1
00:00:03,510 --> 00:00:06,060
So when we have a vector
product of two vectors,

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A cross B equals C, let's
compute that vector product

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00:00:11,430 --> 00:00:13,330
in different coordinate systems.

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00:00:13,330 --> 00:00:17,190
So let's begin by
choosing two vectors.

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I hat, and J hat.

6
00:00:20,730 --> 00:00:23,020
And notice they're
at a right angle,

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and because there
is a unit vector,

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00:00:25,710 --> 00:00:29,400
the area here is equal to and 1.

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00:00:29,400 --> 00:00:37,020
And I want to define K hat to
be equal to I hat cross J hat.

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00:00:37,020 --> 00:00:41,910
Now, our angle theta
here was 90 degrees.

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00:00:41,910 --> 00:00:45,980
And so, if I use the right
hand rule, then I hat,

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00:00:45,980 --> 00:00:52,440
cross J hat is-- the right
handed unit normal is out

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00:00:52,440 --> 00:00:54,600
of the plane of the figure.

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00:00:54,600 --> 00:00:57,930
And so I would write
K hat like that,

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and notice K hat is
out of plane of figure.

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And because I used
my right hand rule,

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this is what we call a right
handed coordinate system.

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Now, there's something very
nice about this cross product

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definition.

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Notice that there's
a cyclic order.

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IJK is a cyclic order.

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And if you interchange any two.

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For example, JIK,
that's anti-cyclic.

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And the cross products satisfy
this cyclic rule, in that,

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J cross K hat is I hat.

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00:01:54,110 --> 00:02:04,490
And notice JKI, maintains
that cyclic order, and K hat

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00:02:04,490 --> 00:02:08,503
cross I hat is J hat.

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KIJ maintains that cyclic
order, but because of the way

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we defined a cross-product
in general A cross B

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is minus B cross A,
because now you're

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using the opposite direction,
so there's a minus sign.

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00:02:29,390 --> 00:02:33,380
And therefore, any anti-cyclic
permutation of these unit

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00:02:33,380 --> 00:02:39,320
vectors, as an example, K
hat cross J hat, has to be--

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00:02:39,320 --> 00:02:45,620
notice I've-- is minus I
hat-- that's anti-communitive

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00:02:45,620 --> 00:02:47,510
property of the cross-product.

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Similarly, I hat cross
K hat is minus J hat.

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And lastly, J hat cross
I hat is minus K hat.

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And so, in fact, you
only need to know one.

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00:03:05,180 --> 00:03:08,750
And this idea of
cyclic, and anti-cyclic,

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to be able to write
down all of the other 6.

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00:03:13,370 --> 00:03:16,640
So we have one, two,
three, four, five, six.

42
00:03:16,640 --> 00:03:20,360
Now, when you want to
compute the cross products

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00:03:20,360 --> 00:03:22,340
in Cartesian coordinates,
for instance,

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it can be a little bit messy.

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There's going to
be a lot of terms.

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00:03:26,780 --> 00:03:29,960
If I write a vector
A as A X I hat

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00:03:29,960 --> 00:03:36,230
plus A Y J hat plus A Z K hat.

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00:03:36,230 --> 00:03:45,335
And I write a vector B as BX
I hat plus BY J hat plus BZ K

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00:03:45,335 --> 00:03:46,270
hat.

50
00:03:46,270 --> 00:03:53,300
And now I want to compute the
cross-product of these vectors

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00:03:53,300 --> 00:03:55,310
to get the new one.

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00:03:55,310 --> 00:04:00,500
Notice that there's going
to be six terms, because I

53
00:04:00,500 --> 00:04:04,610
have I hat-- well, we actually
should say one more thing.

54
00:04:04,610 --> 00:04:09,860
that I hat cross I hat-- the
angle between these two vectors

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00:04:09,860 --> 00:04:11,500
is zero.

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00:04:11,500 --> 00:04:14,330
There's no perpendicular
projection.

57
00:04:14,330 --> 00:04:22,970
So that's zero, as is J hat
cross J hat, K hat cross K hat.

58
00:04:22,970 --> 00:04:28,970
So of these nine parts, when
we take the cross-product,

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three of them will
be 0, by this rule,

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00:04:32,390 --> 00:04:35,030
and we'll we apply our
cyclic or anti-cyclic rules

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00:04:35,030 --> 00:04:36,840
for the others.

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00:04:36,840 --> 00:04:38,930
And so what we have
here-- let's just do it

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00:04:38,930 --> 00:04:48,750
in-- so C equals A cross B.

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00:04:48,750 --> 00:04:51,850
And now let's just
go one by one.

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00:04:51,850 --> 00:04:54,490
I hat cross I hat.

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00:04:54,490 --> 00:04:56,380
That's 0.

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00:04:56,380 --> 00:04:58,240
There's no perpendicular part.

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The area formed by
these two vectors is 0.

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00:05:00,700 --> 00:05:03,600
I hat cross J hat.

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00:05:03,600 --> 00:05:05,110
That's cyclic.

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00:05:05,110 --> 00:05:07,810
IJ is plus K hat.

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00:05:07,810 --> 00:05:14,800
So our first non-zero term is
AXBY K hat I hat cross J hat.

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00:05:14,800 --> 00:05:17,950
And now let's do
I hat cross K hat.

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00:05:17,950 --> 00:05:19,540
Notice that's anti-cyclic.

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00:05:19,540 --> 00:05:28,540
I K minus J. So our next
term is minus AX BZ J hat.

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00:05:28,540 --> 00:05:30,430
So there's the first two.

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00:05:30,430 --> 00:05:32,920
And now let's just
continue this process.

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00:05:32,920 --> 00:05:35,590
J hat cross I hat.

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00:05:35,590 --> 00:05:37,840
That's anti-cyclic.

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00:05:37,840 --> 00:05:45,670
So we have minus AY BX K hat.

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00:05:45,670 --> 00:05:48,310
AY J hat across BX I hat.

82
00:05:48,310 --> 00:05:50,200
J hat cross J hat.

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00:05:50,200 --> 00:05:51,760
That's 0.

84
00:05:51,760 --> 00:05:53,260
So we have no
contribution there.

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00:05:53,260 --> 00:05:56,650
And J hat crossed K
hat, that is cyclic.

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00:05:56,650 --> 00:06:02,650
So that's plus AY BZ K hat.

87
00:06:02,650 --> 00:06:06,990
And now we have our last
two terms K hat cross I hat.

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00:06:06,990 --> 00:06:07,780
That cyclic.

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00:06:07,780 --> 00:06:14,870
So that's plus AZ BX J hat.

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00:06:14,870 --> 00:06:16,360
K hat cross J hat.

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00:06:16,360 --> 00:06:18,130
That's anti-cyclic.

92
00:06:18,130 --> 00:06:19,940
So there's a minus I hat.

93
00:06:19,940 --> 00:06:26,520
So it's minus AZ BY I hat.

94
00:06:26,520 --> 00:06:30,500
And finally K hat cross K
hat, well that's also 0.

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00:06:30,500 --> 00:06:32,230
So we have six terms.

96
00:06:32,230 --> 00:06:42,015
And we can collect them
equal to AX BY minus-- well

97
00:06:42,015 --> 00:06:45,340
let's check this one.

98
00:06:45,340 --> 00:06:48,550
Here we used the
wrong symbol here.

99
00:06:48,550 --> 00:06:51,450
We have to be a little
bit careful here.

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00:06:51,450 --> 00:06:55,860
AY cross BZ is J
hat cross K hat.

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00:06:55,860 --> 00:06:58,870
That's plus I hat.

102
00:06:58,870 --> 00:07:07,600
So we have AXBY
minus AY BX K hat.

103
00:07:07,600 --> 00:07:10,750
And now let's look
at the I hat terms.

104
00:07:10,750 --> 00:07:12,820
I'll just check those off.

105
00:07:12,820 --> 00:07:24,050
We have AY BZ minus AZ BY I hat.

106
00:07:24,050 --> 00:07:26,350
And check those two terms off.

107
00:07:26,350 --> 00:07:31,020
And lastly, we have
AZ BX minus AXPZ.

108
00:07:31,020 --> 00:07:42,120
So we have AZBX minus
AX BZ and that's J hat.

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00:07:42,120 --> 00:07:46,530
And that's how we
calculate the cross-product

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00:07:46,530 --> 00:07:50,178
in Cartesian coordinates.