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Let me just tell you first
about the list of topics.
Basically, the list of topics
is simple.
It is everything.
I mean, everything we have seen
so far is on the exam.
But let me just remind you of
the main topics that we have
seen.
First of all,
we learned about vectors,
how to use them,
and dot-product.
At this point,
you probably should know that
the dot-product of two vectors
is obtained by summing products
of components.
And geometrically it is the
length of A times the length of
B times the cosine of the angle
between them.
And, in particular,
we can use dot-product to
measure angles by solving for
cosine theta in this equality.
And most importantly to detect
whether two vectors are
perpendicular to each other.
Two vectors are perpendicular
when their dot-product is zero.
Any questions about that?
No.
Is everyone reasonably happy
with dot-product by now?
I see a stunned silence.
Nobody happy with dot-product
so far?
OK.
If you want to look at Practice
1A, a good example of a typical
problem with dot-product would
be problem 1.
Let's see.
We are going to go over the
practice exam when I am done
writing this list of topics.
I think probably we actually
will skip this problem because I
think most of you know how to do
it.
And if not then you should run
for help from me or from your
recitation instructor to figure
out how to do it.
The second topic that we saw
was cross-product.
When you have two vectors in
space, you can just form that
cross-product by computing its
determinant.
So, implicitly,
we should also know about
determinants.
By that I mean two by two and
three by three.
Don't worry about larger ones,
even if you are interested,
they won't be on the test.
And applications of
cross-product,
for example,
finding the area of a triangle
or a parallelogram in space.
If you have a triangle in space
with sides A and B then its area
is one-half of the length of A
cross B.
Because the length of A cross B
is length A, length B sine
theta, which is the same as the
area of the parallelogram formed
by these two vectors.
And the other application of
cross-product is to find a
vector that's perpendicular to
two given vectors.
In particular,
to find the vector that is
normal to a plane and then find
the equation of a plane.
Another application is finding
the normal vector to a plane and
using that finding the equation
of a plane.
Basically, remember,
to find the equation of a
plane, ax by cz = d,
what you need is the normal
vector to the plane.
And the components of the
normal vector are exactly the
coefficients that go into this.
And we have seen an argument
for why that happens to be the
case.
To find a normal vector to a
plane typically what we will do
is take two vectors that lie in
the plane and will take their
cross-product.
And the cross-product will
automatically be perpendicular
to both of them.
We are going to see an example
of that when we look at problem
5 in practice 1A.
I think we will try to do that
one.
Another application,
well, we will just mention it
as a topic that goes along with
this one.
We have seen also about
equations of lines and how to
find where a line intersects a
plane.
Just to refresh your memories,
the equation of a line,
well, we will be looking at
parametric equations.
To know the parametric equation
of a line, we need to know a
point on the line and we need to
know a vector that is parallel
to the line.
And, if we know a point on the
line and a vector along the
line,
then we can express the
parametric equations for the
motion of a point that is moving
on the line.
Actually, starting at point,
at time zero,
and moving with velocity v.
To put things in symbolic form,
you will get a position of that
point by starting with a
position of time zero and adding
t times the vector v.
It gives you x,
y and z in terms of t.
And that is how we represent
lines.
We will look at problem 5 in a
bit, but any general questions
about these topics?
No.
Do you have a question?
Do we have to know Taylor
series?
That is a good question.
No, not on the exam.
[APPLAUSE]
Taylor series are something you
should be aware of,
generally speaking.
It will be useful for you in
real life,
probably not when you go to the
supermarket,
but if you solve engineering
problems you will need Taylor
series.
It would be good not to forget
them entirely,
but on the 18.02 exams they
probably won't be there.
Let me continue with more
topics.
And then we can see if you can
think of other topics that
should or should not be on the
exam.
Third topics would be matrices,
linear systems,
inverting matrices.
I know that most of you that
have calculators that can invert
matrices, but still you are
expected at this point to know
how to do it by hand.
If you have looked at the
practice tests,
both of them have a problem
that asks you to invert a matrix
or at least do part of it.
And so it is very likely that
tomorrow there will be a problem
like that as well.
In general, when a kind of
problem is on both practice
tests it's a good indication
that it might be there also on
the actual exam.
Unfortunately,
not with the same matrix so you
cannot learn the answer by
heart.
Another thing that we have
learned about,
well, I should say this is
going to be problem 3 on the
test and will on the practice
test.
On the actual test,
too, I think,
actually.
Anyway, we will come back to it
later.
A couple of things that you
should remember.
If you have a system of the
form AX equals B then there are
two cases.
If a determinant of A is not
zero then that means you can
compute the inverse matrix and
you can just solve by taking A
inverse times B.
And the other case is when the
determinant of A is zero,
and there is either no solution
or there is infinitely many
solutions.
In particular,
if you know that there is a
solution,
for example,
if B is zero there always is an
obvious solution,
X equals zero,
then you will actually have
infinitely many.
In general, we don't really
know how to tell whether it is
no solution or infinitely many.
Questions about that?
Yes?
Will we have to know how to
rotate vectors and so on?
Not in general,
but you might still want to
remember how to rotate a vector
in a plane by 90 degrees because
that has been useful when we
have done problems about
parametric equations,
which is what I am coming to
next.
What we have seen about
rotation matrices,
that was the homework part B
problem,
you are not supposed to
remember by heart everything
that was in part B of your
homework.
It is a good idea to have some
vague knowledge because it is
useful culture,
I would say,
useful background for later in
your lives,
but I won't ask you by heart to
know what is the formation for a
rotation matrix.
And then we come to,
last by not least,
the problem of finding
parametric equations.
And, in particular,
possibly by decomposing the
position vector into a sum of
simpler vectors.
You have seen quite an evil
exam of that on the last problem
set with this picture that maybe
by now you have had some
nightmares about.
Anyway, the one on the exam
will certainly be easier than
that.
But, as you have seen -- I
mean, you should know,
basically,
how to analyze a motion that is
being described to you and
express it in terms of vectors
and then figure out what the
parametric equation will be.
Now, again, it won't be as
complicated on the exam as the
one in the problem set.
But there are a couple of those
on the practice exam,
so that gives you an idea of
what is realistically expected
of you.
And now once we have parametric
equations for motion,
so that means when we know how
to find the position vector as a
function of a parameter maybe of
time,
then we have seen also about
velocity and acceleration,
which the vector is obtained by
taking the first and second
derivatives of a position
vector.
And so one topic that I will
add in there as well is somehow
how to prove things about
motions by differentiating
vector identities.
One example of that,
for example,
is when we try to look at
Kepler's law in class last time.
We look at Kepler's second law
of planetary motion,
and we reduced it to a
calculation about a derivative
of the cross-product R cross v.
Now, on the exam you don't need
to know the details of Kepler's
law,
but you need to be able to
manipulate vector quantities a
bit in the way that we did.
And so on practice exam 1A,
you actually have a variety of
problems on this topics because
you have problems two,
four and six,
all about parametric motions.
Probably tomorrow there will
not be three distinct problems
about parametric motions,
but maybe a couple of them.
I think that is basically the
list of topics.
Anybody spot something that I
have forgotten to put on the
exam or questions about
something that should or should
not be there?
You go first.
Yeah?
How about parametrizing weird
trigonometric functions?
I am not sure what you mean by
that.
Well, parametric curves,
you need to know how to
parameterize motions,
and that involves a little bit
of trigonometrics.
When we have seen these
problems about rotating wheels,
say the cycloid,
for example,
and so on there is a bit of
cosine and sine and so on.
I think not much more on that.
You won't need obscure
trigonometric identities.
You're next.
Any proofs on the exam or just
like problems?
Well, a problem can ask you to
show things.
It is not going to be a
complicated proof.
The proofs are going to be
fairly easy.
If you look at practice 1A,
the last problem does have a
little bit of proof.
6B says that show that blah,
blah, blah.
But, as you will see,
it is not a difficult kind of
proof.
So, about the same.
Yes?
Are there equations of 3D
shapes that we should know at
this point?
We should know definitely a lot
about the equations of planes on
lines.
And you should probably know
that a sphere centered at the
origin is the set of points
where distance to the center is
equal to the radius of the
sphere.
We don't need more at this
point.
As the semester goes on,
we will start seeing cones and
things like that.
But at this point planes, lines.
And maybe you need to know
about circles and spheres,
but nothing beyond that.
More questions?
Yes?
If there is a formula that you
have proved on the homework
then, yes, you can assume it on
the test.
Maybe you want to write on your
test that this is a formula you
have seen in homework just so
that we know that you remember
it from homework and not from
looking over your neighbor's
shoulder or whatever.
Yes, it is OK to use things
that you know general-speaking.
That being said,
for example,
probably there will be a linear
system to solve.
It will say on the exam you are
supposed to solve that using
matrices, not by elimination.
There are things like that.
If a problem says solve by
using vector methods,
things like that,
then try to use at least a
vector somewhere.
But, in general,
you are allowed to use things
that you know.
Yes?
Will we need to go from
parametric equations to xy
equations?
Well, let's say only if it is
very easy.
If I give you a parametric
curve, sin t,
sin t, then you should be able
to observe that it is on the
line y equals x,
not beyond that.
Yes?
Do we have to use -- Yes.
I don't know if you will have
to use it, but certainly you
should know a little bit about
the unit tangent vector.
Just remember the main thing to
know that the unit tangent
vector is velocity divided by
the speed.
I mean there is not much more
to it when you think about it.
Yes?
Kepler's law,
well, you are allowed to use it
if it helps you,
if you find a way to squeeze it
in.
You don't have to know Kepler's
law in detail.
You just have to know how to
reproduce the general steps.
If I tell you R cross v is
constant, you might be expected
to know what to do with that.
I would say -- Basically,
you don't need to know Kepler's
law.
You need to know the kind of
stuff that we saw when we
derived it such as how to take
the derivative of a dot-product
or a cross-product.
That is basically the answer.
I don't see any questions
anymore.
Oh, you are raising your hand.
Yes.
How to calculate the distance
between two lines and the
distance between two planes?
Well, you have seen,
probably recently,
that it is quite painful to do
in general.
And, no, I don't think that
will be on the exam by itself.
You need to know how to compute
the distance between two points.
That certainly you need to know.
And also maybe how to find the
compliment of a vector in a
certain direction.
And that is about it,
I would say.
I mean the more you know about
things the better.
Things that come up on part Bs
of the problem sets are
interesting things,
but they are usually not needed
on the exams.
If you have more questions then
you are not raising your hand
high enough for me to see it.
OK.
Let's try to do a bit of this
practice exam 1A.
Hopefully, everybody has it.
If you don't have it,
hopefully your neighbor has it.
If you don't have it and your
neighbor doesn't have it then
please raise your hand.
I have a couple.
If you neighbor has it then
just follow with them for now.
I think there are a few people
behind you over there.
I will stop handing them out
now.
If you really need one,
it is on the website,
it will be here at the end of
class.
Let's see.
Well, I think we are going to
just skip problems 1 and 2
because they are pretty
straightforward and I hope that
you know how to do them.
I mean I don't know.
Let's see.
How many of you have no problem
with problem 1?
How many of you have trouble
with problem 1?
OK.
How many of you haven't raised
your hands?
OK.
How many of you have trouble
with problem 2?
OK.
Well, if you have questions
about those, maybe you should
just come see me at the end
because that is probably more
efficient that way.
I am going to start right away
with problem 3,
actually.
Problem 3 says we have a matrix
given to us |1 3 2;
2 0 - 1; 1 1 0|.
And it tells us determinant of
A is 2 and inverse equals
something, but we are missing
two values A and B and we are
supposed to find them.
That means we need to do the
steps of the algorithm to find
the inverse of A.
We are told that A inverse is
one-half of |1 ...
...; - 1 - 2 5;
2 2 - 6|.
And here there are two unknown
values.
Remember, to invert a matrix,
first we compute the minors.
Then we flip some signs to get
the cofactors.
Then we transpose.
And, finally,
we divide by the determinant.
Let's try to be smart about
this.
Do we need to compute all nine
minors?
No.
We only need to compute two of
them, right?
Which minors do we need to
compute?
Here and here or here and here?
Yeah, that looks better.
Because, remember,
we need to transpose things so
these two guys will end up here.
I claim we should compute these
two minors.
And we will see if that is good
enough.
If you start doing others and
you find that they don't end up
in the right place then just do
more,
but you don't need to spend
your time computing all nine of
them.
If you are worried about not
doing it right then,
of course, you can maybe
compute one or two more to just
double-check your answers.
But let us just do those that
we think are needed.
The matrix of minors.
The one that goes in the middle
position is obtained by deleting
this row and that column,
and we are left with a
determinant |3 2;1 0|,
3 times 0 minus 1 times 2
should be - 2 should be - 2.
Then the one in the lower left
corner, we delete the last row
and the first column,
we are left with |3 2;
0 - 1|.
3 times (- 1) is negative 3
minus 0.
We are still left with negative
three.
Is that step clear for everyone?
Then we need to go to cofactors.
That means we need to change
signs.
The rule is -- We change signs
in basically these four places.
That means we will be left with
positive 2 and negative 3.
Then we take the transpose.
That means the first column
will copy into the first row,
so this guy we still don't
know,
but here we will have two and
here we will have minus three.
Finally, we have to divide by
the determinant of A.
And here we are actually told
that the determinant of A is
two.
So we will divide by two.
But there is only one-half here
so actually it is done for us.
The values that we will put up
there are going to be 2 and
negative 3.
Now let's see how we use that
to solve a linear system.
If we have to solve a linear
system, Ax equals B,
well, if the matrix is
invertible, its determinant is
not zero,
so we can certainly write x
equals A inverse B.
So we have to multiply,
that is one-half | 1 2 - 3;
- 1 - 2 5; 2 2 - 6|.
Times B [
1, - 2,1].
Remember, to do a matrix
multiplication you take the rows
in here, the columns in here and
you do dot-products.
The first entry will be one
times one plus two times minus
two plus minus three times one,
one minus four minus three
should be negative six,
except I still have,
of course, a one-half in front.
Then minus one plus four plus
five should be 8.
Two minus four minus six should
be -8.
That will simplify to [- 3,4,
- 5].
Any questions about that?
OK.
Now we come to part C which is
the harder part of this problem.
It says let's take this matrix
A and let's replace the two in
the upper right corner by some
other number C.
That means we will look at 1 3
C; 2 0 - 1; 1 1 0|.
And let's call that M.
And it first asks you to find
the value of C for which this
matrix is not invertible.
M is not invertible exactly
when the determinant of M is
zero.
Let's compute the determinant.
Well, we should do one times
that smaller determinant,
which is zero minus negative
one,
which is 1 times 1 minus three
times that determinant,
which is zero plus one is 1.
And then we have plus C times
the lower left determinant which
is two times one minus zero is
2.
That gives us one minus three
is - 2 2C.
That is zero when C equals 1.
For C equals 1,
this matrix is not invertible.
For other values it is
invertible.
It goes on to say let's look at
this value of C and let's look
at the system Mx equals zero.
I am going to put value one in
there.
Now, if we look at Mx equals
zero, well, this has either no
solution or infinitely many
solutions.
But here there is an obvious
solution.
Namely x equals zero is a
solution.
Maybe let me rewrite it more
geometrically.
X 3 y z = 0.
2x - z = 0.
And x y = 0.
You see we have an obvious
solution, (0,0,
0).
But we have more solutions.
How do we find more solutions?
Well, (x, y,
z) is a solution if it is in
all three of these planes.
That is a way to think about it.
Probably we are actually in
this situation where,
in fact, we have three planes
that are all passing through the
origin and all parallel to the
same line.
And so that would be the line
of solutions.
To find it actually we can
think of this as follows.
The first observation is that
actually in this situation we
don't need all three equations.
The fact that the system has
infinitely many solutions means
that actually one of the
equations is redundant.
If you look at it long enough
you will see,
for example,
if you multiply three times
this equation and you subtract
that one then you will get the
first equation.
Three times (x y) - (2x - z)
will be x 3y z.
Now, we don't actually need to
see that to solve a problem.
I am just showing you that is
what happens when you have a
matrix with determinant zero.
One of the equations is somehow
a duplicate of the others.
We don't actually need to
figure out how exactly.
What that means is really we
want to solve,
let's say start with two of the
equations.
To find the solution we can
observe that the first equation
says actually that <x,
y, z> dot-product with
=0.
And the second equation says
dot-product with  is zero.
And the third equation,
if we really want to keep it,
says we should be also having
this.
Now, these equations now
written like this,
they are just saying we want an
x, y, z that is perpendicular to
these vectors.
Let's forget this one and let's
just look at these two.
They are saying we want a
vector that is perpendicular to
these two given vectors.
How do we find that?
We do the cross-product.
To find x, y,
z perpendicular to <1,3,
1> and , we take the
cross-product.
And that will give us
something.
Well, let me just give you the
answer.
I am sure you know how to do
cross-products by now.
I don't have the answer here,
so I guess I have to do it.
That should be <- 3,
probably positive 3,
and then - 6>.
That is the solution.
And any multiple of that is a
solution.
If you like to neatly simplify
them you could say negative one,
one, negative two.
If you like larger numbers you
can multiply that by a million.
That is also a solution.
Any questions about that?
Yes?
That is correct.
If you pick these two guys
instead, you will get the same
solution.
Well, up to a multiple.
It could be if you do the
cross-product of these two guys
you actually get something that
is a multiple -- Actually,
I think if you do the
cross-product of the first and
third one you will get actually
minus one, one,
minus two, the smaller one.
But it doesn't matter.
I mean it is really in the same
direction.
This is all because a plane has
actually normal vectors of all
sizes.
Yes?
I don't think so because -- An
important thing to remember
about cross-product is we
compute for minors,
but then we put a minus sign on
the second component.
The coefficient of j in here,
the second component,
you do one times minus two
times one.
That is negative three indeed.
But then you actually change
that to a positive three.
Yes?
Well, we don't have parametric
equations here.
Oh, solving by elimination.
Well, if it says that you have
to use vector methods then you
should use vector methods.
If it says you should use
vectors and matrices then you
are expected to do it that way.
Yes?
It depends what the problem is
asking.
The question is,
is it enough to find the
components of a vector or do we
have to find the equation of a
line?
Here it says find one solution
using vector operations.
We have found one solution.
If you wanted to find the line
then it would all the things
that are proportional to this.
It would be maybe minus 3t,
3t minus 6t,
all the multiples of that
vector.
We do because (0,0,
0) is an obvious solution.
Maybe I should write that on
the board.
You had another question?
Not quite.
Let me re-explain first how we
get all the solutions and why I
did that cross-product.
First of all,
why did I take that
cross-product again?
I took that cross-product
because I looked at my three
equations and I observed that my
three equations can be
reformulated in terms of these
dot-products saying that x,
y, z is actually perpendicular
these guys and these guys have
normal vectors to the planes.
Remember, to be in all three
planes it has to be
perpendicular to the normal
vectors.
That is how we got here.
And now, if we want something
that is perpendicular to a bunch
of given vectors,
well, to be perpendicular to
two vectors,
an easy way to find one is to
take that cross-product.
And, if you take any two of
them, you will get something
that is the same up to scaling.
Now, what it means
geometrically is that when we
have our three planes and they
all actually contain the same
line -- And we know that is
actually the smae case because
they all pass through the
origin.
They pass through the origin
because the constant terms are
just zero.
What happens is that the normal
vectors to these planes are,
in fact, all perpendicular to
that line.
The normal vectors -- Say this
line is vertical.
The normal vectors are all
going to be horizontal.
Well, it is kind of hard to
draw.
By taking the cross-product
between two normal vectors we
found this direction.
Now, to find actually all the
solutions.
What we know so far is that we
have this direction .
That is going to be parallel to
the line of intersections.
Let me do it here,
for example,
.
Now we have one particular
solution.
0,0, 0.
Actually, we have found another
one, too, which is <- 3,3,
- 6>.
Anyway, if a line of solutions
-- -- has parametric equation x
= - 3t, y = 3t,
z = - 6t, anything proportional
to that.
That is how we would find all
the solutions if we wanted them.
It is almost time.
I think I need to jump ahead to
other problems.
Let's see.
I think problem 4 you can
probably find for yourselves.
It is a reasonably
straightforward parametric
equation problem.
You just have to find the
coordinates of point P.
And for that it is a very
simple trick.
Problem 5.
Find the area of a spaced
triangle.
It sounds like a cross-product.
Find the equation of a plane
also sounds like a
cross-product.
And find the intersection of
this plane with a line means we
find first the parametric
equation of the line and then we
plug that into the equation of
the plane to get where they
intersect.
Does that sound reasonable?
Who is disparate about problem
5?
OK. Let me repeat problem 5.
First part we need to find the
area of a triangle.
And the way to do that is to
just do one-half the length of a
cross-product.
If we have three points,
P0, P1, P2 then maybe we can
form vectors P0P1 and P0P2.
And, if we take that
cross-product and take the
length of that and divide by
two, that will give us the area
of a triangle.
Here it turns out that this guy
is ,
if I look at the solutions,
so you will end up with square
root of 6 over 2.
The second is asking you for
the equation of a plane
containing these three points.
Well, first of all,
we know that a normal vector to
the plane is going to be given
by this cross-product again.
That means that the equation of
plane will be of a form x plus y
plus 2z equals something.
If a coefficient is here it
comes from the normal vector.
And to find what goes in the
right-hand side,
we just plug in any of the
points.
If you plug in P0,
which is (2,1,
0) then two plus one seems like
it is 3.
And, if you want to
double-check your answer,
you can take P1 and P2 and
check that you also get three.
It is a good way to check your
answer.
Then the third part.
We have a line parallel to the
vector v equals one,
one, one through the point S,
which is (- 1,0,
0).
That means you can find its
parametric equation.
X will start at - 1,
increases at rate 1.
Y starts at zero,
increases at rate one.
Z starts at zero,
increases at rate one.
You plug these into the plane
equation, and that will tell you
where they intersect.
Is that clear?
And now, in the last one
minute,
on that side I have one minute,
let me just say very quickly --
Well,
do you want to hear about
problem 6 anyway very quickly?
Yeah. OK.
Problem 6 is one of these like
vector calculations.
It says we have a position
vector R.
And it asks you how do we find
the derivative of R dot R?
Well, remember we have a
product rule for taking the
derivative.
UV prime is U prime V plus UV
prime.
It also applies for dot-product.
That is dR by dt dot R plus R
dot dR by dt.
And these are both the same
thing.
You get two R dot dR/dt,
but dR/dt is v for velocity
vector.
Hopefully you have seen things
like that.
Now, it says show that if R has
constant length then they are
perpendicular.
All you need to write basically
is we assume length R is
constant.
That is what it says,
R has constant length.
Well, how do we get to,
say, something we probably want
to reduce to that?
Well, if R is constant in
length then R dot R is also
constant.
And so that means d by dt of R
dot R is zero.
That is what it means to be
constant.
And so that means R dot v is
zero.
That means R is perpendicular
to v.
That is a proof.
It is not a scary proof.
And then the last question of
the exam says let's continue to
assume that R has constant
length, and let's try to find R
dot v.
If there is acceleration then
probably we should bring it in
somewhere, maybe by taking a
derivative of something.
If we know that R dot v equals
zero, let's take the derivative
of that.
That is still zero.
But now, using the product
rule, dR/dt is v dot v plus R
dot dv/dt is going to be zero.
That means that you are asked
about R dot A.
Well, that is equal to minus V
dot V.
And that is it.
