
and again with the right hand[br]screw rule I cross K will give

0:16:39.092,0:16:43.652
you a direction which is in this[br]time in the sense of minus J we

0:16:43.652,0:16:47.604
just look at that for a minute.[br]If you imagine to curling your

0:16:47.604,0:16:50.644
fingers in the sense of round[br]from my towards K.

0:16:52.410,0:16:58.434
Then the thumb will point in[br]the direction of minus J, so

0:16:58.434,0:17:04.960
we've got I Cross K is minus[br]J, or equivalently K Cross I

0:17:04.960,0:17:05.964
equals J.

0:17:08.240,0:17:10.580
So again important results.

0:17:11.420,0:17:16.110
Now, it's important that you[br]can remember how you calculate

0:17:16.110,0:17:22.207
vector product of the eyes in[br]the Jays. In the case that I'm

0:17:22.207,0:17:27.835
going to suggest a way that[br]you might remember this if you

0:17:27.835,0:17:32.994
write down the IJ&K in a[br]cyclic order like that, and

0:17:32.994,0:17:37.215
imagine moving around this[br]cycle in a clockwise sense.

0:17:38.370,0:17:43.323
And if you want to workout, I[br]cross Jay. The result will be

0:17:43.323,0:17:45.228
the next vector round K.

0:17:45.230,0:17:51.854
If you want to[br]workout J Cross K.

0:17:52.390,0:17:57.370
The result will be the next[br]vector round when we move around

0:17:57.370,0:17:59.030
clockwise, which is I.

0:18:00.780,0:18:04.510
And finally came across I.

0:18:04.620,0:18:06.000
Same argument.

0:18:07.040,0:18:11.192
Next vector round is Jay, so[br]that's that's an easy way of

0:18:11.192,0:18:14.998
remembering the vector products[br]of the eyes and Jason case. And

0:18:14.998,0:18:18.804
clearly when you reverse the[br]order of any of these, you

0:18:18.804,0:18:21.918
introduce a minus sign.[br]Alternatively, if you wanted for

0:18:21.918,0:18:26.416
example K Cross J, you'd realize[br]that to move from kata ju moving

0:18:26.416,0:18:30.914
anticlockwise, so K Cross Jay is[br]minus I, that's the way that I

0:18:30.914,0:18:32.644
find helpful to remember these

0:18:32.644,0:18:38.960
results. We're now in a position[br]to calculate the vector product

0:18:38.960,0:18:41.915
of two vectors given in

0:18:41.915,0:18:49.049
Cartesian form. Suppose we start[br]with the Vector A and let's

0:18:49.049,0:18:55.661
suppose this is a general 3[br]dimensional vector, a one I plus

0:18:55.661,0:19:01.722
A2J plus a 3K where A1A2A3 or[br]any numbers we choose.

0:19:02.580,0:19:09.543
Suppose our second vector B,[br]again arbitrary is be one. I

0:19:09.543,0:19:12.708
be 2 J and B3K.

0:19:12.820,0:19:17.930
And we set about trying to[br]calculate the vector product

0:19:17.930,0:19:25.190
across be. So we[br]want the first one A1I

0:19:25.190,0:19:28.990
Plus A2J plus a 3K.

0:19:28.990,0:19:36.150
I'm going to cross[br]it with B1IB2J and

0:19:36.150,0:19:39.786
B3K. Now this is a little bit[br]laborious and it's going to take

0:19:39.786,0:19:43.206
a little bit of time to develop[br]it at the end of the day, will

0:19:43.206,0:19:45.486
end up with a formula which is[br]relatively simple for

0:19:45.486,0:19:49.144
calculating the vector product.[br]So we start to remove these

0:19:49.144,0:19:53.038
brackets in the way that we've[br]learned about previously. We use

0:19:53.038,0:19:57.286
the distributive law to say that[br]the first component here a one

0:19:57.286,0:20:02.810
I. And then the 2nd and then[br]the third distributes themselves

0:20:02.810,0:20:08.725
over the second vector here. So[br]will get a one. I crossed with

0:20:08.725,0:20:11.000
all of this second vector.

0:20:11.030,0:20:17.820
The One I plus[br]B2J plus B3K.

0:20:18.540,0:20:21.864
Then added to the second vector

0:20:21.864,0:20:25.390
here. A2J Crossed with.

0:20:26.040,0:20:27.380
This whole vector here.

0:20:27.910,0:20:35.260
The One I[br]plus B2J plus

0:20:35.260,0:20:41.904
B3K. And finally, the third[br]one here, a 3K.

0:20:41.910,0:20:48.438
Crossed with the whole[br]of this vector B1

0:20:48.438,0:20:51.702
I add B2J ad

0:20:51.702,0:20:56.380
D3K. So at that stage we've used[br]the distributive law wants to

0:20:56.380,0:20:57.780
expand this first bracket.

0:20:58.830,0:21:03.081
We can use it again because now[br]we've got a vector here crossed

0:21:03.081,0:21:06.351
with three vectors added[br]together and we can use the

0:21:06.351,0:21:09.621
distributive law to distribute[br]this. A one I cross product

0:21:09.621,0:21:11.583
across each of these three terms

0:21:11.583,0:21:16.561
here. Then again, to distribute[br]this across these three terms

0:21:16.561,0:21:21.709
and to distribute this across[br]these three terms. So if we do

0:21:21.709,0:21:24.712
that, we'll get a one. I crossed

0:21:24.712,0:21:31.397
with B1I. Added to a[br]one I cross B2J.

0:21:32.630,0:21:36.326
Do you want I cross B2J?

0:21:37.310,0:21:40.490
Added to a one I.

0:21:40.490,0:21:44.030
Crossed with B3K.

0:21:44.030,0:21:47.306
So that's taken care of[br]removing the brackets from

0:21:47.306,0:21:48.762
this first term here.

0:21:50.010,0:21:55.600
Let's move to the second term.[br]We've got a 2 J crossed with

0:21:55.600,0:22:02.380
B1I. A2 J[br]crossed with

0:22:02.380,0:22:09.638
B2J. And[br]a two J

0:22:09.638,0:22:13.364
crossed with B3K.

0:22:13.370,0:22:19.859
And that's taking care[br]of this term.

0:22:21.100,0:22:24.250
And finally we use the[br]distributive law once more to

0:22:24.250,0:22:28.345
remove the brackets over. Here[br]will get a 3K cross be one I.

0:22:29.350,0:22:35.606
Plus a[br]3K cross

0:22:35.606,0:22:38.734
be 2

0:22:38.734,0:22:46.055
J. And finally,[br]a 3K cross

0:22:46.055,0:22:51.650
be 3K. So we get all[br]these nine terms. In fact, here

0:22:51.650,0:22:56.405
now it's not as bad as it looks,[br]because some of this is going to

0:22:56.405,0:22:59.892
cancel out, and in particular[br]one of the things you may

0:22:59.892,0:23:03.379
remember we said was that if you[br]have two parallel vectors.

0:23:04.000,0:23:05.860
Their vector product is 0.

0:23:06.650,0:23:12.149
Now these two vectors A1I and B1[br]I a parallel because both of

0:23:12.149,0:23:16.802
them are pointing in the[br]direction of Vector I. So these

0:23:16.802,0:23:22.301
are these are parallel and so[br]the vector product is 0, so that

0:23:22.301,0:23:23.570
will become zero.

0:23:24.550,0:23:29.980
For the same reason, the A2 J[br]Crosby to Jay is 0 because A2 J

0:23:29.980,0:23:32.152
is parallel to be 2 J.

0:23:33.480,0:23:40.080
And a 3K Crosby 3K is 0 because[br]a 3K and B3K apparel vectors so

0:23:40.080,0:23:43.786
they disappear. So that's[br]reduced it to six terms.

0:23:44.550,0:23:50.190
Now, what about this term here?[br]Let's look at a one I cross be 2

0:23:50.190,0:23:54.570
J. If you work out the vector[br]product of these, we've got a

0:23:54.570,0:23:57.990
vector in the direction of. I[br]crossed with a vector in the

0:23:57.990,0:24:01.695
direction of Jay, and we've seen[br]already that if you have an I

0:24:01.695,0:24:05.685
cross OJ the result is K. So[br]when we workout a one across be

0:24:05.685,0:24:09.675
2, J will write down the length[br]of the first, the length of the

0:24:09.675,0:24:14.892
2nd. And then the direction is[br]going to be such that it's at

0:24:14.892,0:24:19.585
right angles to I and TJ in a[br]sense defined by the right

0:24:19.585,0:24:23.556
hand screw rule. And that[br]sense is K. So when we

0:24:23.556,0:24:27.166
simplify this first term here,[br]it'll just simplify to A1B2K.

0:24:29.460,0:24:30.920
What about this one here?

0:24:31.850,0:24:35.516
This direct this factor here a[br]one is in the direction of I.

0:24:36.690,0:24:41.149
This ones in the direction of K[br]and we've already seen that if

0:24:41.149,0:24:45.608
we workout I cross K let me[br]remind you of our little diagram

0:24:45.608,0:24:50.067
we had. I JK this cycle of[br]vectors here. If we want a

0:24:50.067,0:24:53.497
vector in the direction of[br]across with vector in the

0:24:53.497,0:24:56.241
direction of K were coming[br]anticlockwise around this

0:24:56.241,0:25:00.014
diagram and I Cross K is going[br]to be minus J.

0:25:00.600,0:25:05.528
So when we come to workout this[br]term, we want the length of the

0:25:05.528,0:25:10.104
first term A1 length of the[br]second one which is B3. But I

0:25:10.104,0:25:13.028
Cross K. Is going to be minus J.

0:25:13.620,0:25:17.164
So that start with

0:25:17.164,0:25:22.110
that. What about this one?[br]Again, length of the first one

0:25:22.110,0:25:27.080
is A2. Length of the second one[br]is B1, and a Jake Ross I.

0:25:28.440,0:25:32.160
For the diagram again Jake Ross[br]I moving anticlockwise around

0:25:32.160,0:25:36.252
this circle, J Cross Eye is[br]going to be minus K.

0:25:36.270,0:25:42.000
And also this one will have a 2[br]J Crosby 3K length of the first

0:25:42.000,0:25:47.348
one is A2 length of the second[br]one is B3 and Jake Ross K.

0:25:48.110,0:25:51.750
Clockwise now Jake Ross K is I.

0:25:52.800,0:25:59.100
We dealt with that one and the[br]last two terms, a 3K cross be

0:25:59.100,0:26:05.850
one. I will be a 3B One and[br]K Cross. I came across. I will

0:26:05.850,0:26:13.576
be J. And last of all,[br]a 3K cross B2J will be a

0:26:13.576,0:26:19.802
3B2K Cross JK cross. Jay going[br]anticlockwise will be minus I.

0:26:21.110,0:26:25.543
And these six times if we study[br]them now, you realize that two

0:26:25.543,0:26:29.976
of the terms of involved K2 of[br]the terms involved Jay and two

0:26:29.976,0:26:34.068
of the terms involved I. So we[br]can collect those like terms

0:26:34.068,0:26:37.819
together and in terms of eyes,[br]there will be a 2B3I

0:26:37.860,0:26:45.504
And and minus a 3B2I. So[br]just the items will give you

0:26:45.504,0:26:47.415
this term here.

0:26:49.430,0:26:53.238
Just the Jay terms will be a

0:26:53.238,0:26:56.655
3B one. Minus

0:26:56.655,0:27:00.410
A1B3? There the Jay terms.

0:27:01.280,0:27:04.635
And the Kay terms, there's

0:27:04.635,0:27:08.290
an A1B2. Minus an

0:27:08.290,0:27:13.006
A2B one. And those are[br]the key terms.

0:27:13.600,0:27:17.550
So All in all, with now reduced[br]this complicated calculation

0:27:17.550,0:27:22.290
down to one which just gives us[br]a formula for calculating a

0:27:22.290,0:27:26.240
cross be in terms of the[br]components of the original

0:27:26.240,0:27:30.190
vectors, and that's an important[br]formula that will now illustrate

0:27:30.190,0:27:31.770
in the following example.

0:27:32.440,0:27:39.425
OK, so to illustrate this[br]formula with an example, let me

0:27:39.425,0:27:46.410
write down the formula again[br]across be is given by a

0:27:46.410,0:27:50.220
two B 3 - 8 three

0:27:50.220,0:27:56.470
B2. Plus a[br]3B one minus

0:27:56.470,0:27:59.830
a one V3J.

0:28:00.900,0:28:07.985
Plus A1 B 2 - 8, two[br]B1K. So that's the formula will

0:28:07.985,0:28:11.255
use. Let's look at a specific

0:28:11.255,0:28:18.556
example. And let's choose[br]A to be the Vector

0:28:18.556,0:28:26.176
4I Plus 3J plus 7K.[br]And let's suppose that B

0:28:26.176,0:28:33.034
is the vector to I[br]plus 5J plus 4K.

0:28:33.720,0:28:39.510
So to use the formula we[br]need to identify A1A2A3B1B2B3

0:28:39.510,0:28:42.984
the components, so a one will

0:28:42.984,0:28:50.033
be 4. A2 will be 3 and[br]a three will be 7B. One will be

0:28:50.033,0:28:56.850
two, B2 will be 5 and be three,[br]will be 4 and all we need to do

0:28:56.850,0:29:01.261
now is to take these numbers and[br]substitute them in the

0:29:01.261,0:29:05.672
appropriate place in the[br]formula. So will do that and see

0:29:05.672,0:29:08.880
what we get across be. We want a

0:29:08.880,0:29:12.380
2B3. Which is 3 * 4.

0:29:13.060,0:29:16.342
Which is 12

0:29:16.342,0:29:23.520
subtract A3B2. Which is[br]7 * 5, which is 35 and that

0:29:23.520,0:29:26.642
will give us the I component of

0:29:26.642,0:29:33.640
the answer. Plus a 3B[br]One which is 7 *

0:29:33.640,0:29:40.590
2 which is 14. Subtract[br]A1B3 which is 4 *

0:29:40.590,0:29:43.370
4 which is 16.

0:29:44.510,0:29:48.162
And that will give us the J[br]component of the answer.

0:29:48.940,0:29:52.650
And finally. A1B2?

0:29:53.470,0:30:00.295
Which is 4 * 5, which is 20.[br]Subtract a 2B one which is 3

0:30:00.295,0:30:06.665
* 2, which is 6 and that will[br]give us the key component of

0:30:06.665,0:30:07.575
the answer.

0:30:08.750,0:30:14.606
So just tidying this up 12.[br]Subtract 35 is minus 23 I.

0:30:15.430,0:30:18.800
14 subtract 16 is minus

0:30:18.800,0:30:26.104
2 J. And 20[br]subtract 6 is plus 14K.

0:30:26.950,0:30:31.592
That's the result of calculating[br]the vector product of these two

0:30:31.592,0:30:36.656
vectors A&B and you'll notice[br]that the answer we get is indeed

0:30:36.656,0:30:41.283
another vector. Now that example[br]that we've just seen shows that

0:30:41.283,0:30:45.100
it's a little bit cumbersome to[br]try to workout of vector

0:30:45.100,0:30:48.223
product, and for those of you[br]familiar with mathematical

0:30:48.223,0:30:51.346
objects, called determinants,[br]there's another way which we can

0:30:51.346,0:30:54.816
use to calculate the vector[br]product and I'll illustrate it.

0:30:54.816,0:30:58.286
Now, if you've never seen a[br]determinant before, it doesn't

0:30:58.286,0:31:02.103
really matter because you should[br]still get the Gist of what's

0:31:02.103,0:31:05.920
going on here. If we want to[br]calculate a cross be.

0:31:06.660,0:31:10.477
We can evaluate this as a[br]determinant, which is an object

0:31:10.477,0:31:14.641
with two straight lines down the[br]size like this along the first

0:31:14.641,0:31:18.111
line will write the three unit[br]vectors I Jane K.

0:31:18.640,0:31:22.710
On the second line, will write[br]the components of the first

0:31:22.710,0:31:26.780
vector, the first vector being a[br]as components A1A2 and A3.

0:31:27.940,0:31:32.698
And on the last line, the line[br]below will write the cover 3

0:31:32.698,0:31:35.992
components of the vector be[br]which is B1B 2B3.

0:31:36.770,0:31:39.604
Now, as I say, when you evaluate[br]the determinant, you do it like

0:31:39.604,0:31:42.002
this, and if you've never seen[br]it before, it doesn't matter.

0:31:42.002,0:31:44.618
Just watch what I do and we'll[br]see how to do it.

0:31:45.930,0:31:50.012
Imagine first of all, that we[br]cover up the row and column with

0:31:50.012,0:31:51.268
the I in it.

0:31:52.560,0:31:55.376
This first entry here corrupt[br]their own column and look at

0:31:55.376,0:31:59.380
what's left. We've got four[br]entries left and what we do is

0:31:59.380,0:32:02.968
we calculate the product of the[br]entries from the top left to the

0:32:02.968,0:32:06.503
bottom right. And subtract the[br]product of the entries from the

0:32:06.503,0:32:08.045
top right to the bottom left.

0:32:08.780,0:32:11.419
In other words,[br]we workout a 2B3.

0:32:12.460,0:32:19.441
Subtract A3B2.[br]That's 82B3, subtract

0:32:19.441,0:32:26.828
A3B2. So that Operation A2[br]B 3 - 8 three B2 is what will

0:32:26.828,0:32:30.572
give us the eye components of[br]the vector product.

0:32:32.270,0:32:35.469
Then we come to the J component.

0:32:36.080,0:32:41.405
And again, we cover up the row[br]and the column with a J in it,

0:32:41.405,0:32:45.310
and we do the same thing. We[br]calculate the product A1B3.

0:32:46.500,0:32:49.269
Subtract A3B, One.

0:32:49.830,0:32:56.664
And that will give us a one B 3[br]minus a 3B One J. But when we

0:32:56.664,0:33:01.086
work a determinant out, the[br]convention is that we change the

0:33:01.086,0:33:03.498
sign of this middle term here.

0:33:05.600,0:33:09.296
Finally, we moved to the last[br]entry here on the 1st Row.

0:33:09.850,0:33:14.166
And we cover up their own[br]column with a K in and do

0:33:14.166,0:33:17.154
the same thing again,[br]A1B2 subtract a 2B one.

0:33:18.760,0:33:23.258
A1B2 subtract 8 two B1 and that[br]will give us the key component

0:33:23.258,0:33:27.064
and this formula is equivalent[br]to the formula that we just

0:33:27.064,0:33:30.524
developed earlier on. The only[br]difference is there's a minus

0:33:30.524,0:33:35.022
sign here, but when you apply[br]the minus sign to these terms in

0:33:35.022,0:33:39.174
here, this will swap them around[br]and you'll get the same formula

0:33:39.174,0:33:40.558
as we have before.

0:33:41.470,0:33:46.690
So let's repeat the previous[br]example, doing it using these

0:33:46.690,0:33:53.476
determinants we had that a was[br]the vector 4I Plus 3J plus 7K,

0:33:53.476,0:34:00.784
and B was the vector to I[br]plus 5J Plus 4K, so will find

0:34:00.784,0:34:06.004
the vector product. But this[br]time will use the determinant.

0:34:06.560,0:34:13.526
Always first row right down[br]the unit vectors IJ&K.

0:34:14.490,0:34:18.726
Always in the 2nd row, the three[br]components of the first vector,

0:34:18.726,0:34:22.609
which is A the three components[br]being 4, three and Seven.

0:34:22.630,0:34:28.119
And the last line, the three[br]components of the second vector.

0:34:28.119,0:34:29.616
25 and four.

0:34:29.620,0:34:34.760
And then we evaluate this.[br]As I said before, imagine

0:34:34.760,0:34:39.900
crossing out the row and[br]the column with the IN.

0:34:41.930,0:34:47.780
And calculating the product 3[br]* 4 - 7 * 5, which is 3, four

0:34:47.780,0:34:53.240
12 - 7, five 35 and that will[br]give you the I component of

0:34:53.240,0:34:54.020
the answer.

0:34:55.350,0:34:57.654
Cross out their own column[br]with the Jays in.

0:34:58.990,0:35:06.080
4 * 4 - 7[br]* 2, which is four

0:35:06.080,0:35:09.639
416-7214. That will give you the[br]J component and as always with

0:35:09.639,0:35:11.970
determinants, we change the sign[br]of this middle term.

0:35:12.720,0:35:16.200
And finally cross out the row[br]and column with a K in.

0:35:16.940,0:35:18.540
We want 4 * 5.

0:35:19.040,0:35:23.618
Which is 20. Subtract 3 * 2[br]which is 6. So we've got 20

0:35:23.618,0:35:27.542
subtract 6 and that will give[br]you the cake component of the

0:35:27.542,0:35:32.120
answer and just to tidy it all[br]up, 12 subtract 35 will give you

0:35:32.120,0:35:38.140
minus 23 I. 16 subtract 14 is 2[br]with the minus sign. There is

0:35:38.140,0:35:39.310
minus 2 J.

0:35:39.820,0:35:44.045
And 20 subtract 6 is 14K and[br]that's the answer we got before

0:35:44.045,0:35:47.620
this method that we've used to[br]expand the determinant is called

0:35:47.620,0:35:51.195
expansion along the first row[br]and those of you that know

0:35:51.195,0:35:54.445
determinants will find this very[br]straightforward and those of you

0:35:54.445,0:35:58.020
that haven't. All you need to[br]know about determinants is what

0:35:58.020,0:35:59.320
we've just done here.

0:36:00.510,0:36:06.395
We now want to look at some[br]applications of the vector

0:36:06.395,0:36:09.825
product. And the first[br]application I want to

0:36:09.825,0:36:13.400
introduce you to is how to[br]find a vector which is

0:36:13.400,0:36:15.025
perpendicular to two given[br]vectors.

0:36:35.040,0:36:40.584
And the easiest way to[br]illustrate this is by using an

0:36:40.584,0:36:45.624
example. So let's suppose are[br]two given vectors. Are these

0:36:45.624,0:36:50.664
supposing the vector A is I plus[br]3J minus 2K?

0:36:50.690,0:36:57.350
The second given vector B is[br]5. I minus 3K.

0:36:59.110,0:37:04.193
Now you remember when we defined[br]the vector product of the two of

0:37:04.193,0:37:08.103
two vectors A&B, the direction[br]of the result was perpendicular

0:37:08.103,0:37:09.276
both to a.

0:37:09.780,0:37:13.882
And to be, and indeed the plane[br]which contains A and be. So if

0:37:13.882,0:37:17.398
we want to find a vector which[br]is perpendicular to these two

0:37:17.398,0:37:20.914
given vectors or we have to do[br]is find the vector product

0:37:20.914,0:37:28.098
across be. So we do that a[br]cross B and again will use the

0:37:28.098,0:37:31.521
determinants, firstrow being the[br]unit vectors IJ&K.

0:37:32.520,0:37:37.850
The 2nd row being the components[br]of A the first vector, which are

0:37:37.850,0:37:39.900
1, three and minus 2.

0:37:39.910,0:37:44.145
And the last drove being the[br]components of the second vector,

0:37:44.145,0:37:49.150
which is B and those components[br]are 5 zero because there are no

0:37:49.150,0:37:50.305
JS in here.

0:37:50.840,0:37:53.120
And minus three from the case.

0:37:54.130,0:37:57.730
So let's workout this[br]determinant. So when we work it

0:37:57.730,0:38:01.690
out, we cross out the row and a[br]column containing I.

0:38:02.410,0:38:06.062
And we calculate the products of[br]these entries that are left

0:38:06.062,0:38:09.050
three times, minus three[br]subtract minus two times. Not.

0:38:09.050,0:38:13.034
So we want three times minus[br]three, which is minus 9 subtract

0:38:13.034,0:38:16.686
minus two times. Not. So we're[br]subtracting zero, and that will

0:38:16.686,0:38:19.010
give you the I component of the

0:38:19.010,0:38:24.716
result. Then we want to move to[br]the J component, cross out the

0:38:24.716,0:38:29.132
role in the column with the Jays[br]in, and again evaluate the

0:38:29.132,0:38:33.548
product's one times. Minus three[br]is minus 3, subtract minus 2 *

0:38:33.548,0:38:37.964
5, so we're subtracting minus[br]10, which is the same as adding

0:38:37.964,0:38:41.554
10. And you remember we change[br]the sign of the middle term.

0:38:43.090,0:38:46.480
Finally, last of all the

0:38:46.480,0:38:51.828
K component. Cross out the row[br]in the column with a K in.

0:38:52.760,0:38:57.828
And the products that are left[br]are 1 * 0, which is 0. Subtract

0:38:57.828,0:39:02.896
3 five 15, so it's 0 subtract[br]15. And if we tidy it, what

0:39:02.896,0:39:05.068
we've got, they'll be minus nine

0:39:05.068,0:39:12.428
I. This 10 subtract 3 is[br]7 so it will be minus Seven

0:39:12.428,0:39:16.250
J. Minus 15.

0:39:16.250,0:39:23.450
And this vector that we've found[br]here is perpendicular to both A

0:39:23.450,0:39:25.250
and to be.

0:39:25.800,0:39:27.480
And we've solved the problem.

0:39:28.280,0:39:31.540
Sometimes you asked to find a[br]unit vector which is

0:39:31.540,0:39:34.800
perpendicular to two given[br]vectors. So if this problem had

0:39:34.800,0:39:38.712
been find a unit vector[br]perpendicular to a into B, or we

0:39:38.712,0:39:43.276
have to do is calculate a Crosby[br]and then find a unit vector in

0:39:43.276,0:39:48.380
this direction. Now to find a[br]unit vector in the direction of

0:39:48.380,0:39:53.392
any given vector or we have to[br]do is divide the vector by its

0:39:53.392,0:39:57.330
modulus. It's a general result,[br]the unit vector in the direction

0:39:57.330,0:40:01.984
of any vector is found by taking[br]the vector and dividing it by

0:40:01.984,0:40:06.582
its modulus. So if we want a[br]unit vector in the direction of

0:40:06.582,0:40:07.638
a cross be.

0:40:07.850,0:40:13.362
All we have to do is divide[br]minus nine. I minus Seven J

0:40:13.362,0:40:18.874
minus 15 K by the modulus of[br]that vector and the modulus of

0:40:18.874,0:40:23.962
this vector is found by finding[br]the square root of minus 9

0:40:23.962,0:40:27.700
squared. Add 2 - 7 squared.

0:40:28.230,0:40:33.261
Added 2 - 15 squared and if you[br]do that calculation you'll find

0:40:33.261,0:40:37.905
out that this number at the[br]bottom is the square root of

0:40:37.905,0:40:43.323
355, so I can write my unit[br]vector in this form one over the

0:40:43.323,0:40:44.871
square root of 355.

0:40:45.550,0:40:52.690
Minus 9 - 7 J minus 15[br]K, so that's now a unit vector.

0:40:53.480,0:40:57.040
Which is perpendicular to the[br]two given factors.

0:40:59.370,0:41:05.970
A second application that I want[br]to introduce you to is a

0:41:05.970,0:41:12.020
geometrical one. We can use the[br]vector product to calculate the

0:41:12.020,0:41:14.220
area of a parallelogram.

0:41:14.880,0:41:22.560
Let's suppose[br]we have

0:41:22.560,0:41:26.400
a parallelogram.

0:41:29.740,0:41:36.520
And let[br]me denote.

0:41:37.570,0:41:40.860
A vector along one of the sides

0:41:40.860,0:41:44.825
as be. And along this side here

0:41:44.825,0:41:49.959
as see. And this angle in the[br]parallelogram here is theater.

0:41:50.590,0:41:54.922
Now this is a parallelogram so[br]that sides parallel to this side

0:41:54.922,0:41:58.893
and this sides parallel to that[br]side. The area of a

0:41:58.893,0:42:00.698
parallelogram is the length of

0:42:00.698,0:42:03.735
the base. Times the[br]perpendicular height. Let

0:42:03.735,0:42:06.080
me put this perpendicular[br]height in here.

0:42:08.990,0:42:11.982
So there's a perpendicular[br]and let's call this

0:42:11.982,0:42:13.104
perpendicular height age.

0:42:14.380,0:42:18.916
Now if we focus our attention on[br]this right angle triangle in

0:42:18.916,0:42:23.074
here, we can do a bit of[br]trigonometry in here to

0:42:23.074,0:42:24.208
calculate this perpendicular

0:42:24.208,0:42:29.620
height H. In particular, if we[br]find the sign of Theta, remember

0:42:29.620,0:42:33.899
that the sign is the opposite[br]over the hypotenuse. We can

0:42:33.899,0:42:35.844
write down that sign Theta.

0:42:35.920,0:42:39.450
Is the opposite side, which is[br]H, the perpendicular height

0:42:39.450,0:42:42.980
divided by the hypotenuse.[br]That's the length of this side.

0:42:43.490,0:42:48.586
And the length of this side is[br]just the length of this vector,

0:42:48.586,0:42:51.330
see, so that's the modulus of C.

0:42:52.080,0:42:57.408
If we rearrange this formula, we[br]can get a formula for the

0:42:57.408,0:43:02.736
perpendicular height age and we[br]can write it as modulus of C

0:43:02.736,0:43:07.470
sign theater. We're now in a[br]position to write down the area

0:43:07.470,0:43:10.530
of the parallelogram. The area[br]of the parallelogram is the

0:43:10.530,0:43:13.590
length of the base times the[br]perpendicular height and the

0:43:13.590,0:43:17.568
length of the base is just the[br]modulus of this vector. Be now

0:43:17.568,0:43:19.098
it's just modulus of be.

0:43:20.350,0:43:24.490
What's the length of the base[br]when we multiply it by the

0:43:24.490,0:43:26.905
perpendicular height, the[br]perpendicular height is the

0:43:26.905,0:43:28.630
modulus of C sign theater.

0:43:29.350,0:43:30.918
If you look at what we've got

0:43:30.918,0:43:35.330
here now. We've got the modulus[br]of a vector, the modulus of

0:43:35.330,0:43:39.482
another vector times the sign of[br]the angle in between the two

0:43:39.482,0:43:43.634
vectors, and this is just the[br]definition of the modulus of the

0:43:43.634,0:43:47.786
vector product be crossy, so[br]that is just the modulus of the

0:43:47.786,0:43:52.012
cross, see. So this is an[br]important result. If we ever

0:43:52.012,0:43:55.468
want to find the area of a[br]parallelogram and we know that

0:43:55.468,0:43:59.212
two of the sides are represented[br]by vector being vector C or we

0:43:59.212,0:44:02.956
have to do to find the area of[br]the parallelogram is find the

0:44:02.956,0:44:06.124
vector product be crossy and[br]take the modulus of the answer

0:44:06.124,0:44:09.004
that we get? That's a very[br]straightforward way of finding

0:44:09.004,0:44:10.444
the area of a parallelogram.

0:44:11.460,0:44:17.999
The final application I want to[br]look at is to finding the volume

0:44:17.999,0:44:22.023
of the parallelepiped now[br]parallelepiped such as that

0:44:22.023,0:44:28.562
shown here is A6 faced solid[br]where each of the faces is a

0:44:28.562,0:44:32.582
parallelogram. And the opposite[br]faces are identical

0:44:32.582,0:44:36.346
parallelograms. Now, if we want[br]the volume of this

0:44:36.346,0:44:39.613
parallelepiped, we want to find[br]the area of the base and

0:44:39.613,0:44:42.880
multiply it by the perpendicular[br]height. So let's put that in.

0:44:43.960,0:44:46.640
Extend the top here.

0:44:47.180,0:44:50.210
So that we can see what the[br]perpendicular height is.

0:44:53.840,0:44:57.656
And let's call that[br]perpendicular height H. Now in

0:44:57.656,0:45:02.320
fact, this perpendicular height[br]is the component of a which is

0:45:02.320,0:45:06.136
in the direction which is normal[br]to the base.

0:45:06.980,0:45:12.102
Now we've seen that a direction[br]which is normal to the base can

0:45:12.102,0:45:15.648
be obtained by finding the[br]vector product be crossy.

0:45:15.648,0:45:21.164
Remember when we find be cross C[br]we get a vector which is normal,

0:45:21.164,0:45:26.286
so the component of a in the[br]direction which is normal to the

0:45:26.286,0:45:31.014
base is given by a dotted with a[br]unit vector in this

0:45:31.014,0:45:34.560
perpendicular direction, which[br]is a unit vector in the

0:45:34.560,0:45:36.924
direction be across see so this

0:45:36.924,0:45:39.210
formula. Will give us this[br]perpendicular height.

0:45:40.770,0:45:45.148
Now we want to multiply this[br]perpendicular height by the area

0:45:45.148,0:45:49.924
of the base. We've already seen[br]that the base area in the

0:45:49.924,0:45:53.904
previous application was just[br]the modulus of be cross, see.

0:45:54.800,0:45:59.398
So In other words, the volume of[br]the parallelepiped, let's call

0:45:59.398,0:46:05.250
that V is going to be a dot B[br]Cross SI unit vector multiplied

0:46:05.250,0:46:08.176
by the modulus of be cross, see.

0:46:10.020,0:46:15.259
Now remember when you find the[br]unit vector, one way of doing it

0:46:15.259,0:46:19.692
is to find the cross product and[br]divide by the length.

0:46:19.700,0:46:23.756
So you'll see when we recognize[br]it in that form, you'll see

0:46:23.756,0:46:26.798
there's a bee crossy modulus[br]here, Annabi, Crossy modulus

0:46:26.798,0:46:28.488
here, and those will cancel.

0:46:29.840,0:46:34.415
And what we're left with is that[br]if we want to find the volume of

0:46:34.415,0:46:37.770
the parallelepiped, all we have[br]to do is evaluate the dot

0:46:37.770,0:46:42.040
product of A with the vector be[br]cross. See now that may or may

0:46:42.040,0:46:45.700
not give you a positive or[br]negative answer. So what we do

0:46:45.700,0:46:49.665
normally is say that if we want[br]the volume we want the modulus

0:46:49.665,0:46:53.935
of a dotted with be Cross C and[br]that is the formula for the

0:46:53.935,0:46:55.155
volume of the parallelepiped.

0:46:55.170,0:46:59.620
We look at one final example[br]which illustrates the previous

0:46:59.620,0:47:04.515
formula that the volume of the[br]parallelepiped is given by the

0:47:04.515,0:47:08.965
modulus of a dotted with the[br]vector product of BNC.

0:47:08.970,0:47:16.310
Let's suppose that a is[br]the vector 3I Plus 2J

0:47:16.310,0:47:23.524
Plus K. B is the[br]vector, two I plus J

0:47:23.524,0:47:30.653
Plus K. And see[br]is the vector I plus

0:47:30.653,0:47:32.780
2J plus 4K.

0:47:32.780,0:47:36.320
So those three vectors represent[br]the three edges of the

0:47:36.320,0:47:40.787
parallelepiped. OK, first of[br]all, to apply this formula we

0:47:40.787,0:47:43.771
want to workout the vector[br]product of B&C.

0:47:43.820,0:47:50.255
And I'll use the determinants[br]again, be crossy first row. As

0:47:50.255,0:47:52.010
always I JK.

0:47:52.550,0:47:57.410
The 2nd row we want the three[br]components of the first vector,

0:47:57.410,0:47:58.625
which are 211.

0:47:58.630,0:48:02.338
And then the last row we want[br]the three components of the

0:48:02.338,0:48:04.192
second vector, which are 1, two

0:48:04.192,0:48:07.404
and four. And then we proceed to[br]evaluate the determinant

0:48:07.404,0:48:11.603
crossing out the row and column[br]with the eyes in will get ones,

0:48:11.603,0:48:16.125
four is 4. Subtract ones too is[br]2, four subtract 2 is 2, and

0:48:16.125,0:48:20.324
that will give you the number of[br]eyes in the solution to I.

0:48:20.990,0:48:25.792
Then we come to the Jays. We[br]want 248 subtract 1 is one is

0:48:25.792,0:48:30.937
one, so it's 8. Subtract 1 is 7[br]and then we change the signs of

0:48:30.937,0:48:35.053
the middle term. So we'll get[br]minus Seven J and finally for

0:48:35.053,0:48:41.570
the case two tubes of 4 - 1 is[br]one is one 4 - 1 is 3. So will

0:48:41.570,0:48:43.285
end up with plus 3.

0:48:44.360,0:48:46.262
So that's the vector product of

0:48:46.262,0:48:50.508
BNC. And we now need to find the[br]scalar product the dot product

0:48:50.508,0:48:52.461
of A with this result that we've

0:48:52.461,0:48:57.270
just obtained. And I'll use the[br]column vector notation to do

0:48:57.270,0:49:01.780
this because it's easy to[br]identify the terms we need to

0:49:01.780,0:49:06.290
multiply together the vector a[br]as a column vector is 321.

0:49:06.310,0:49:11.014
We're going to dot it with be[br]cross, see which we've just

0:49:11.014,0:49:13.758
found is 2 - 7 and three.

0:49:14.470,0:49:17.680
And you'll remember that to[br]calculate this dot product we

0:49:17.680,0:49:19.927
multiply corresponding[br]components together and then add

0:49:19.927,0:49:22.174
up the results. So we want 3

0:49:22.174,0:49:29.470
twos at 6. Two times minus 7[br]- 14 and 1. Three is

0:49:29.470,0:49:35.590
3. So we've got 9 subtract 14,[br]which is minus 5.

0:49:35.590,0:49:40.067
Finally, to find the volume of[br]the parallelepiped, we find the

0:49:40.067,0:49:45.765
modulus of this answer, so V[br]will be simply 5, so five is its

0:49:45.765,0:49:49.175
volume. And that's an[br]application of both scalar

0:49:49.175,0:49:50.950
product and the vector product.