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Let me just tell you first
about the list of topics.

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Basically, the list of topics
is simple.

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It is everything.
I mean, everything we have seen

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so far is on the exam.
But let me just remind you of

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the main topics that we have
seen.

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First of all,
we learned about vectors,

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how to use them,
and dot-product.

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00:01:03 --> 00:01:07
At this point,
you probably should know that

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the dot-product of two vectors
is obtained by summing products

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of components.
And geometrically it is the

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length of A times the length of
B times the cosine of the angle

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between them.
And, in particular,

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we can use dot-product to
measure angles by solving for

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cosine theta in this equality.
And most importantly to detect

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whether two vectors are
perpendicular to each other.

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Two vectors are perpendicular
when their dot-product is zero.

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00:02:00 --> 00:02:07
Any questions about that?
No.

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Is everyone reasonably happy
with dot-product by now?

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I see a stunned silence.
Nobody happy with dot-product

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so far?
OK.

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00:02:28 --> 00:02:36
If you want to look at Practice
1A, a good example of a typical

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00:02:36 --> 00:02:41
problem with dot-product would
be problem 1.

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00:02:41 --> 00:02:43
Let's see.
We are going to go over the

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00:02:43 --> 00:02:46
practice exam when I am done
writing this list of topics.

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00:02:46 --> 00:02:49
I think probably we actually
will skip this problem because I

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00:02:49 --> 00:02:51
think most of you know how to do
it.

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00:02:51 --> 00:02:57
And if not then you should run
for help from me or from your

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recitation instructor to figure
out how to do it.

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00:03:02 --> 00:03:10
The second topic that we saw
was cross-product.

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00:03:10 --> 00:03:19
When you have two vectors in
space, you can just form that

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00:03:19 --> 00:03:25
cross-product by computing its
determinant.

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00:03:25 --> 00:03:31
So, implicitly,
we should also know about

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determinants.
By that I mean two by two and

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00:03:35 --> 00:03:38
three by three.
Don't worry about larger ones,

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00:03:38 --> 00:03:43
even if you are interested,
they won't be on the test.

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00:03:43 --> 00:03:49
And applications of
cross-product,

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for example,
finding the area of a triangle

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or a parallelogram in space.
If you have a triangle in space

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00:04:04 --> 00:04:10
with sides A and B then its area
is one-half of the length of A

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00:04:10 --> 00:04:14
cross B.
Because the length of A cross B

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00:04:14 --> 00:04:19
is length A, length B sine
theta, which is the same as the

48
00:04:19 --> 00:04:24
area of the parallelogram formed
by these two vectors.

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00:04:24 --> 00:04:28
And the other application of
cross-product is to find a

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vector that's perpendicular to
two given vectors.

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In particular,
to find the vector that is

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00:04:36 --> 00:04:42
normal to a plane and then find
the equation of a plane.

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00:04:42 --> 00:04:57
Another application is finding
the normal vector to a plane and

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using that finding the equation
of a plane.

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00:05:08 --> 00:05:16
Basically, remember,
to find the equation of a

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plane, ax by cz = d,
what you need is the normal

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vector to the plane.
And the components of the

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normal vector are exactly the
coefficients that go into this.

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And we have seen an argument
for why that happens to be the

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case.
To find a normal vector to a

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plane typically what we will do
is take two vectors that lie in

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the plane and will take their
cross-product.

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And the cross-product will
automatically be perpendicular

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to both of them.
We are going to see an example

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of that when we look at problem
5 in practice 1A.

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I think we will try to do that
one.

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Another application,
well, we will just mention it

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as a topic that goes along with
this one.

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We have seen also about
equations of lines and how to

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find where a line intersects a
plane.

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Just to refresh your memories,
the equation of a line,

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well, we will be looking at
parametric equations.

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To know the parametric equation
of a line, we need to know a

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point on the line and we need to
know a vector that is parallel

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to the line.
And, if we know a point on the

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line and a vector along the
line,

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00:07:06 --> 00:07:11
then we can express the
parametric equations for the

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motion of a point that is moving
on the line.

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Actually, starting at point,
at time zero,

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and moving with velocity v.

81
00:07:21 --> 00:07:38

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00:07:38 --> 00:07:44
To put things in symbolic form,
you will get a position of that

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00:07:44 --> 00:07:48
point by starting with a
position of time zero and adding

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00:07:48 --> 00:07:53
t times the vector v.
It gives you x,

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y and z in terms of t.
And that is how we represent

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lines.
We will look at problem 5 in a

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00:08:06 --> 00:08:15
bit, but any general questions
about these topics?

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00:08:15 --> 00:08:20
No.
Do you have a question?

89
00:08:20 --> 00:08:26
Do we have to know Taylor
series?

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00:08:26 --> 00:08:34
That is a good question.
No, not on the exam.

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00:08:34 --> 00:08:36
[APPLAUSE]
Taylor series are something you

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00:08:36 --> 00:08:38
should be aware of,
generally speaking.

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00:08:38 --> 00:08:40
It will be useful for you in
real life,

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00:08:40 --> 00:08:44
probably not when you go to the
supermarket,

95
00:08:44 --> 00:08:48
but if you solve engineering
problems you will need Taylor

96
00:08:48 --> 00:08:52
series.
It would be good not to forget

97
00:08:52 --> 00:08:58
them entirely,
but on the 18.02 exams they

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00:08:58 --> 00:09:03
probably won't be there.
Let me continue with more

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00:09:03 --> 00:09:06
topics.
And then we can see if you can

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00:09:06 --> 00:09:11
think of other topics that
should or should not be on the

101
00:09:11 --> 00:09:22
exam.
Third topics would be matrices,

102
00:09:22 --> 00:09:34
linear systems,
inverting matrices.

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00:09:34 --> 00:09:37
I know that most of you that
have calculators that can invert

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00:09:37 --> 00:09:40
matrices, but still you are
expected at this point to know

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00:09:40 --> 00:09:42
how to do it by hand.
If you have looked at the

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00:09:42 --> 00:09:44
practice tests,
both of them have a problem

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00:09:44 --> 00:09:47
that asks you to invert a matrix
or at least do part of it.

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00:09:47 --> 00:09:53
And so it is very likely that
tomorrow there will be a problem

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00:09:53 --> 00:09:57
like that as well.
In general, when a kind of

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00:09:57 --> 00:10:00
problem is on both practice
tests it's a good indication

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00:10:00 --> 00:10:04
that it might be there also on
the actual exam.

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00:10:04 --> 00:10:08
Unfortunately,
not with the same matrix so you

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00:10:08 --> 00:10:12
cannot learn the answer by
heart.

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00:10:12 --> 00:10:17
Another thing that we have
learned about,

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00:10:17 --> 00:10:24
well, I should say this is
going to be problem 3 on the

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00:10:24 --> 00:10:29
test and will on the practice
test.

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00:10:29 --> 00:10:33
On the actual test,
too, I think,

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00:10:33 --> 00:10:37
actually.
Anyway, we will come back to it

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00:10:37 --> 00:10:39
later.
A couple of things that you

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00:10:39 --> 00:10:42
should remember.
If you have a system of the

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00:10:42 --> 00:10:46
form AX equals B then there are
two cases.

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00:10:46 --> 00:10:51
If a determinant of A is not
zero then that means you can

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00:10:51 --> 00:10:56
compute the inverse matrix and
you can just solve by taking A

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00:10:56 --> 00:11:02
inverse times B.
And the other case is when the

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00:11:02 --> 00:11:11
determinant of A is zero,
and there is either no solution

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00:11:11 --> 00:11:18
or there is infinitely many
solutions.

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00:11:18 --> 00:11:20
In particular,
if you know that there is a

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00:11:20 --> 00:11:21
solution,
for example,

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00:11:21 --> 00:11:24
if B is zero there always is an
obvious solution,

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00:11:24 --> 00:11:28
X equals zero,
then you will actually have

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00:11:28 --> 00:11:30
infinitely many.
In general, we don't really

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00:11:30 --> 00:11:33
know how to tell whether it is
no solution or infinitely many.

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00:11:33 --> 00:11:43
 
 

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00:11:43 --> 00:11:52
Questions about that?
Yes?

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00:11:52 --> 00:11:58
Will we have to know how to
rotate vectors and so on?

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00:11:58 --> 00:12:01
Not in general, 
but you might still want to

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00:12:01 --> 00:12:04
remember how to rotate a vector
in a plane by 90 degrees because

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00:12:04 --> 00:12:07
that has been useful when we
have done problems about

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00:12:07 --> 00:12:10
parametric equations,
which is what I am coming to

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00:12:10 --> 00:12:12
next.
What we have seen about

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00:12:12 --> 00:12:15
rotation matrices,
that was the homework part B

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00:12:15 --> 00:12:17
problem,
you are not supposed to

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00:12:17 --> 00:12:21
remember by heart everything
that was in part B of your

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00:12:21 --> 00:12:24
homework.
It is a good idea to have some

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00:12:24 --> 00:12:27
vague knowledge because it is
useful culture,

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00:12:27 --> 00:12:28
I would say,
useful background for later in

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00:12:28 --> 00:12:33
your lives,
but I won't ask you by heart to

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00:12:33 --> 00:12:40
know what is the formation for a
rotation matrix.

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00:12:40 --> 00:12:49
And then we come to,
last by not least,

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00:12:49 --> 00:13:00
the problem of finding
parametric equations.

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00:13:00 --> 00:13:03
And, in particular,
possibly by decomposing the

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00:13:03 --> 00:13:06
position vector into a sum of
simpler vectors.

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00:13:06 --> 00:13:13
You have seen quite an evil
exam of that on the last problem

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00:13:13 --> 00:13:19
set with this picture that maybe
by now you have had some

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00:13:19 --> 00:13:23
nightmares about.
Anyway, the one on the exam

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00:13:23 --> 00:13:27
will certainly be easier than
that.

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00:13:27 --> 00:13:36
But, as you have seen -- I
mean, you should know,

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00:13:36 --> 00:13:40
basically,
how to analyze a motion that is

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00:13:40 --> 00:13:44
being described to you and
express it in terms of vectors

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00:13:44 --> 00:13:48
and then figure out what the
parametric equation will be.

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00:13:48 --> 00:13:52
Now, again, it won't be as
complicated on the exam as the

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00:13:52 --> 00:13:57
one in the problem set.
But there are a couple of those

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00:13:57 --> 00:14:01
on the practice exam,
so that gives you an idea of

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00:14:01 --> 00:14:04
what is realistically expected
of you.

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00:14:04 --> 00:14:09
And now once we have parametric
equations for motion,

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00:14:09 --> 00:14:13
so that means when we know how
to find the position vector as a

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00:14:13 --> 00:14:15
function of a parameter maybe of
time,

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00:14:15 --> 00:14:24
then we have seen also about
velocity and acceleration,

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00:14:24 --> 00:14:28
which the vector is obtained by
taking the first and second

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00:14:28 --> 00:14:31
derivatives of a position
vector.

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00:14:31 --> 00:14:41
And so one topic that I will
add in there as well is somehow

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how to prove things about
motions by differentiating

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00:14:50 --> 00:14:55
vector identities.
One example of that,

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00:14:55 --> 00:14:58
for example,
is when we try to look at

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00:14:58 --> 00:15:03
Kepler's law in class last time.
We look at Kepler's second law

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00:15:03 --> 00:15:06
of planetary motion,
and we reduced it to a

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00:15:06 --> 00:15:10
calculation about a derivative
of the cross-product R cross v.

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00:15:10 --> 00:15:14
Now, on the exam you don't need
to know the details of Kepler's

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00:15:14 --> 00:15:16
law,
but you need to be able to

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00:15:16 --> 00:15:22
manipulate vector quantities a
bit in the way that we did.

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00:15:22 --> 00:15:27
And so on practice exam 1A, 
you actually have a variety of

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00:15:27 --> 00:15:30
problems on this topics because
you have problems two,

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00:15:30 --> 00:15:36
four and six,
all about parametric motions.

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00:15:36 --> 00:15:42
Probably tomorrow there will
not be three distinct problems

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00:15:42 --> 00:15:48
about parametric motions,
but maybe a couple of them.

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00:15:48 --> 00:15:51
I think that is basically the
list of topics.

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00:15:51 --> 00:15:54
Anybody spot something that I
have forgotten to put on the

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00:15:54 --> 00:15:58
exam or questions about
something that should or should

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00:15:58 --> 00:16:03
not be there?
You go first.

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00:16:03 --> 00:16:11
Yeah?
How about parametrizing weird

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00:16:11 --> 00:16:18
trigonometric functions?
I am not sure what you mean by

192
00:16:18 --> 00:16:21
that.
Well, parametric curves,

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00:16:21 --> 00:16:25
you need to know how to
parameterize motions,

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00:16:25 --> 00:16:30
and that involves a little bit
of trigonometrics.

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00:16:30 --> 00:16:33
When we have seen these
problems about rotating wheels,

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00:16:33 --> 00:16:34
say the cycloid,
for example,

197
00:16:34 --> 00:16:38
and so on there is a bit of
cosine and sine and so on.

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00:16:38 --> 00:16:41
I think not much more on that.
You won't need obscure

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00:16:41 --> 00:16:48
trigonometric identities.
You're next.

200
00:16:48 --> 00:16:51
Any proofs on the exam or just
like problems?

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00:16:51 --> 00:16:55
Well, a problem can ask you to
show things.

202
00:16:55 --> 00:16:58
It is not going to be a
complicated proof.

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00:16:58 --> 00:17:00
The proofs are going to be
fairly easy.

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00:17:00 --> 00:17:04
If you look at practice 1A,
the last problem does have a

205
00:17:04 --> 00:17:08
little bit of proof.
6B says that show that blah,

206
00:17:08 --> 00:17:11
blah, blah.
But, as you will see,

207
00:17:11 --> 00:17:13
it is not a difficult kind of
proof.

208
00:17:13 --> 00:17:24
So, about the same.
Yes?

209
00:17:24 --> 00:17:29
Are there equations of 3D
shapes that we should know at

210
00:17:29 --> 00:17:32
this point?
We should know definitely a lot

211
00:17:32 --> 00:17:34
about the equations of planes on
lines.

212
00:17:34 --> 00:17:37
And you should probably know
that a sphere centered at the

213
00:17:37 --> 00:17:40
origin is the set of points
where distance to the center is

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00:17:40 --> 00:17:43
equal to the radius of the
sphere.

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00:17:43 --> 00:17:45
We don't need more at this
point.

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00:17:45 --> 00:17:48
As the semester goes on,
we will start seeing cones and

217
00:17:48 --> 00:17:52
things like that.
But at this point planes, lines.

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00:17:52 --> 00:17:58
And maybe you need to know
about circles and spheres,

219
00:17:58 --> 00:18:03
but nothing beyond that.
More questions?

220
00:18:03 --> 00:18:11
Yes?
If there is a formula that you

221
00:18:11 --> 00:18:14
have proved on the homework
then, yes, you can assume it on

222
00:18:14 --> 00:18:17
the test.
Maybe you want to write on your

223
00:18:17 --> 00:18:20
test that this is a formula you
have seen in homework just so

224
00:18:20 --> 00:18:24
that we know that you remember
it from homework and not from

225
00:18:24 --> 00:18:27
looking over your neighbor's
shoulder or whatever.

226
00:18:27 --> 00:18:31
Yes, it is OK to use things
that you know general-speaking.

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00:18:31 --> 00:18:33
That being said,
for example,

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00:18:33 --> 00:18:37
probably there will be a linear
system to solve.

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00:18:37 --> 00:18:40
It will say on the exam you are
supposed to solve that using

230
00:18:40 --> 00:18:43
matrices, not by elimination.
There are things like that.

231
00:18:43 --> 00:18:47
If a problem says solve by
using vector methods,

232
00:18:47 --> 00:18:51
things like that,
then try to use at least a

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00:18:51 --> 00:18:54
vector somewhere.
But, in general,

234
00:18:54 --> 00:18:58
you are allowed to use things
that you know.

235
00:18:58 --> 00:19:07
Yes?
Will we need to go from

236
00:19:07 --> 00:19:09
parametric equations to xy
equations?

237
00:19:09 --> 00:19:16
Well, let's say only if it is
very easy.

238
00:19:16 --> 00:19:19
If I give you a parametric
curve, sin t,

239
00:19:19 --> 00:19:25
sin t, then you should be able
to observe that it is on the

240
00:19:25 --> 00:19:29
line y equals x,
not beyond that.

241
00:19:29 --> 00:19:37
Yes?
Do we have to use -- Yes.

242
00:19:37 --> 00:19:40
I don't know if you will have
to use it, but certainly you

243
00:19:40 --> 00:19:43
should know a little bit about
the unit tangent vector.

244
00:19:43 --> 00:19:51
Just remember the main thing to
know that the unit tangent

245
00:19:51 --> 00:19:57
vector is velocity divided by
the speed.

246
00:19:57 --> 00:20:04
I mean there is not much more
to it when you think about it.

247
00:20:04 --> 00:20:11
Yes?
Kepler's law,

248
00:20:11 --> 00:20:13
well, you are allowed to use it
if it helps you,

249
00:20:13 --> 00:20:16
if you find a way to squeeze it
in.

250
00:20:16 --> 00:20:20
You don't have to know Kepler's
law in detail.

251
00:20:20 --> 00:20:22
You just have to know how to
reproduce the general steps.

252
00:20:22 --> 00:20:28
If I tell you R cross v is
constant, you might be expected

253
00:20:28 --> 00:20:33
to know what to do with that.
I would say -- Basically,

254
00:20:33 --> 00:20:36
you don't need to know Kepler's
law.

255
00:20:36 --> 00:20:38
You need to know the kind of
stuff that we saw when we

256
00:20:38 --> 00:20:41
derived it such as how to take
the derivative of a dot-product

257
00:20:41 --> 00:20:52
or a cross-product.
That is basically the answer.

258
00:20:52 --> 00:20:55
I don't see any questions
anymore.

259
00:20:55 --> 00:21:03
Oh, you are raising your hand.
Yes.

260
00:21:03 --> 00:21:06
How to calculate the distance
between two lines and the

261
00:21:06 --> 00:21:08
distance between two planes?
Well, you have seen,

262
00:21:08 --> 00:21:10
probably recently,
that it is quite painful to do

263
00:21:10 --> 00:21:13
in general.
And, no, I don't think that

264
00:21:13 --> 00:21:16
will be on the exam by itself.
You need to know how to compute

265
00:21:16 --> 00:21:19
the distance between two points.
That certainly you need to know.

266
00:21:19 --> 00:21:24
And also maybe how to find the
compliment of a vector in a

267
00:21:24 --> 00:21:28
certain direction.
And that is about it,

268
00:21:28 --> 00:21:33
I would say.
I mean the more you know about

269
00:21:33 --> 00:21:38
things the better.
Things that come up on part Bs

270
00:21:38 --> 00:21:43
of the problem sets are
interesting things,

271
00:21:43 --> 00:21:48
but they are usually not needed
on the exams.

272
00:21:48 --> 00:21:53
If you have more questions then
you are not raising your hand

273
00:21:53 --> 00:21:55
high enough for me to see it.
OK.

274
00:21:55 --> 00:22:00
Let's try to do a bit of this
practice exam 1A.

275
00:22:00 --> 00:22:03
Hopefully, everybody has it.
If you don't have it,

276
00:22:03 --> 00:22:07
hopefully your neighbor has it.
If you don't have it and your

277
00:22:07 --> 00:22:09
neighbor doesn't have it then
please raise your hand.

278
00:22:09 --> 00:22:10
I have a couple.

279
00:22:10 --> 00:22:57

280
00:22:57 --> 00:23:02
If you neighbor has it then
just follow with them for now.

281
00:23:02 --> 00:23:05
I think there are a few people
behind you over there.

282
00:23:05 --> 00:23:06
I will stop handing them out
now.

283
00:23:06 --> 00:23:11
If you really need one,
it is on the website,

284
00:23:11 --> 00:23:15
it will be here at the end of
class.

285
00:23:15 --> 00:23:23
Let's see.
Well, I think we are going to

286
00:23:23 --> 00:23:27
just skip problems 1 and 2
because they are pretty

287
00:23:27 --> 00:23:32
straightforward and I hope that
you know how to do them.

288
00:23:32 --> 00:23:37
I mean I don't know.
Let's see.

289
00:23:37 --> 00:23:42
How many of you have no problem
with problem 1?

290
00:23:42 --> 00:23:46
How many of you have trouble
with problem 1?

291
00:23:46 --> 00:23:50
OK.
How many of you haven't raised

292
00:23:50 --> 00:23:52
your hands?
OK.

293
00:23:52 --> 00:23:56
How many of you have trouble
with problem 2?

294
00:23:56 --> 00:23:58
OK.
Well, if you have questions

295
00:23:58 --> 00:24:01
about those, maybe you should
just come see me at the end

296
00:24:01 --> 00:24:04
because that is probably more
efficient that way.

297
00:24:04 --> 00:24:11
I am going to start right away
with problem 3,

298
00:24:11 --> 00:24:18
actually.
Problem 3 says we have a matrix

299
00:24:18 --> 00:24:25
given to us |1 3 2;
2 0 - 1; 1 1 0|.

300
00:24:25 --> 00:24:30
And it tells us determinant of
A is 2 and inverse equals

301
00:24:30 --> 00:24:34
something, but we are missing
two values A and B and we are

302
00:24:34 --> 00:24:41
supposed to find them.
That means we need to do the

303
00:24:41 --> 00:24:51
steps of the algorithm to find
the inverse of A.

304
00:24:51 --> 00:25:00
We are told that A inverse is
one-half of |1 ...

305
00:25:00 --> 00:25:06
...; - 1 - 2 5;
2 2 - 6|.

306
00:25:06 --> 00:25:10
And here there are two unknown
values.

307
00:25:10 --> 00:25:15
Remember, to invert a matrix,
first we compute the minors.

308
00:25:15 --> 00:25:17
Then we flip some signs to get
the cofactors.

309
00:25:17 --> 00:25:19
Then we transpose.
And, finally,

310
00:25:19 --> 00:25:23
we divide by the determinant.
Let's try to be smart about

311
00:25:23 --> 00:25:26
this.
Do we need to compute all nine

312
00:25:26 --> 00:25:27
minors?
No.

313
00:25:27 --> 00:25:30
We only need to compute two of
them, right?

314
00:25:30 --> 00:25:34
Which minors do we need to
compute?

315
00:25:34 --> 00:25:39
Here and here or here and here?
Yeah, that looks better.

316
00:25:39 --> 00:25:43
Because, remember,
we need to transpose things so

317
00:25:43 --> 00:25:48
these two guys will end up here.
I claim we should compute these

318
00:25:48 --> 00:25:51
two minors.
And we will see if that is good

319
00:25:51 --> 00:25:52
enough.
If you start doing others and

320
00:25:52 --> 00:25:56
you find that they don't end up
in the right place then just do

321
00:25:56 --> 00:25:58
more,
but you don't need to spend

322
00:25:58 --> 00:26:00
your time computing all nine of
them.

323
00:26:00 --> 00:26:03
If you are worried about not
doing it right then,

324
00:26:03 --> 00:26:06
of course, you can maybe
compute one or two more to just

325
00:26:06 --> 00:26:11
double-check your answers.
But let us just do those that

326
00:26:11 --> 00:26:16
we think are needed.
The matrix of minors.

327
00:26:16 --> 00:26:22
The one that goes in the middle
position is obtained by deleting

328
00:26:22 --> 00:26:26
this row and that column,
and we are left with a

329
00:26:26 --> 00:26:32
determinant |3 2;1 0|,
3 times 0 minus 1 times 2

330
00:26:32 --> 00:26:40
should be - 2 should be - 2.
Then the one in the lower left

331
00:26:40 --> 00:26:45
corner, we delete the last row
and the first column,

332
00:26:45 --> 00:26:47
we are left with |3 2;
0 - 1|.

333
00:26:47 --> 00:26:50
3 times (- 1) is negative 3
minus 0.

334
00:26:50 --> 00:26:58
We are still left with negative
three.

335
00:26:58 --> 00:27:08
Is that step clear for everyone?
Then we need to go to cofactors.

336
00:27:08 --> 00:27:10
That means we need to change
signs.

337
00:27:10 --> 00:27:24
The rule is -- We change signs
in basically these four places.

338
00:27:24 --> 00:27:32
That means we will be left with
positive 2 and negative 3.

339
00:27:32 --> 00:27:44
Then we take the transpose.
That means the first column

340
00:27:44 --> 00:27:49
will copy into the first row,
so this guy we still don't

341
00:27:49 --> 00:27:52
know,
but here we will have two and

342
00:27:52 --> 00:27:59
here we will have minus three.
Finally, we have to divide by

343
00:27:59 --> 00:28:05
the determinant of A.
And here we are actually told

344
00:28:05 --> 00:28:08
that the determinant of A is
two.

345
00:28:08 --> 00:28:12
So we will divide by two.
But there is only one-half here

346
00:28:12 --> 00:28:18
so actually it is done for us.
The values that we will put up

347
00:28:18 --> 00:28:22
there are going to be 2 and
negative 3.

348
00:28:22 --> 00:28:30
 
 

349
00:28:30 --> 00:28:38
Now let's see how we use that
to solve a linear system.

350
00:28:38 --> 00:28:46
If we have to solve a linear
system, Ax equals B,

351
00:28:46 --> 00:28:50
well, if the matrix is
invertible, its determinant is

352
00:28:50 --> 00:28:54
not zero,
so we can certainly write x

353
00:28:54 --> 00:29:01
equals A inverse B.
So we have to multiply,

354
00:29:01 --> 00:29:09
that is one-half | 1 2 - 3;
- 1 - 2 5; 2 2 - 6|.

355
00:29:09 --> 00:29:14
Times B [
1, - 2,1].

356
00:29:14 --> 00:29:17
Remember, to do a matrix
multiplication you take the rows

357
00:29:17 --> 00:29:21
in here, the columns in here and
you do dot-products.

358
00:29:21 --> 00:29:25
The first entry will be one
times one plus two times minus

359
00:29:25 --> 00:29:30
two plus minus three times one,
one minus four minus three

360
00:29:30 --> 00:29:35
should be negative six,
except I still have,

361
00:29:35 --> 00:29:43
of course, a one-half in front.
Then minus one plus four plus

362
00:29:43 --> 00:29:50
five should be 8.
Two minus four minus six should

363
00:29:50 --> 00:29:57
be -8.
That will simplify to [- 3,4,

364
00:29:57 --> 00:30:04
- 5].
Any questions about that?

365
00:30:04 --> 00:30:08
OK.
Now we come to part C which is

366
00:30:08 --> 00:30:13
the harder part of this problem.
It says let's take this matrix

367
00:30:13 --> 00:30:17
A and let's replace the two in
the upper right corner by some

368
00:30:17 --> 00:30:25
other number C.
That means we will look at 1 3

369
00:30:25 --> 00:30:34
C; 2 0 - 1; 1 1 0|.
And let's call that M.

370
00:30:34 --> 00:30:40
And it first asks you to find
the value of C for which this

371
00:30:40 --> 00:30:48
matrix is not invertible.
M is not invertible exactly

372
00:30:48 --> 00:30:53
when the determinant of M is
zero.

373
00:30:53 --> 00:31:05
Let's compute the determinant.
Well, we should do one times

374
00:31:05 --> 00:31:12
that smaller determinant,
which is zero minus negative

375
00:31:12 --> 00:31:16
one,
which is 1 times 1 minus three

376
00:31:16 --> 00:31:23
times that determinant,
which is zero plus one is 1.

377
00:31:23 --> 00:31:28
And then we have plus C times
the lower left determinant which

378
00:31:28 --> 00:31:31
is two times one minus zero is
2.

379
00:31:31 --> 00:31:39
That gives us one minus three
is - 2 2C.

380
00:31:39 --> 00:31:47
That is zero when C equals 1.
For C equals 1,

381
00:31:47 --> 00:31:53
this matrix is not invertible.
For other values it is

382
00:31:53 --> 00:31:58
invertible.
It goes on to say let's look at

383
00:31:58 --> 00:32:04
this value of C and let's look
at the system Mx equals zero.

384
00:32:04 --> 00:32:14
I am going to put value one in
there.

385
00:32:14 --> 00:32:20
Now, if we look at Mx equals
zero, well, this has either no

386
00:32:20 --> 00:32:24
solution or infinitely many
solutions.

387
00:32:24 --> 00:32:26
But here there is an obvious
solution.

388
00:32:26 --> 00:32:30
Namely x equals zero is a
solution.

389
00:32:30 --> 00:32:35
Maybe let me rewrite it more
geometrically.

390
00:32:35 --> 00:32:44
X 3 y z = 0.
2x - z = 0.

391
00:32:44 --> 00:32:52
And x y = 0.
You see we have an obvious

392
00:32:52 --> 00:32:54
solution, (0,0,
0).

393
00:32:54 --> 00:32:57
But we have more solutions.
How do we find more solutions?

394
00:32:57 --> 00:33:01
Well, (x, y,
z) is a solution if it is in

395
00:33:01 --> 00:33:06
all three of these planes.
That is a way to think about it.

396
00:33:06 --> 00:33:13
Probably we are actually in
this situation where,

397
00:33:13 --> 00:33:20
in fact, we have three planes
that are all passing through the

398
00:33:20 --> 00:33:25
origin and all parallel to the
same line.

399
00:33:25 --> 00:33:28
And so that would be the line
of solutions.

400
00:33:28 --> 00:33:32
To find it actually we can
think of this as follows.

401
00:33:32 --> 00:33:37
The first observation is that
actually in this situation we

402
00:33:37 --> 00:33:41
don't need all three equations.
The fact that the system has

403
00:33:41 --> 00:33:45
infinitely many solutions means
that actually one of the

404
00:33:45 --> 00:33:49
equations is redundant.
If you look at it long enough

405
00:33:49 --> 00:33:51
you will see,
for example,

406
00:33:51 --> 00:33:55
if you multiply three times
this equation and you subtract

407
00:33:55 --> 00:33:58
that one then you will get the
first equation.

408
00:33:58 --> 00:34:05
Three times (x y) - (2x - z)
will be x 3y z.

409
00:34:05 --> 00:34:08
Now, we don't actually need to
see that to solve a problem.

410
00:34:08 --> 00:34:10
I am just showing you that is
what happens when you have a

411
00:34:10 --> 00:34:14
matrix with determinant zero.
One of the equations is somehow

412
00:34:14 --> 00:34:19
a duplicate of the others.
We don't actually need to

413
00:34:19 --> 00:34:24
figure out how exactly.
What that means is really we

414
00:34:24 --> 00:34:28
want to solve,
let's say start with two of the

415
00:34:28 --> 00:34:33
equations.
To find the solution we can

416
00:34:33 --> 00:34:41
observe that the first equation
says actually that <x,

417
00:34:41 --> 00:34:45
y, z> dot-product with

418
00:34:45 --> 00:34:46

419
00:34:46 --> 00:34:51
=0.
And the second equation says

420
00:34:51 --> 00:34:52
 
 

421
00:34:52 --> 00:34:57
dot-product with <2,0,-
1> is zero.

422
00:34:57 --> 00:35:03
And the third equation,
if we really want to keep it,

423
00:35:03 --> 00:35:08
says we should be also having
this.

424
00:35:08 --> 00:35:11
Now, these equations now
written like this,

425
00:35:11 --> 00:35:15
they are just saying we want an
x, y, z that is perpendicular to

426
00:35:15 --> 00:35:19
these vectors.
Let's forget this one and let's

427
00:35:19 --> 00:35:23
just look at these two.
They are saying we want a

428
00:35:23 --> 00:35:27
vector that is perpendicular to
these two given vectors.

429
00:35:27 --> 00:35:35
How do we find that?
We do the cross-product.

430
00:35:35 --> 00:35:43
To find x, y,
z perpendicular to <1,3,

431
00:35:43 --> 00:35:51
1> and <2,0,
- 1>, we take the

432
00:35:51 --> 00:35:56
cross-product.
And that will give us

433
00:35:56 --> 00:35:58
something.
Well, let me just give you the

434
00:35:58 --> 00:36:00
answer.
I am sure you know how to do

435
00:36:00 --> 00:36:07
cross-products by now.
I don't have the answer here,

436
00:36:07 --> 00:36:18
so I guess I have to do it.
That should be <- 3,

437
00:36:18 --> 00:36:26
probably positive 3,
and then - 6>.

438
00:36:26 --> 00:36:29
That is the solution.
And any multiple of that is a

439
00:36:29 --> 00:36:31
solution.
If you like to neatly simplify

440
00:36:31 --> 00:36:34
them you could say negative one,
one, negative two.

441
00:36:34 --> 00:36:37
If you like larger numbers you
can multiply that by a million.

442
00:36:37 --> 00:36:50
That is also a solution.
Any questions about that?

443
00:36:50 --> 00:36:56
Yes?
That is correct.

444
00:36:56 --> 00:37:00
If you pick these two guys
instead, you will get the same

445
00:37:00 --> 00:37:02
solution.
Well, up to a multiple.

446
00:37:02 --> 00:37:06
It could be if you do the
cross-product of these two guys

447
00:37:06 --> 00:37:10
you actually get something that
is a multiple -- Actually,

448
00:37:10 --> 00:37:14
I think if you do the
cross-product of the first and

449
00:37:14 --> 00:37:17
third one you will get actually
minus one, one,

450
00:37:17 --> 00:37:20
minus two, the smaller one.
But it doesn't matter.

451
00:37:20 --> 00:37:22
I mean it is really in the same
direction.

452
00:37:22 --> 00:37:27
This is all because a plane has
actually normal vectors of all

453
00:37:27 --> 00:37:32
sizes.
Yes?

454
00:37:32 --> 00:37:35
I don't think so because -- An
important thing to remember

455
00:37:35 --> 00:37:38
about cross-product is we
compute for minors,

456
00:37:38 --> 00:37:40
but then we put a minus sign on
the second component.

457
00:37:40 --> 00:37:44
The coefficient of j in here,
the second component,

458
00:37:44 --> 00:37:47
you do one times minus two
times one.

459
00:37:47 --> 00:37:51
That is negative three indeed.
But then you actually change

460
00:37:51 --> 00:37:54
that to a positive three.
Yes?

461
00:37:54 --> 00:38:14
 
 

462
00:38:14 --> 00:38:21
Well, we don't have parametric
equations here.

463
00:38:21 --> 00:38:25
Oh, solving by elimination.
Well, if it says that you have

464
00:38:25 --> 00:38:29
to use vector methods then you
should use vector methods.

465
00:38:29 --> 00:38:32
If it says you should use
vectors and matrices then you

466
00:38:32 --> 00:38:41
are expected to do it that way.
Yes?

467
00:38:41 --> 00:38:43
It depends what the problem is
asking.

468
00:38:43 --> 00:38:45
The question is,
is it enough to find the

469
00:38:45 --> 00:38:47
components of a vector or do we
have to find the equation of a

470
00:38:47 --> 00:38:50
line?
Here it says find one solution

471
00:38:50 --> 00:38:55
using vector operations.
We have found one solution.

472
00:38:55 --> 00:38:58
If you wanted to find the line
then it would all the things

473
00:38:58 --> 00:39:04
that are proportional to this.
It would be maybe minus 3t,

474
00:39:04 --> 00:39:09
3t minus 6t,
all the multiples of that

475
00:39:09 --> 00:39:12
vector.
We do because (0,0,

476
00:39:12 --> 00:39:17
0) is an obvious solution.
Maybe I should write that on

477
00:39:17 --> 00:39:19
the board.
You had another question?

478
00:39:19 --> 00:39:28
 
 

479
00:39:28 --> 00:39:31
Not quite.
Let me re-explain first how we

480
00:39:31 --> 00:39:35
get all the solutions and why I
did that cross-product.

481
00:39:35 --> 00:40:08
 
 

482
00:40:08 --> 00:40:09
First of all,
why did I take that

483
00:40:09 --> 00:40:12
cross-product again?
I took that cross-product

484
00:40:12 --> 00:40:17
because I looked at my three
equations and I observed that my

485
00:40:17 --> 00:40:21
three equations can be
reformulated in terms of these

486
00:40:21 --> 00:40:25
dot-products saying that x,
y, z is actually perpendicular

487
00:40:25 --> 00:40:29
these guys and these guys have
normal vectors to the planes.

488
00:40:29 --> 00:40:33
Remember, to be in all three
planes it has to be

489
00:40:33 --> 00:40:36
perpendicular to the normal
vectors.

490
00:40:36 --> 00:40:40
That is how we got here.
And now, if we want something

491
00:40:40 --> 00:40:43
that is perpendicular to a bunch
of given vectors,

492
00:40:43 --> 00:40:45
well, to be perpendicular to
two vectors,

493
00:40:45 --> 00:40:48
an easy way to find one is to
take that cross-product.

494
00:40:48 --> 00:40:51
And, if you take any two of
them, you will get something

495
00:40:51 --> 00:40:54
that is the same up to scaling.
Now, what it means

496
00:40:54 --> 00:41:00
geometrically is that when we
have our three planes and they

497
00:41:00 --> 00:41:06
all actually contain the same
line -- And we know that is

498
00:41:06 --> 00:41:11
actually the smae case because
they all pass through the

499
00:41:11 --> 00:41:16
origin.
They pass through the origin

500
00:41:16 --> 00:41:20
because the constant terms are
just zero.

501
00:41:20 --> 00:41:26
What happens is that the normal
vectors to these planes are,

502
00:41:26 --> 00:41:29
in fact, all perpendicular to
that line.

503
00:41:29 --> 00:41:40
The normal vectors -- Say this
line is vertical.

504
00:41:40 --> 00:41:45
The normal vectors are all
going to be horizontal.

505
00:41:45 --> 00:41:49
Well, it is kind of hard to
draw.

506
00:41:49 --> 00:41:53
By taking the cross-product
between two normal vectors we

507
00:41:53 --> 00:42:01
found this direction.
Now, to find actually all the

508
00:42:01 --> 00:42:06
solutions.
What we know so far is that we

509
00:42:06 --> 00:42:11
have this direction <-3 3 -
6>.

510
00:42:11 --> 00:42:14
That is going to be parallel to
the line of intersections.

511
00:42:14 --> 00:42:19
Let me do it here,
for example,

512
00:42:19 --> 00:42:29
.
Now we have one particular

513
00:42:29 --> 00:42:34
solution.
0,0, 0.

514
00:42:34 --> 00:42:39
Actually, we have found another
one, too, which is <- 3,3,

515
00:42:39 --> 00:42:46
- 6>.
Anyway, if a line of solutions

516
00:42:46 --> 00:42:55
-- -- has parametric equation x
= - 3t, y = 3t,

517
00:42:55 --> 00:43:04
z = - 6t, anything proportional
to that.

518
00:43:04 --> 00:43:07
That is how we would find all
the solutions if we wanted them.

519
00:43:07 --> 00:43:18
 
 

520
00:43:18 --> 00:43:22
It is almost time.
I think I need to jump ahead to

521
00:43:22 --> 00:43:24
other problems.
Let's see.

522
00:43:24 --> 00:43:29
I think problem 4 you can
probably find for yourselves.

523
00:43:29 --> 00:43:32
It is a reasonably
straightforward parametric

524
00:43:32 --> 00:43:35
equation problem.
You just have to find the

525
00:43:35 --> 00:43:39
coordinates of point P.
And for that it is a very

526
00:43:39 --> 00:43:41
simple trick.
Problem 5.

527
00:43:41 --> 00:43:43
Find the area of a spaced
triangle.

528
00:43:43 --> 00:43:47
It sounds like a cross-product.
Find the equation of a plane

529
00:43:47 --> 00:43:50
also sounds like a
cross-product.

530
00:43:50 --> 00:43:53
And find the intersection of
this plane with a line means we

531
00:43:53 --> 00:43:56
find first the parametric
equation of the line and then we

532
00:43:56 --> 00:43:59
plug that into the equation of
the plane to get where they

533
00:43:59 --> 00:44:02
intersect.
Does that sound reasonable?

534
00:44:02 --> 00:44:06
Who is disparate about problem
5?

535
00:44:06 --> 00:44:15
OK. Let me repeat problem 5.
First part we need to find the

536
00:44:15 --> 00:44:19
area of a triangle.
And the way to do that is to

537
00:44:19 --> 00:44:22
just do one-half the length of a
cross-product.

538
00:44:22 --> 00:44:32
If we have three points,
P0, P1, P2 then maybe we can

539
00:44:32 --> 00:44:38
form vectors P0P1 and P0P2.
And, if we take that

540
00:44:38 --> 00:44:42
cross-product and take the
length of that and divide by

541
00:44:42 --> 00:44:46
two, that will give us the area
of a triangle.

542
00:44:46 --> 00:44:52
Here it turns out that this guy
is ,

543
00:44:52 --> 00:44:59
if I look at the solutions,
so you will end up with square

544
00:44:59 --> 00:45:06
root of 6 over 2.
The second is asking you for

545
00:45:06 --> 00:45:13
the equation of a plane
containing these three points.

546
00:45:13 --> 00:45:23
Well, first of all,
we know that a normal vector to

547
00:45:23 --> 00:45:34
the plane is going to be given
by this cross-product again.

548
00:45:34 --> 00:45:40
That means that the equation of
plane will be of a form x plus y

549
00:45:40 --> 00:45:45
plus 2z equals something.
If a coefficient is here it

550
00:45:45 --> 00:45:49
comes from the normal vector.
And to find what goes in the

551
00:45:49 --> 00:45:51
right-hand side,
we just plug in any of the

552
00:45:51 --> 00:45:55
points.
If you plug in P0,

553
00:45:55 --> 00:46:03
which is (2,1,
0) then two plus one seems like

554
00:46:03 --> 00:46:06
it is 3.
And, if you want to

555
00:46:06 --> 00:46:08
double-check your answer,
you can take P1 and P2 and

556
00:46:08 --> 00:46:15
check that you also get three.
It is a good way to check your

557
00:46:15 --> 00:46:18
answer.
Then the third part.

558
00:46:18 --> 00:46:25
We have a line parallel to the
vector v equals one,

559
00:46:25 --> 00:46:31
one, one through the point S,
which is (- 1,0,

560
00:46:31 --> 00:46:34
0).
That means you can find its

561
00:46:34 --> 00:46:37
parametric equation.
X will start at - 1,

562
00:46:37 --> 00:46:41
increases at rate 1.
Y starts at zero,

563
00:46:41 --> 00:46:44
increases at rate one.
Z starts at zero,

564
00:46:44 --> 00:46:48
increases at rate one.
You plug these into the plane

565
00:46:48 --> 00:46:52
equation, and that will tell you
where they intersect.

566
00:46:52 --> 00:46:58
Is that clear?
And now, in the last one

567
00:46:58 --> 00:47:03
minute,
on that side I have one minute, 

568
00:47:03 --> 00:47:08
let me just say very quickly --
Well,

569
00:47:08 --> 00:47:10
do you want to hear about
problem 6 anyway very quickly?

570
00:47:10 --> 00:47:20
Yeah. OK.
Problem 6 is one of these like

571
00:47:20 --> 00:47:24
vector calculations.
It says we have a position

572
00:47:24 --> 00:47:28
vector R.
And it asks you how do we find

573
00:47:28 --> 00:47:32
the derivative of R dot R?
Well, remember we have a

574
00:47:32 --> 00:47:34
product rule for taking the
derivative.

575
00:47:34 --> 00:47:37
UV prime is U prime V plus UV
prime.

576
00:47:37 --> 00:47:44
It also applies for dot-product.
That is dR by dt dot R plus R

577
00:47:44 --> 00:47:49
dot dR by dt.
And these are both the same

578
00:47:49 --> 00:47:52
thing.
You get two R dot dR/dt,

579
00:47:52 --> 00:47:57
but dR/dt is v for velocity
vector.

580
00:47:57 --> 00:48:00
Hopefully you have seen things
like that.

581
00:48:00 --> 00:48:04
Now, it says show that if R has
constant length then they are

582
00:48:04 --> 00:48:10
perpendicular.
All you need to write basically

583
00:48:10 --> 00:48:15
is we assume length R is
constant.

584
00:48:15 --> 00:48:18
That is what it says,
R has constant length.

585
00:48:18 --> 00:48:21
Well, how do we get to,
say, something we probably want

586
00:48:21 --> 00:48:26
to reduce to that?
Well, if R is constant in

587
00:48:26 --> 00:48:32
length then R dot R is also
constant.

588
00:48:32 --> 00:48:39
And so that means d by dt of R
dot R is zero.

589
00:48:39 --> 00:48:41
That is what it means to be
constant.

590
00:48:41 --> 00:48:46
And so that means R dot v is
zero.

591
00:48:46 --> 00:48:52
That means R is perpendicular
to v.

592
00:48:52 --> 00:48:58
That is a proof.
It is not a scary proof.

593
00:48:58 --> 00:49:04
And then the last question of
the exam says let's continue to

594
00:49:04 --> 00:49:10
assume that R has constant
length, and let's try to find R

595
00:49:10 --> 00:49:13
dot v.
If there is acceleration then

596
00:49:13 --> 00:49:17
probably we should bring it in
somewhere, maybe by taking a

597
00:49:17 --> 00:49:21
derivative of something.
If we know that R dot v equals

598
00:49:21 --> 00:49:24
zero, let's take the derivative
of that.

599
00:49:24 --> 00:49:32
That is still zero.
But now, using the product

600
00:49:32 --> 00:49:40
rule, dR/dt is v dot v plus R
dot dv/dt is going to be zero.

601
00:49:40 --> 00:49:43
That means that you are asked
about R dot A.

602
00:49:43 --> 00:49:48
Well, that is equal to minus V
dot V.

603
00:49:48 --> 00:49:50
And that is it.
 

604
00:49:50 --> 00:49:55