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The topic for today is going to
be equations of planes,
8
00:00:30 --> 00:00:39
and how they relate to linear
systems and matrices as we have
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seen during Tuesday's lecture.
So, let's start again with
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00:00:49 --> 00:00:57
equations of planes.
Remember, we've seen briefly
11
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that an equation for a plane is
of the form ax by cz = d,
12
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where a, b, c,
and d are just numbers.
13
00:01:16 --> 00:01:21
This expresses the condition
for a point at coordinates x,
14
00:01:21 --> 00:01:29
y, z, to be in the plane.
An equation of this form
15
00:01:29 --> 00:01:38
defines a plane.
Let's see how that works, again.
16
00:01:38 --> 00:01:45
Let's start with an example.
Let's say that we want to find
17
00:01:45 --> 00:01:56
the equation of a plane through
the origin with normal vector --
18
00:01:56 --> 00:02:04
-- let's say vector N equals the
vector <1,5,
19
00:02:04 --> 00:02:09
10>.
How do we find an equation of
20
00:02:09 --> 00:02:16
this plane?
Remember that we can get an
21
00:02:16 --> 00:02:23
equation by thinking
geometrically.
22
00:02:23 --> 00:02:27
So, what's our thinking going
to be?
23
00:02:27 --> 00:02:41
Well, we have the x, y, z axes.
And, we have this vector N:
24
00:02:41 --> 00:02:46
.
It's supposed to be
25
00:02:46 --> 00:02:49
perpendicular to our plane.
And, our plane passes through
26
00:02:49 --> 00:02:54
the origin here.
So, we want to think of the
27
00:02:54 --> 00:03:00
plane that's perpendicular to
this vector.
28
00:03:00 --> 00:03:03
Well, when is a point in that
plane?
29
00:03:03 --> 00:03:11
Let's say we have a point,
P -- -- at coordinates x,
30
00:03:11 --> 00:03:15
y, z.
Well, the condition for P to be
31
00:03:15 --> 00:03:20
in the plane should be that we
have a right angle here.
32
00:03:20 --> 00:03:34
OK, so P is in the plane
whenever OP dot N is 0.
33
00:03:34 --> 00:03:38
And, if we write that
explicitly, the vector OP has
34
00:03:38 --> 00:03:40
components x,
y, z;
35
00:03:40 --> 00:03:48
N has components 1,5, 10.
So that will give us x 5y 10z =
36
00:03:48 --> 00:03:55
0.
That's the equation of our
37
00:03:55 --> 00:04:00
plane.
Now, let's think about a
38
00:04:00 --> 00:04:04
slightly different problem.
So, let's do another problem.
39
00:04:04 --> 00:04:11
Let's try to find the equation
of the plane through the point
40
00:04:11 --> 00:04:17
P0 with coordinates,
say, (2,1,-1),
41
00:04:17 --> 00:04:26
with normal vector,
again, the same N = <1,5,
42
00:04:26 --> 00:04:34
10>.
How do we find an equation of
43
00:04:34 --> 00:04:39
this thing?
Well, we're going to use the
44
00:04:39 --> 00:04:44
same method.
In fact, let's think for a
45
00:04:44 --> 00:04:47
second.
I said we have our normal
46
00:04:47 --> 00:04:52
vector, N, and it's going to be
perpendicular to both planes at
47
00:04:52 --> 00:04:53
the same time.
So, in fact,
48
00:04:53 --> 00:04:56
our two planes will be parallel
to each other.
49
00:04:56 --> 00:04:58
The difference is,
well, before,
50
00:04:58 --> 00:05:00
we had a plane that was
perpendicular to N,
51
00:05:00 --> 00:05:05
and passing through the origin.
And now, we have a new plane
52
00:05:05 --> 00:05:10
that's going to pass not through
the origin but through this
53
00:05:10 --> 00:05:13
point, P0.
I don't really know where it
54
00:05:13 --> 00:05:15
is, but let's say,
for example,
55
00:05:15 --> 00:05:21
that P0 is here.
Then, I will just have to shift
56
00:05:21 --> 00:05:26
my plane so that,
instead of passing through the
57
00:05:26 --> 00:05:32
origin, it passes through this
new point.
58
00:05:32 --> 00:05:36
How am I going to do that?
Well, now, for a point P to be
59
00:05:36 --> 00:05:41
in our new plane,
we need the vector no longer OP
60
00:05:41 --> 00:05:44
but P0P to be perpendicular to
N.
61
00:05:44 --> 00:05:57
So P is in this new plane if
the vector P0P is perpendicular
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00:05:57 --> 00:06:01
to N.
And now, let's think,
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00:06:01 --> 00:06:05
what's the vector P0P?
Well, we take the coordinates
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00:06:05 --> 00:06:08
of P, and we subtract those of
P0.
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00:06:08 --> 00:06:15
So, that should be x-2,
y-1, and z 1,
66
00:06:15 --> 00:06:23
dot product with <1,5,
10> equals 0.
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00:06:23 --> 00:06:41
Let's expand this.
We get (x-2) 5(y-1) 10(z 1) = 0.
68
00:06:41 --> 00:06:45
Let's put the constants on the
other side.
69
00:06:45 --> 00:06:50
We get: x 5y 10z equals -- here
minus two becomes two,
70
00:06:50 --> 00:06:56
minus five becomes five,
ten becomes minus ten.
71
00:06:56 --> 00:07:01
I think we end up with negative
three.
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00:07:01 --> 00:07:05
So, the only thing that changes
between these two equations is
73
00:07:05 --> 00:07:07
the constant term on the
right-hand side,
74
00:07:07 --> 00:07:11
the thing that I called d.
The other common feature is
75
00:07:11 --> 00:07:13
that the coefficients of x,
y, and z: one,
76
00:07:13 --> 00:07:15
five, and ten,
correspond exactly to the
77
00:07:15 --> 00:07:19
normal vector.
That's something you should
78
00:07:19 --> 00:07:22
remember about planes.
These coefficients here
79
00:07:22 --> 00:07:27
correspond exactly to a normal
vector and, well,
80
00:07:27 --> 00:07:33
this constant term here roughly
measures how far you move
81
00:07:33 --> 00:07:35
from...
I f you have a plane through
82
00:07:35 --> 00:07:37
the origin, the right-hand side
will be zero.
83
00:07:37 --> 00:07:41
And, if you move to a parallel
plane, then this number will
84
00:07:41 --> 00:07:44
become something else.
Actually, how could we have
85
00:07:44 --> 00:07:48
found that -3 more quickly?
Well, we know that the first
86
00:07:48 --> 00:07:51
part of the equation is like
this.
87
00:07:51 --> 00:07:55
And we know something else.
We know that the point P0 is in
88
00:07:55 --> 00:08:00
the plane.
So, if we plug the coordinates
89
00:08:00 --> 00:08:05
of P0 into this,
well, x is 2 5 times 1 10 times
90
00:08:05 --> 00:08:08
-1.
We get -3.
91
00:08:08 --> 00:08:12
So, in fact,
the number we should have here
92
00:08:12 --> 00:08:16
should be minus three so that P0
is a solution.
93
00:08:16 --> 00:08:25
Let me point out -- (I'll put a
1 here again) -- these three
94
00:08:25 --> 00:08:31
numbers: 1,5,
10, are exactly the normal
95
00:08:31 --> 00:08:37
vector.
And one way that we can get
96
00:08:37 --> 00:08:45
this number here is by computing
the value of the left-hand side
97
00:08:45 --> 00:08:51
at the point P0.
We plug in the point P0 into
98
00:08:51 --> 00:08:56
the left hand side.
OK, any questions about that?
99
00:08:56 --> 00:09:07
100
00:09:07 --> 00:09:10
By the way, of course,
a plane doesn't have just one
101
00:09:10 --> 00:09:12
equation.
It has infinitely many
102
00:09:12 --> 00:09:18
equations because if instead,
say, I multiply everything by
103
00:09:18 --> 00:09:23
two, 2x 10y 20z = -6 is also an
equation for this plane.
104
00:09:23 --> 00:09:32
That's because we have normal
vectors of all sizes -- we can
105
00:09:32 --> 00:09:40
choose how big we make it.
Again, the single most
106
00:09:40 --> 00:09:49
important thing here:
in the equation ax by cz = d,
107
00:09:49 --> 00:09:57
the coefficients,
a, b, c, give us a normal
108
00:09:57 --> 00:10:03
vector to the plane.
So, that's why,
109
00:10:03 --> 00:10:07
in fact, what matters to us the
most is finding the normal
110
00:10:07 --> 00:10:08
vector.
In particular,
111
00:10:08 --> 00:10:11
if you remember,
last time I explained something
112
00:10:11 --> 00:10:14
about how we can find a normal
vector to a plane if we know
113
00:10:14 --> 00:10:17
points in the plane.
Namely, we can take the cross
114
00:10:17 --> 00:10:20
product of two vectors contained
in the plane.
115
00:10:20 --> 00:10:48
116
00:10:48 --> 00:10:54
Let's just do an example to see
if we completely understand
117
00:10:54 --> 00:10:59
what's going on.
Let's say that I give you the
118
00:10:59 --> 00:11:03
vector with components
,
119
00:11:03 --> 00:11:08
and I give you the plane x y 3z
= 5.
120
00:11:08 --> 00:11:12
So, do you think that this
vector is parallel to the plane,
121
00:11:12 --> 00:11:15
perpendicular to it,
neither?
122
00:11:15 --> 00:11:24
123
00:11:24 --> 00:11:43
I'm starting to see a few votes.
OK, I see that most of you are
124
00:11:43 --> 00:11:48
answering number two:
this vector is perpendicular to
125
00:11:48 --> 00:11:51
the plane.
There are some other answers
126
00:11:51 --> 00:11:59
too.
Well, let's try to figure it
127
00:11:59 --> 00:12:02
out.
Let's do the example.
128
00:12:02 --> 00:12:12
Say v is <1,2,
-1> and the plane is x y 3z
129
00:12:12 --> 00:12:15
= 5.
Let's just draw that plane
130
00:12:15 --> 00:12:18
anywhere -- it doesn't really
matter.
131
00:12:18 --> 00:12:21
Let's first get a normal vector
out of it.
132
00:12:21 --> 00:12:28
Well, to get a normal vector to
the plane, what I will do is
133
00:12:28 --> 00:12:33
take the coefficients of x,
y, and z.
134
00:12:33 --> 00:12:36
So, that's .
So
135
00:12:36 --> 00:12:40
is perpendicular to the plane.
How do we get all the other
136
00:12:40 --> 00:12:43
vectors that are perpendicular
to the plane?
137
00:12:43 --> 00:12:47
Well, all the perpendicular
vectors are parallel to each
138
00:12:47 --> 00:12:50
other.
That means that they are just
139
00:12:50 --> 00:12:54
obtained by multiplying this guy
by some number.
140
00:12:54 --> 00:12:55
141
00:12:55 --> 00:12:59
for example,
would still be perpendicular to
142
00:12:59 --> 00:13:00
the plane.
143
00:13:00 --> 00:13:01
144
00:13:01 --> 00:13:04
is also perpendicular to the
plane.
145
00:13:04 --> 00:13:07
But now, see,
these guys are not proportional
146
00:13:07 --> 00:13:18
to each other.
So, V is not perpendicular to
147
00:13:18 --> 00:13:28
the plane.
So it's not perpendicular to
148
00:13:28 --> 00:13:33
the plane.
Being perpendicular to the
149
00:13:33 --> 00:13:37
plane is the same as being
parallel to its normal vector.
150
00:13:37 --> 00:13:41
Now, what about testing if v
is, instead, parallel to the
151
00:13:41 --> 00:13:43
plane?
Well, it's parallel to the
152
00:13:43 --> 00:13:46
plane if it's perpendicular to
N.
153
00:13:46 --> 00:13:46
Let's check.
154
00:13:46 --> 00:13:56
155
00:13:56 --> 00:14:04
So, let's try to see if v is
perpendicular to N.
156
00:14:04 --> 00:14:11
Well, let's do v.N.
That's <1,2,
157
00:14:11 --> 00:14:15
- 1> dot <1,1,
3>.
158
00:14:15 --> 00:14:25
You get 1 2 - 3=0.
So, yes.
159
00:14:25 --> 00:14:37
If it's perpendicular to N,
it means -- It's actually going
160
00:14:37 --> 00:14:43
to be parallel to the plane.
161
00:14:43 --> 00:14:56
162
00:14:56 --> 00:15:00
OK, any questions?
Yes?
163
00:15:00 --> 00:15:03
[QUESTION FROM STUDENT:]
When you plug the vector into
164
00:15:03 --> 00:15:05
the plane equation,
you get zero.
165
00:15:05 --> 00:15:13
What does that mean?
Let's see.
166
00:15:13 --> 00:15:18
If I plug the vector into the
plane equation:
167
00:15:18 --> 00:15:23
1 2-3, well,
the left hand side becomes
168
00:15:23 --> 00:15:30
zero.
So, it's not a solution of the
169
00:15:30 --> 00:15:34
plane equation.
There's two different things
170
00:15:34 --> 00:15:38
here.
One is that the point with
171
00:15:38 --> 00:15:44
coordinates (1,2,- 1) is not in
the plane.
172
00:15:44 --> 00:15:52
What that tells us is that,
if I put my vector V at the
173
00:15:52 --> 00:16:01
origin, then its head is not
going to be in the plane.
174
00:16:01 --> 00:16:03
On the other hand,
you're right,
175
00:16:03 --> 00:16:06
the left hand side evaluates to
zero.
176
00:16:06 --> 00:16:09
What that means is that,
if instead I had taken the
177
00:16:09 --> 00:16:12
plane x y 3z = 0,
then it would be inside.
178
00:16:12 --> 00:16:21
The plane is x y 3z = 5,
so x y 3z = 0 would be a plane
179
00:16:21 --> 00:16:27
parallel to it,
but through the origin.
180
00:16:27 --> 00:16:30
So, that would be another way
to see that the vector is
181
00:16:30 --> 00:16:33
parallel to the plane.
If we move the plane to a
182
00:16:33 --> 00:16:37
parallel plane through the
origin, then the endpoint of the
183
00:16:37 --> 00:16:44
vector is in the plane.
OK, that's another way to
184
00:16:44 --> 00:16:56
convince ourselves.
Any other questions?
185
00:16:56 --> 00:17:04
OK, let's move on.
So, last time we learned about
186
00:17:04 --> 00:17:08
matrices and linear systems.
So, let's try to think,
187
00:17:08 --> 00:17:12
now, about linear systems in
terms of equations of planes and
188
00:17:12 --> 00:17:15
intersections of planes.
Remember that a linear system
189
00:17:15 --> 00:17:19
is a bunch of equations -- say,
a 3x3 linear system is three
190
00:17:19 --> 00:17:23
different equations.
Each of them is the equation of
191
00:17:23 --> 00:17:25
a plane.
So, in fact,
192
00:17:25 --> 00:17:29
if we try to solve a system of
equations, that means actually
193
00:17:29 --> 00:17:33
we are trying to find a point
that is on several planes at the
194
00:17:33 --> 00:17:46
same time.
So...
195
00:17:46 --> 00:17:52
Let's say that we have a 3x3
linear system.
196
00:17:52 --> 00:18:04
Just to take an example -- it
doesn't really matter what I
197
00:18:04 --> 00:18:14
give you, but let's say I give
you x z = 1, x y = 2,
198
00:18:14 --> 00:18:21
x 2y 3z = 3.
What does it mean to solve this?
199
00:18:21 --> 00:18:27
It means we want to find x,
y, z which satisfy all of these
200
00:18:27 --> 00:18:30
conditions.
Let's just look at the first
201
00:18:30 --> 00:18:33
equation, first.
Well, the first equation says
202
00:18:33 --> 00:18:37
our point should be on the plane
which has this equation.
203
00:18:37 --> 00:18:42
Then, the second equation says
that our point should also be on
204
00:18:42 --> 00:18:46
that plane.
So, if you just look at the
205
00:18:46 --> 00:18:50
first two equations,
you have two planes.
206
00:18:50 --> 00:19:08
And the solutions -- these two
equations determine for you two
207
00:19:08 --> 00:19:22
planes, and two planes intersect
in a line.
208
00:19:22 --> 00:19:27
Now, what happens with the
third equation?
209
00:19:27 --> 00:19:30
That's actually going to be a
third plane.
210
00:19:30 --> 00:19:33
So, if we want to solve the
first two equations,
211
00:19:33 --> 00:19:37
we have to be on this line.
And if we want to solve the
212
00:19:37 --> 00:19:41
third one, we also need to be on
another plane.
213
00:19:41 --> 00:19:52
And, in general,
the three planes intersect in a
214
00:19:52 --> 00:20:02
point because this line of
intersection...
215
00:20:02 --> 00:20:04
Three planes intersect in a
point,
216
00:20:04 --> 00:20:09
and one way to think about it
is that the line where the first
217
00:20:09 --> 00:20:14
two planes intersect meets the
third plane in a point.
218
00:20:14 --> 00:20:21
And, that point is the solution
to the linear system.
219
00:20:21 --> 00:20:28
The line -- this is
mathematical notation for the
220
00:20:28 --> 00:20:36
intersection between the first
two planes -- intersects the
221
00:20:36 --> 00:20:46
third plane in a point,
which is going to be the
222
00:20:46 --> 00:20:53
solution.
So, how do we find the solution?
223
00:20:53 --> 00:20:58
One way is to draw pictures and
try to figure out where the
224
00:20:58 --> 00:21:03
solution is, but that's not how
we do it in practice if we are
225
00:21:03 --> 00:21:07
given the equations.
Let me use matrix notation.
226
00:21:07 --> 00:21:17
Remember, we saw on Tuesday
that the solution to AX = B is
227
00:21:17 --> 00:21:23
given by X = A inverse B.
We got from here to there by
228
00:21:23 --> 00:21:26
multiplying on the left by A
inverse.
229
00:21:26 --> 00:21:32
A inverse AX simplifies to X
equals A inverse B.
230
00:21:32 --> 00:21:35
And, once again,
it's A inverse B and not BA
231
00:21:35 --> 00:21:37
inverse.
If you try to set up the
232
00:21:37 --> 00:21:39
multiplication,
BA inverse doesn't work.
233
00:21:39 --> 00:21:47
The sizes are not compatible,
you can't multiply the other
234
00:21:47 --> 00:21:54
way around.
OK, that's pretty good --
235
00:21:54 --> 00:22:03
unless it doesn't work that way.
What could go wrong?
236
00:22:03 --> 00:22:07
Well, let's say that our first
two planes do intersect nicely
237
00:22:07 --> 00:22:10
in a line, but let's think about
the third plane.
238
00:22:10 --> 00:22:13
Maybe the third plane does not
intersect that line nicely in a
239
00:22:13 --> 00:22:19
point.
Maybe it's actually parallel to
240
00:22:19 --> 00:22:26
that line.
Let's try to think about this
241
00:22:26 --> 00:22:33
question for a second.
Let's say that the set of
242
00:22:33 --> 00:22:39
solutions to a 3x3 linear system
is not just one point.
243
00:22:39 --> 00:22:43
So, we don't have a unique
solution that we can get this
244
00:22:43 --> 00:22:53
way.
What do you think could happen?
245
00:22:53 --> 00:22:58
OK, I see that answers number
three and five seem to be
246
00:22:58 --> 00:23:03
dominating.
There's also a bit of answer
247
00:23:03 --> 00:23:06
number one.
In fact, these are pretty good
248
00:23:06 --> 00:23:08
answers.
I see that some of you figured
249
00:23:08 --> 00:23:12
out that you can answer one and
three at the same time,
250
00:23:12 --> 00:23:15
or three and five at the same
time.
251
00:23:15 --> 00:23:18
I yet have to see somebody with
three hands answer all three
252
00:23:18 --> 00:23:20
numbers at the same time.
OK.
253
00:23:20 --> 00:23:26
Indeed, we'll see very soon
that we could have either no
254
00:23:26 --> 00:23:29
solution, a line,
or a plane.
255
00:23:29 --> 00:23:33
The other answers:
"two points"
256
00:23:33 --> 00:23:35
(two solutions),
we will see,
257
00:23:35 --> 00:23:37
is actually not a possibility
because if you have two
258
00:23:37 --> 00:23:40
different solutions,
then the entire line through
259
00:23:40 --> 00:23:44
these two points is also going
to be made of solutions.
260
00:23:44 --> 00:23:47
"A tetrahedron"
is just there to amuse you,
261
00:23:47 --> 00:23:51
it's actually not a good answer
to the question.
262
00:23:51 --> 00:23:54
It's not very likely that you
will get a tetrahedron out of
263
00:23:54 --> 00:23:56
intersecting planes.
"A plane"
264
00:23:56 --> 00:23:58
is indeed possible,
and "I don't know"
265
00:23:58 --> 00:24:00
is still OK for a few more
minutes,
266
00:24:00 --> 00:24:04
but we're going to get to the
bottom of this,
267
00:24:04 --> 00:24:09
and then we will know.
OK, let's try to figure out
268
00:24:09 --> 00:24:16
what can happen.
Let me go back to my picture.
269
00:24:16 --> 00:24:20
I had my first two planes;
they determine a line.
270
00:24:20 --> 00:24:23
And now I have my third plane.
Maybe my third plane is
271
00:24:23 --> 00:24:29
actually parallel to the line
but doesn't pass through it.
272
00:24:29 --> 00:24:32
Well, then, there's no
solutions because,
273
00:24:32 --> 00:24:37
to solve the system of
equations, I need to be in the
274
00:24:37 --> 00:24:40
first two planes.
So, that means I need to be in
275
00:24:40 --> 00:24:43
that vertical line.
(That line was supposed to be
276
00:24:43 --> 00:24:47
red, but I guess it doesn't
really show up as red).
277
00:24:47 --> 00:24:49
And it also needs to be in the
third plane.
278
00:24:49 --> 00:24:52
But the line and the plane are
parallel to each other.
279
00:24:52 --> 00:24:55
There's just no place where
they intersect.
280
00:24:55 --> 00:24:59
So there's no way to solve all
the equations.
281
00:24:59 --> 00:25:03
On the other hand,
the other thing that could
282
00:25:03 --> 00:25:07
happen is that actually the line
is contained in the plane.
283
00:25:07 --> 00:25:13
And then, any point on that
line will automatically solve
284
00:25:13 --> 00:25:19
the third equation.
So if you try solving a system
285
00:25:19 --> 00:25:23
that looks like this by hand,
if you do substitutions,
286
00:25:23 --> 00:25:25
eliminations,
and so on,
287
00:25:25 --> 00:25:28
what you will notice is that,
after you have dealt with two
288
00:25:28 --> 00:25:31
of the equations,
the third one would actually
289
00:25:31 --> 00:25:35
turn out to be the same as what
you got out of the first two.
290
00:25:35 --> 00:25:36
It doesn't give you any
additional information.
291
00:25:36 --> 00:25:41
It's as if you had only two
equations.
292
00:25:41 --> 00:25:45
The previous case would be when
actually the third equation
293
00:25:45 --> 00:25:49
contradicts something that you
can get out of the first two.
294
00:25:49 --> 00:25:51
For example,
maybe out of the first two,
295
00:25:51 --> 00:25:54
you got that x plus z equals
one, and the third equation is x
296
00:25:54 --> 00:25:57
plus z equals two.
Well, it can't be one and two
297
00:25:57 --> 00:26:00
at the same time.
Another way to say it is that
298
00:26:00 --> 00:26:04
this picture is one where you
can get out of the equations
299
00:26:04 --> 00:26:07
that a number equals a different
number.
300
00:26:07 --> 00:26:10
That's impossible.
And, that picture is one where
301
00:26:10 --> 00:26:12
out of the equations you get
zero equals zero,
302
00:26:12 --> 00:26:15
which is certainly true,
but isn't a very useful
303
00:26:15 --> 00:26:19
equation.
So, you can't actually finish
304
00:26:19 --> 00:26:27
solving.
OK, let me write that down.
305
00:26:27 --> 00:26:48
unless the third plane is
parallel to the line where P1
306
00:26:48 --> 00:26:58
and P2 intersect.
Then there's two subcases.
307
00:26:58 --> 00:27:11
If the line of intersections of
P1 and P2 is actually contained
308
00:27:11 --> 00:27:22
in P3 (the third plane),
then we have infinitely many
309
00:27:22 --> 00:27:26
solutions.
Namely, any point on the line
310
00:27:26 --> 00:27:29
will automatically solve the
third equation.
311
00:27:29 --> 00:27:49
312
00:27:49 --> 00:28:05
The other subcase is if the
line of the intersection of P1
313
00:28:05 --> 00:28:19
and P2 is parallel to P3 and not
contained in it.
314
00:28:19 --> 00:28:35
Then we get no solutions.
Just to show you the pictures
315
00:28:35 --> 00:28:38
once again: when we have the
first two planes,
316
00:28:38 --> 00:28:42
they give us a line.
And now, depending on what
317
00:28:42 --> 00:28:45
happens to that line in relation
to the third plane,
318
00:28:45 --> 00:28:50
various situations can happen.
If the line hits the third
319
00:28:50 --> 00:28:55
plane in a point,
then that's going to be our
320
00:28:55 --> 00:28:58
solution.
If that line,
321
00:28:58 --> 00:29:01
instead, is parallel to the
third plane, well,
322
00:29:01 --> 00:29:05
if it's parallel and outside of
it, then we have no solution.
323
00:29:05 --> 00:29:16
If it's parallel and contained
in it, then we have infinitely
324
00:29:16 --> 00:29:23
many solutions.
So, going back to our list of
325
00:29:23 --> 00:29:29
possibilities,
let's see what can happen.
326
00:29:29 --> 00:29:32
No solution:
we've seen that it happens when
327
00:29:32 --> 00:29:37
the line where the first two
planes intersect is parallel to
328
00:29:37 --> 00:29:40
the third one.
Two points: well,
329
00:29:40 --> 00:29:45
that didn't come up.
As I said, the problem is that,
330
00:29:45 --> 00:29:49
if the line of intersections of
the first two planes has two
331
00:29:49 --> 00:29:52
points that are in the third
plane,
332
00:29:52 --> 00:29:55
then that means the entire line
must actually be in the third
333
00:29:55 --> 00:29:58
plane.
So, if you have two solutions,
334
00:29:58 --> 00:30:03
then you have more than two.
In fact, you have infinitely
335
00:30:03 --> 00:30:05
many, and we've seen that can
happen.
336
00:30:05 --> 00:30:10
A tetrahedron:
still doesn't look very
337
00:30:10 --> 00:30:13
promising.
What about a plane?
338
00:30:13 --> 00:30:17
Well, that's a case that I
didn't explain because I've been
339
00:30:17 --> 00:30:20
assuming that P1 and P2 are
different planes and they
340
00:30:20 --> 00:30:23
intersect in a line.
But, in fact,
341
00:30:23 --> 00:30:26
they could be parallel,
in which case we already have
342
00:30:26 --> 00:30:28
no solution to the first two
equations;
343
00:30:28 --> 00:30:32
or they could be the same plane.
And now, if the third plane is
344
00:30:32 --> 00:30:36
also the same plane -- if all
three planes are the same plane,
345
00:30:36 --> 00:30:38
then you have a plane of
solutions.
346
00:30:38 --> 00:30:40
If I give you three times the
same equation,
347
00:30:40 --> 00:30:44
that is a linear system.
It's not a very interesting
348
00:30:44 --> 00:30:50
one, but it's a linear system.
And "I don't know"
349
00:30:50 --> 00:30:58
is no longer a solution either.
OK, any questions?
350
00:30:58 --> 00:31:01
[STUDENT QUESTION:]
What's the geometric
351
00:31:01 --> 00:31:04
significance of the plane x y z
equals 1, as opposed to 2,
352
00:31:04 --> 00:31:07
or 3?
That's a very good question.
353
00:31:07 --> 00:31:10
The question is,
what is the geometric
354
00:31:10 --> 00:31:14
significance of an equation like
x y z equals to 1,2,
355
00:31:14 --> 00:31:19
3, or something else?
Well, if the equation is x y z
356
00:31:19 --> 00:31:23
equals zero, it means that our
plane is passing through the
357
00:31:23 --> 00:31:25
origin.
And then, if we change the
358
00:31:25 --> 00:31:28
constant, it means we move to a
parallel plane.
359
00:31:28 --> 00:31:31
So, the first guess that you
might have is that this number
360
00:31:31 --> 00:31:35
on the right-hand side is the
distance between the origin and
361
00:31:35 --> 00:31:37
the plane.
It tells us how far from the
362
00:31:37 --> 00:31:42
origin we are.
That is not quite true.
363
00:31:42 --> 00:31:47
In fact, that would be true if
the coefficients here formed a
364
00:31:47 --> 00:31:50
unit vector.
Then this would just be the
365
00:31:50 --> 00:31:55
distance to the origin.
Otherwise, you have to actually
366
00:31:55 --> 00:31:57
scale by the length of this
normal vector.
367
00:31:57 --> 00:32:01
And, I think there's a problem
in the Notes that will show you
368
00:32:01 --> 00:32:05
exactly how this works.
You should think of it roughly
369
00:32:05 --> 00:32:09
as how much we have moved the
plane away from the origin.
370
00:32:09 --> 00:32:13
That's the meaning of the last
term, D, in the right-hand side
371
00:32:13 --> 00:32:14
of the equation.
372
00:32:14 --> 00:32:29
373
00:32:29 --> 00:32:34
So, let's try to think about
what exactly these cases are --
374
00:32:34 --> 00:32:38
how do we detect in which
situation we are?
375
00:32:38 --> 00:32:43
It's all very nice in the
picture, but it's difficult to
376
00:32:43 --> 00:32:46
draw planes.
In fact, when I draw these
377
00:32:46 --> 00:32:48
pictures, I'm always very
careful not to actually pretend
378
00:32:48 --> 00:32:51
to draw an actual plane given by
an equation.
379
00:32:51 --> 00:32:56
When I do, then it's blatantly
false -- it's difficult to draw
380
00:32:56 --> 00:32:58
a plane correctly.
So, instead,
381
00:32:58 --> 00:33:02
let's try to think about it in
terms of matrices.
382
00:33:02 --> 00:33:04
In particular,
what's wrong with this?
383
00:33:04 --> 00:33:09
Why can't we always say the
solution is X = A inverse B?
384
00:33:09 --> 00:33:19
Well, the point is that,
actually, you cannot always
385
00:33:19 --> 00:33:26
invert a matrix.
Recall we've seen this formula:
386
00:33:26 --> 00:33:32
A inverse is one over
determinant of A times the
387
00:33:32 --> 00:33:36
adjoint matrix.
And we've learned how to
388
00:33:36 --> 00:33:39
compute this thing:
remember, we had to take
389
00:33:39 --> 00:33:43
minors, then flip some signs,
and then transpose.
390
00:33:43 --> 00:33:46
That step we can always do.
We can always do these
391
00:33:46 --> 00:33:48
calculations.
But then, at the end,
392
00:33:48 --> 00:33:51
we have to divide by the
determinant.
393
00:33:51 --> 00:33:53
That's fine if the determinant
is not zero.
394
00:33:53 --> 00:34:00
But, if the determinant is
zero, then certainly we cannot
395
00:34:00 --> 00:34:05
do that.
What I didn't mention last time
396
00:34:05 --> 00:34:11
is that the matrix is invertible
-- that means it has an inverse
397
00:34:11 --> 00:34:16
-- exactly when its determinant
is not zero.
398
00:34:16 --> 00:34:20
That's something we should
remember.
399
00:34:20 --> 00:34:24
So, if the determinant is not
zero, then we can use our method
400
00:34:24 --> 00:34:28
to find the inverse.
And then we can solve using
401
00:34:28 --> 00:34:31
this method.
If not, then not.
402
00:34:31 --> 00:34:33
Yes?
[STUDENT QUESTION:]
403
00:34:33 --> 00:34:36
Sorry, can you reexplain that?
You can invert A if the
404
00:34:36 --> 00:34:38
determinant of A is not equal to
zero?
405
00:34:38 --> 00:34:41
That's correct.
We can invert the matrix A if
406
00:34:41 --> 00:34:46
the determinant is not zero.
If you look again at the method
407
00:34:46 --> 00:34:49
that we saw last time:
first we had to compute the
408
00:34:49 --> 00:34:52
adjoint matrix.
And, these are operations we
409
00:34:52 --> 00:34:54
can always do.
If we are given a 3x3 matrix,
410
00:34:54 --> 00:34:56
we can always compute the
adjoint.
411
00:34:56 --> 00:34:59
And then, the last step to find
the inverse was to divide by the
412
00:34:59 --> 00:35:02
determinant.
And that we can only do if the
413
00:35:02 --> 00:35:06
determinant is not zero.
So, if we have a matrix whose
414
00:35:06 --> 00:35:09
determinant is not zero,
then we know how to find the
415
00:35:09 --> 00:35:11
inverse.
If the determinant is zero,
416
00:35:11 --> 00:35:14
then of course this method
doesn't work.
417
00:35:14 --> 00:35:17
I'm actually saying even more:
there isn't an inverse at all.
418
00:35:17 --> 00:35:19
It's not just that our method
fails.
419
00:35:19 --> 00:35:27
I cannot take the inverse of a
matrix with determinant zero.
420
00:35:27 --> 00:35:30
Geometrically,
the situation where the
421
00:35:30 --> 00:35:34
determinant is not zero is
exactly this nice usual
422
00:35:34 --> 00:35:39
situation where the three planes
intersect in a point,
423
00:35:39 --> 00:35:45
while the situation where the
determinant is zero is this
424
00:35:45 --> 00:35:52
situation here where the line
determined by the first two
425
00:35:52 --> 00:35:56
planes is parallel to the third
plane.
426
00:35:56 --> 00:36:06
Let me emphasize this again,
and let's see again what
427
00:36:06 --> 00:36:19
happens.
Let's start with an easier case.
428
00:36:19 --> 00:36:21
It's called the case of a
homogeneous system.
429
00:36:21 --> 00:36:27
It's called homogeneous because
it's the situation where the
430
00:36:27 --> 00:36:31
equations are invariant under
scaling.
431
00:36:31 --> 00:36:35
So, a homogeneous system is one
where the right hand side is
432
00:36:35 --> 00:36:38
zero -- there's no B.
If you want,
433
00:36:38 --> 00:36:42
the constant terms here are all
zero: 0,0, 0.
434
00:36:42 --> 00:36:46
OK, so this one is not
homogenous.
435
00:36:46 --> 00:36:57
So, let's see what happens
there.
436
00:36:57 --> 00:37:02
Let's take an example.
Instead of this system,
437
00:37:02 --> 00:37:10
we could take x z = 0,
x y = 0, and x 2y 3z also
438
00:37:10 --> 00:37:16
equals zero.
Can we solve these equations?
439
00:37:16 --> 00:37:20
I think actually you already
know a very simple solution to
440
00:37:20 --> 00:37:23
these equations.
Yeah, you can just take x,
441
00:37:23 --> 00:37:34
y, and z all to be zero.
So, there's always an obvious
442
00:37:34 --> 00:37:44
solution -- -- namely,
(0,0, 0).
443
00:37:44 --> 00:37:53
And, in mathematical jargon,
this is called the trivial
444
00:37:53 --> 00:37:57
solution.
There's always this trivial
445
00:37:57 --> 00:37:59
solution.
What's the geometric
446
00:37:59 --> 00:38:01
interpretation?
Well, having zeros here means
447
00:38:01 --> 00:38:04
that all three planes pass
through the origin.
448
00:38:04 --> 00:38:07
So, certainly the origin is
always a solution.
449
00:38:07 --> 00:38:21
450
00:38:21 --> 00:38:35
The origin is always a solution
because the three planes -- --
451
00:38:35 --> 00:38:45
pass through the origin.
Now there's two subcases.
452
00:38:45 --> 00:38:52
One case is if the determinant
of the matrix A is nonzero.
453
00:38:52 --> 00:39:01
That means that we can invert A.
So, if we can invert A,
454
00:39:01 --> 00:39:07
then we can solve the system by
multiplying by A inverse.
455
00:39:07 --> 00:39:13
If we multiply by A inverse,
we'll get X equals A inverse
456
00:39:13 --> 00:39:21
times zero, which is zero.
That's the only solution
457
00:39:21 --> 00:39:24
because,
if AX is zero,
458
00:39:24 --> 00:39:27
then let's multiply by A
inverse: we get that A inverse
459
00:39:27 --> 00:39:29
AX, which is X,
equals A inverse zero,
460
00:39:29 --> 00:39:32
which is zero.
We get that X equals zero.
461
00:39:32 --> 00:39:42
We've solved it,
there's no other solution.
462
00:39:42 --> 00:39:55
To go back to these pictures
that we all enjoy,
463
00:39:55 --> 00:40:03
it's this case.
Now the other case,
464
00:40:03 --> 00:40:13
if the determinant of A equals
zero, then this method doesn't
465
00:40:13 --> 00:40:18
quite work.
What does it mean that the
466
00:40:18 --> 00:40:22
determinant of A is zero?
Remember, the entries in A are
467
00:40:22 --> 00:40:25
the coefficients in the
equations.
468
00:40:25 --> 00:40:29
But now, the coefficients in
the equations are exactly the
469
00:40:29 --> 00:40:36
normal vectors to the planes.
So, that's the same thing as
470
00:40:36 --> 00:40:47
saying that the determinant of
the three normal vectors to our
471
00:40:47 --> 00:40:54
three planes is 0.
That means that N1,
472
00:40:54 --> 00:41:02
N2, and N3 are actually in a
same plane -- they're coplanar.
473
00:41:02 --> 00:41:06
These three vectors are
coplanar.
474
00:41:06 --> 00:41:14
So, let's see what happens.
I claim it will correspond to
475
00:41:14 --> 00:41:20
this situation here.
Let's draw the normal vectors
476
00:41:20 --> 00:41:27
to these three planes.
(Well, it's not very easy to
477
00:41:27 --> 00:41:33
see, but I've tried to draw the
normal vectors to my planes.)
478
00:41:33 --> 00:41:37
They are all in the direction
that's perpendicular to the line
479
00:41:37 --> 00:41:40
of intersection.
They are all in the same plane.
480
00:41:40 --> 00:41:44
So, if I try to form a
parallelepiped with these three
481
00:41:44 --> 00:41:47
normal vectors,
well, I will get something
482
00:41:47 --> 00:41:50
that's completely flat,
and has no volume,
483
00:41:50 --> 00:42:04
has volume zero.
So the parallelepiped -- -- has
484
00:42:04 --> 00:42:11
volume 0.
And the fact that the normal
485
00:42:11 --> 00:42:19
vectors are coplanar tells us
that, in fact -- (well,
486
00:42:19 --> 00:42:25
let me start a new blackboard).
Let's say that our normal
487
00:42:25 --> 00:42:28
vectors, N1, N2,
N3, are all in the same plane.
488
00:42:28 --> 00:42:32
And let's think about the
direction that's perpendicular
489
00:42:32 --> 00:42:35
to N1, N2, and N3 at the same
time.
490
00:42:35 --> 00:42:37
I claim that it will be the
line of intersection.
491
00:42:37 --> 00:43:08
492
00:43:08 --> 00:43:12
So, let me try to draw that
picture again.
493
00:43:12 --> 00:43:26
We have three planes -- (now
you see why I prepared a picture
494
00:43:26 --> 00:43:31
in advance.
It's easier to draw it
495
00:43:31 --> 00:43:37
beforehand).
And I said their normal vectors
496
00:43:37 --> 00:43:41
are all in the same plane.
What else do I know?
497
00:43:41 --> 00:43:45
I know that all these planes
pass through the origin.
498
00:43:45 --> 00:43:50
So the origin is somewhere in
the intersection of the three
499
00:43:50 --> 00:43:59
planes.
Now, I said that the normal
500
00:43:59 --> 00:44:13
vectors to my three planes are
all actually coplanar.
501
00:44:13 --> 00:44:23
So N1, N2, N3 determine a plane.
Now, if I look at the line
502
00:44:23 --> 00:44:27
through the origin that's
perpendicular to N1,
503
00:44:27 --> 00:44:33
N2, and N3,
so, perpendicular to this red
504
00:44:33 --> 00:44:39
plane here,
it's supposed to be in all the
505
00:44:39 --> 00:44:44
planes.
(You can see that better on the
506
00:44:44 --> 00:44:47
side screens).
And why is that?
507
00:44:47 --> 00:44:51
Well, that's because my line is
perpendicular to the normal
508
00:44:51 --> 00:44:54
vectors, so it's parallel to the
planes.
509
00:44:54 --> 00:44:58
It's parallel to all the planes.
Now, why is it in the planes
510
00:44:58 --> 00:45:01
instead of parallel to them?
Well, that's because my line
511
00:45:01 --> 00:45:03
goes through the origin,
and the origin is on the
512
00:45:03 --> 00:45:07
planes.
So, certainly my line has to be
513
00:45:07 --> 00:45:11
contained in the planes,
not parallel to them.
514
00:45:11 --> 00:45:26
So the line through the origin
and perpendicular to the plane
515
00:45:26 --> 00:45:39
of N1, N2, N3 -- -- is parallel
to all three planes.
516
00:45:39 --> 00:45:47
And, because the planes go
through the origin,
517
00:45:47 --> 00:45:58
it's contained in them.
So what happens here is I have,
518
00:45:58 --> 00:46:06
in fact, infinitely many
solutions.
519
00:46:06 --> 00:46:09
How do I find these solutions?
Well, if I want to find
520
00:46:09 --> 00:46:13
something that's perpendicular
to N1, N2, and N3 -- if I just
521
00:46:13 --> 00:46:16
want to be perpendicular to N1
and N2,
522
00:46:16 --> 00:46:29
I can take their cross product.
So, for example,
523
00:46:29 --> 00:46:38
N1 cross N2 is perpendicular to
N1 and to N2,
524
00:46:38 --> 00:46:43
and also to N3,
because N3 is in the same plane
525
00:46:43 --> 00:46:46
as N1 and N2,
so, if you're perpendicular to
526
00:46:46 --> 00:46:49
N1 and N2, you are also
perpendicular to N3.
527
00:46:49 --> 00:47:03
It's automatic.
So, it's a nontrivial solution.
528
00:47:03 --> 00:47:09
This vector goes along the line
of intersections.
529
00:47:09 --> 00:47:13
OK, that's the case of
homogeneous systems.
530
00:47:13 --> 00:47:24
And then, let's finish with the
other case, the general case.
531
00:47:24 --> 00:47:32
If we look at a system,
AX = B, with B now anything,
532
00:47:32 --> 00:47:41
there's two cases.
If the determinant of A is not
533
00:47:41 --> 00:47:51
zero, then there is a unique
solution -- -- namely,
534
00:47:51 --> 00:47:59
X equals A inverse B.
If the determinant of A is
535
00:47:59 --> 00:48:02
zero,
then it means we have the
536
00:48:02 --> 00:48:06
situation with planes that are
all parallel to a same line,
537
00:48:06 --> 00:48:18
and then we have either no
solution or infinitely many
538
00:48:18 --> 00:48:23
solutions.
It cannot be a single solution.
539
00:48:23 --> 00:48:26
Now, whether you have no
solutions or infinitely many
540
00:48:26 --> 00:48:30
solutions, we haven't actually
developed the tools to answer
541
00:48:30 --> 00:48:32
that.
But, if you try solving the
542
00:48:32 --> 00:48:34
system by hand,
by elimination,
543
00:48:34 --> 00:48:37
you will see that you end up
maybe with something that says
544
00:48:37 --> 00:48:40
zero equals zero,
and you have infinitely many
545
00:48:40 --> 00:48:42
solutions.
Actually, if you can find one
546
00:48:42 --> 00:48:45
solution, then you know that
there's infinitely many.
547
00:48:45 --> 00:48:48
On the other hand,
if you end up with something
548
00:48:48 --> 00:48:51
that's a contradiction,
like one equals two,
549
00:48:51 --> 00:48:54
then you know there's no
solutions.
550
00:48:54 --> 00:48:58
That's the end for today.
Tomorrow, we will learn about
551
00:48:58 --> 00:49:01
parametric equations for lines
and curves.
552
00:49:01 --> 00:49:02