Mathematics - Multivariable Calculus - Lecture 5

Uploaded by UCBerkeley on 17.11.2009

So, what are we doing here?

We're studying vectors in three-dimensional space.
So, last time, we talked about what vectors are.
And what we can do with them.
So what can we do with vectors?
We can take the sum of two vectors.
We get another vector.
It is on.

Let me see if I can adjust it.
By the way I did find batteries after I
finished my last lecture.
They were hidden in the cabinet.
So, don't make me do this.
You don't want that.

Let's see.
Like this?
But tell me if you lose me.
If I lose batteries.

All right.
We can take the sum of two vectors.
And that's a vector.
Or we can multiply a vector by a number.
That's a vector.
But, in addition, there are also more sophisticated
operations we can do on vectors.
For example, dot product.
Now, a dot product is a whole different deal.
Because the input of a dot product is two
different vectors.
But the output is a number.
The output is a number.
And I explained last time that there are two ways to define
the dot product of two vectors.
The first way is geometric.

We talk about the magnitudes of the two vectors and the cosine
of the angle between them.
Which we call theta.
So it's just a product of these three numbers.
But there is also a second way which is very
convenient in calculations.
Which is defined in terms of the components of the vectors.
And those components will be called x, y, and z.
So I put in this one for the first three components.
And and it's 2 for the components of the
second vector.
So the formula looks like this.
Up Now, we can play with this formula.
And we can derive useful information about vectors this
by using the fact that both formulas give us
the same answer.
For example, we can find the angle theta, the angle between
the two vectors, by combining these two formulas.
From the first formula, we learned that the cosine of this
angle is equal to the dot product divided by
the magnitude.
And now, we use the second formula.
So this is the first formula.
And now we use the second one two write this in the second
form, x, y, x1, x2 plus y1 y2 plus z1 z2 divided
by the length.
But the length can also be found from the components.
For example, the length of the first one is given
by this expression.
And likewise, for the length of the second expression.
So we can write this also very explicitly.
Using the components, or coordinates, if you will.
So you see, this is an unexpected formula in the
following sense: you have two vectors.
Which are given by these coordinates.
x1 y1, z1, and x2 y2 z2.
When we just look at these coordinates, we don't know
where exactly the vectors are.
We'd have to plot them in space to really understand
their relative position.
So it looks like it would be very difficult to find out what
the angle is between them.
But in fact, there is a very explicit formula, which uses
nothing but the expression for a and b in terms of components.
So this formula only uses x1, y1 z1, which are the components
of a, and x2, y2, and z2, which are the components of b.
And a priori, there's no reason to believe that there
would be such a formula.
But we were able to derive it by combining two different
definitions of dot product.
It's not like it's very simple, but not too complicated either.
It involves some square roots and so on, but
it is very explicit.
So, for example, you can program it on your computer and
then every time you, so your program on the computer would
be like that black box.
The dot product, for example, you can program your computer
to calcluate the dot product to get a number.
But you can also program your computer to calculate the
cosine of the angle, by simply calculating this.
And that's very simple.
Which can be done very quickly.
So that's the power of this method that we
are talking about.
That using this method, we could learn a lot about the
geometry of vectors, say, or other geometric objects.
By using, as input, al information such as the
components of this vector with respect to the
xyz coordinate system.
One obligation of this is the following.
There is a very special angle for which, special angle
theta, for which the cosine is equal to 0.
So, cosine theta is equal to 0 if theta is equal to 5/2.
So, actually, there's not just one but two different values.
5/2 and negative 5/2.

s So, if you have an angle like this, or like this, the
projection onto the x plane, which is the cosine,
is equal to 0.
So in this case, the dot product formula tells us
that the dot product is 0, in this case.
So, we can use it, you can use this formula, to determine
whether the vectors a and b are perpendicular to
each other or not.

When I say perpendicular, I mean that the angle between
them is 90 degrees.
But to say that the angle between them is 90 degrees,
well, it means that they are either like this or like this.
It doesn't matter.
In both cases they are perpendicular.

And in this case, a dot b is equal to 0.
So if you are given two vectors, it's very easy
to find out if they're perpendicular or not.
You simply take the dot product even by this very simple
formula, #2, and you see whether if it's 0 they
are perpendicular.
If it's not non-0 they are not.
So that's a useful application, with all these formulas.
There is one more operation that we will need for vectors.
And that's called a cross-product.
Now, the terminology here really doesn't make much, how
should I say, it's not very imaginative in some ways.
It's just that, in the first case, we denote it by dots,
so we call it dot product.
In the second case, it is denoted by a cross, it's
called a cross product.
There's no underlying reason for calling them this way,
it's just a tradition.
So, don't try to read too much into this terminology.
It's just the way it is.
Because of the kind of notation that, historically, people have
gotten used to over the years.
So the second operation's the cross product.
And, again, I would like to think of it as a black box
which has certain input and certain output.
But now, the input will be the same, as in the
case of a dot product.
The input will consist of two vectors.
So vector number 1 and vector number 2.
But the output will be, instead of a number, in the case of a
dot product, a vector itself.
A vector as well.
So, notation for this is a cross b.
And, again, there are two different definitions.
Just as in the case of a dot product.
The first definition being geometric and the second
definition being algebraic.
So, what's the first definition?
To explain the first definition, I have
to draw a picture.
So, here I have two vectors, a and b.
And we should think, we should think, that they don't
necessarily lie on this plane but in fact there's some plane
which might be sticking out.
So I want to draw, I would like to imagine this
piece on this plane.
Which is kind of floating somewhere.
It's not inside the blackboard, but it's sticking out.
In fact, I can use this blackboard to illustrate it.
So that's the kind of plane I'm talking about.
So, in fact, I might use it.
Never thought of that before.
It's a very nice three-dimensional
visual aid here.
So that gives me a lot of flexibility.
In fact, I can do all possible planes by turning the slope
here and then rotating.
And so on.
No, really, I'm amused by it.
It's nice to have good surprises.
Okay So here are the two vectors.
Now that I don't have explain to you, OK,
this is sticking out.
So here's a and here's b.
Now, and that's the plane which contains them.
So, the result of the cross-product of these two
vectors is going to be a vector which will be perpendicular
to this plane.
And here I need a big stick which I don't have.
I have a small stick, I have this.
I have my pen.
So it's going to be the eraser.

I'm doing really well on the visual aids today.
I'm very proud.
So, to people watching, this is how it's done.
This is the cross product of a and b.
So it's a vector which is perpendicular to this plane.
Which is good because I give you the direction
of the vector.
Not exactly, because if I just say that it is
perpendicular, there are two possibilities for it.
It could go like this, or it could go like this.
But, that's where I use the corkscrew rule.
And the corkscrew rule tells me that, I use the corkscrew
rule in the following way.
That this vector and this vector should form a
coordinate system.
In other words, if I turn my corkscrew from a to
b, it should move in the direction of the center.
Which means that this is the correct position,
not like this.
So, I explained to you what the direction of this vector is.
But it's not enough.
To define the vector, I have to tell you what the direction is,
and I have to tell you what the magnitude is.
So, the magnitude of this vector is going to be the
area of this parallelogram.
So, in words, the cross-product is the vector perpendicular.
I will say the word oriented.
Where oriented means that a b on this vector satisfy the
corkscrew rule or the right-hand rule,
whatever you like.
And this is a vector perpendicular, I can erase this
now because I have a much better illustration.
Perpendicular to the plane.
I will say planes spanned by a and b.
Spanned by a and b means a plane which contains
both of them.
It is, in fact, a unique plane.
Unless, of course, these two vectors are proportional
to each other.
If the two vectors are proportional to each
other, the answer's going to be a zero vector.
So, let me assume that they are not proportional to
each other like this.
So, they are not pointing in the same direction.
Then they do span a plane.
There is a unique plane which contains them.
Let me qualify this.
There is a unique plane which contains them, as soon as I
specify a particular point.
In in other words, as soon as I say to which point I'm
applying these two vectors.
As I explained to you last time, the representation of
a vector is not unique.
In fact, I could parallel transform it.
So, I could apply it to a different point.
Which would simply mean moving this blackboard.
I'm not going to do this.
Now, it's perpendicular to the plane spanned by a and b.
And by a plane spanned by a and b, I mean
what I just explained.
Of magnitude equal to the area of a parallelogram.

Which is drawn here.
I'm not going to write it any more.
Just to save time.

So, what is this area?
We can find it very easily because we know that the area
of a parallelogram can be computed by dropping
a perpendicular.
A perpendicular line like this.
So this is 90 degrees.
So let's call this h.
And so this area is the length of a times h.
I have to be careful not to shift it.

a times h.
But h, in turn, is the length of b times the
sine of this angle.
Mister Which I call theta, because that's the
angle between a and b.
Because you've got here a triangle.
For which is is 90 degrees.
So you can find this side by taking this side, which is the
length of b, and multiplying by the sine of this angle.
So h is this.
And therefore, the length of a plus b is the
length of a times h.
Which is the length of a times the length of b times
the sine of this angle.
So, to repeat, the cross-product of two vectors is
a vector perpendicular to of the plane spanned by these two
vectors, which satisfies the orientation rule, with respect
to those two vectors.
And whose length is equal to the length of a length of b
times the sine of the angle between them.
This is the analog of the first definition of a dot product,
which also involves nothing but the lengths and the
cosine of the angle.
In this case there is a little more, because I also specify
a particular direction which is determined by the plane.
Now, what about the second definition?
In the second definition, I'm only allowed to use the
components x, y, x1, y1, z1 of the first vector, and x2 y2 and
z2 of the second vector.
So, a priori it's not clear what the formula should be.
So I'll just give you the answer and then, it's an
interesting problem which you can read about in the book.
To see that, indeed, the two definitions are equivalent.
Last time I explained to you how to show that one
definition implies the other.
I did not explain how the first implies the second, which you
can also read in the book.
So the second definition is the following.
It involves a new piece of notation.

Which is called the determinant.

The determinant is given by a certain formula.
You apply the determinant to this table, or 3x3
table, of symbols.
Which you can multiply and add to each other,
as is the case here.
And specifically, the formula in this case, more concretely,
the formula is the following.
There's a simple rule, how to calculate it.
And the rule is that you have to go along the first row so
you have i, then you have j, then you have k.
So the formula will have three terms.
One corresponding to i, one corresponding to j, and
one corresponding to k.
The term corresponding to i is obtained as follows.
You simply cross the colon and the row where you have i.
So what remains is this 2x2 square.
And in this 2x2 square, you take the product along this
diagonal line minus the product along this diagonal.
This minus this.
In other words, you have this i multiplied by
y1 z2 minus z1 y2.

This is the first term.
For the second term, we take j.
And to get the corresponding term, we cross out the row
and the colon where's j is positioned.
So what you end up with, again, is a 2x2 square.
You can put them next to each other.
So you get x1, z1, x2 and x2.
And you take this diagonal product minus this
diagonal product.
So the result is j times x1 z2 minus z1 x.
And finally, you do the same to k.
So the result is this square and you get x1 y2 minus y1 x2.
And there's one more thing you have to remember.
You have to remember to alternate the sign So you put
plus here, minus and plus here.
Well, at least you don't have to put it at the first.
So I put an equals.
Meaning that I'm continuing that formula.

So, very explicit, something you have to
play with it to memorize.
It's not too difficult.
The question is, what about some other rule
I couldn't hear.
But you can use whatever rule you like.
As long as you get the right answer.
That's fine.
I'm sure some people can just look at it and
tell you the answer.
I think it's much easier than doing a Rubik's
cube, to be honest.
Some people can do it in 20 seconds.
Not me.

So I'm sure each of you will have their own
way to calculate this.
I just told you mine.

The one I use, and which I think most people use.
So that's the cross-product.
And so again, you have two different definitions.
One is geometric, it's about the vectors and their positions
and the angle between them and the lengths and so on,
geometric characteristics.
And the second one is algebraic in terms of just the
components of these vectors.

And so what it means is that you can also apply this to
calculate various things.
For example, you can calculate this way the area of the
parallelogram, which is spanned by these two vectors.
Because the area of the parallelogram is just the
magnitude of this vector.
So we can use it to compute the area of the parallelogram.

By simply taking the magnitude of this vector.
Which is to say, taking the square of this coordinate plus
the square of this coordinate plus the square of this
coordinate and taking the square root of the result.
And let's do some calculation.

Let's say, for example, find area of the parallelogram.
Let me just, to save time, write our parallelogram
like this.
Expand by the vectors a, which is equal to 1, 2, 3.
And b, which is equal to when x is 2, 3, and 1.
Just random numbers.
OK, very easy.
So, first of all we calculate the cross-product between
these two vectors.
By using this formula.
So we put i, j, k, 1, 2, 3.
Negative 2, 3, 1.
It would be embarrassing if the answer was 0.
They would actually be parallel.
Although that would be a good illustration.
Of that case.
So, i is equal to, in front of i, we have this one.
So it's 2 minus 9.
And then you have a minus, remember there is a minus,
maybe I should emphasize it.
There's a minus here, in the middle.
I put minus.
And then you have j, so you have 1 plus 6.

And finally you have, in front of here, you
have this 3 plus 4.
So that's minus 7, negative 7 i minus 7j plus 7k.

So that's the cross-product.
And you want to calculate the area.
To calculate the area, I have to take the
magnitude of this vector.

Please check my calculations, because who knows.

So now, the magnitude is just the square root or the sum of
squares, which is 7 squared plus 7 squared plus 7 squared,
which is 3 times 7 squared, which is 7 times square root.
So that's the final answer.
Any questions?

So, that's the cross-product.

And that gives us the area of that parallelogram.
Does it look like the distance formula?
Well, it looks like the distance formula for a reason.
Because what we're calculating is the length of this vector.
And the length of the vector is given by the same formula as
the distance formula for the distance between the initial
points and the end-point.
So in this case the initial point is 0, 0, 0.
And the end-point is negative 7, negative 7, 7.
So that's why we simply have to apply the distance formula.
We are applying the distance formula.
So, let me just say, the question was asked, if this Is
america And just any vector, not necessarily coming from
this problem, it could be from any problem.
If I have a vector, the length of this vector is just the
distance between the initial point and the end-point.
So if this is point a and this is point b, it will be the
distance between a and b.

And so, it would be equal to the square root of the squares
of the distance, of differences of the coordinates.

In the case of a dot product, there was a special case
when the cosine was 0.
And the cosine is 0 when two vectors are perpendicular.
You case of a cross-product, there is also a special
case when sine is 0.

But, unlike the cosine, sine is equal to 0 if theta
is equal to 0 or pi.
Which is to say, that the two vectors either point in the
same direction, so that the angle between them is 0.
Or they point in opposite directions.
So, in this case, the cross-product, according to the
first formula, is the 0 vector.
Now, it's because the first formula tells us that you get
the vector whose magnitude is 0 There's only one vector
whose magnitude is 0, and that's the 0 vector.

And for a good reason.
Because in this case, if the two vectors a and b are
parallel to other, there is no plane which they span.
So we cannot speak of a plane, we cannot speak of a vector
perpendicular to this plane.
Therefore the only answer that we could possibly
get is the answer 0.
Which incidentally is a very good opportunity to talk
about the vector 0.
Because this is something which people might find confusing.
Partly, the confusion is due to a unfortunate
choice of notation.
I'd like to explain this once for all.
There are three different objects which are denoted
in almost the same way.
You have to know the difference between them.

One is 0.
One is 0 like this.
And there is also O.
So, how do you distinguish between this?
It's all kind of, just to emphasize.
These are numbers 0, and this is O.

So what are these objects?
This is the simplest one.
This is just number 0.
This is something which, to understand this, you don't need
to take multivariable calculus.
You don't need to take calculus at all.
We all know what 0 means.
Now, this is something which is very much to the point here.
Amongst all the vectors, we have vector 0.
And the easiest way to explain what vector 0 is, is to think
about vectors the way I explained last time.
To think about vectors as a shift.
Think of a vector as a shift at in three-dimensional space, as
an operation of a parallel transport of all
points in space.
Applied to each point, it gives you another point.
Which we illustrate, in general we illustrate
by an arrow like this.
So, this vector is equal to this vector to this
vector and so on.
Because we shift the entire blackboard this way.
So, each point gets shifted in the same direction,
with the same magnitude.
But once we start looking at shift, we have to include
a very particular shift.
Namely, a shift by nothing.
Identical transformation.
We do nothing.
If we talk about, if we are allowed to take any shift
whatsoever, we certainly should be allowed to
take a trivial shift.
A shift by 0.
That's the vector which is represented in this way.
Now, we can't really draw it.
Because, see, to draw it, we use the initial point
and the final point.
And then we connect them by segment, and then we place an
arrow to indicate from where, to where, we are going.
If there is no shift, we start and end at the same point.
And there's no place to put an arrow.
So it starts looking like a point.
Which this is not.
Because as we discussed, a point is just a static object
in space, which doesn't know about anything else.
Just about itself.
And a vector is a totally different concept.
A vector is not a point.
A vector, rather, is an operation which you apply to
all points at the same time.
And that's why we can represent each vector by an arrow
starting at each point.
So, this is a special vector which corresponds to that
information by nothing.
That's what it is.
Now you can ask, why do we need it.
Why do we need such a transformation.
If it's nothing, then it's not going to help us.
It is important because we have to include it for consistency.
Because we know that vectors, we want to have various
operations on vectors.
For example, addition of vectors.
Multiplication by scalar.
So, for instance, we can take two vectors which are pointing
in opposite directions.
And which have the same length.
Such vectors, we can call a and negative a.
Negative a simply means that it has the same magnitude,
but it points in the opposite direction.
Now, we would like to have a consistent system of vectors in
the sense that if we take the sum of two vectors, we are
going to get another vector.
So, in particular, we should be able to take the sum
of these two vectors.
What's the answer?
We have to have a vector, which is the sum of them.
But the sum of them is precisely this information
which I'm talking about.
You first shift the blackboard this way.
But then somebody comes and say no, no, no, put it back.
So you shift it back.
The net result is if you did nothing.
But nothing came out as the result of something.
So to be able to have the sum of these two vectors, we have
to have this vector on our books.
That's why the vector's important.
It's one of the reasons.
We would not have a consistent system of vectors if we
did not include this.
But, note that this is not the number 0.
Because the sum of two vectors is not a number.
In fact, if we want to represent it, in coordinates,
we would have to write it like this.
And, again, I have to emphasize that this is
not letter but number 0.
So that's the difference.
This is just one 0 and these are three 0's.
So it's different.
But, of course, the three 0's, the meaning of these three
0's is very special.
Because for a general vector, like vector a, this 1, 2,
3, would correspond to the components of this vector.
Here, this is a vector which has component 0 in each place.
So I hope that the difference between these two is clear.
And finally, we also have this, as if this was not enough.
We have one more object which almost looks the same.
Or notation for which looks almost the same.
And that is actually the point, which is the origin.
When we draw our coordinate system, we oftentimes
need some notation for this special point.
Because there's this very special point in the
three-dimensional space, once we introduce the
coordinate system.
Which is at the center of the coordinate system.
We call it the origin, of the coordinate system.
So we oftentimes need a letter for it.
And because it's called origin, it's natural to
call it by the letter O.
Which is very unfortunate, because it looks like 0.
But it's not 0.
The point is, however, that you can represent,
this is the letter O.
But you can, just like every point b, you can write
it in coordinates.
And the coordinates would be 0, 0, 0.

So, now it looks very much like this.
But not the same.
Because if we use the round brackets, we are
talking about points.
And if we use the angle brackets, we are
talking about vectors.
So this corresponds to the actual point.
Which just sits here and it doesn't know anything
about other points.
Whereas this represents a vector which you can think of
as a transformation of the entire space, which you
can apply to any point.
It's just that the result will be that same point.
You can apply to this point, but you can also apply
it to this point.
And that's what these difference between these two,
we indicate by different methods of notation.
So this is something to chew on, I guess, if you're still
confused about, just think about it.
I think it's fairly self-explanatory.
But I think you will not be able to make much progress in
this area which we are discussing now, if you don't
clearly understand the distinction between
these three objects.
So this is a very good test.
How well you understand vector analysis and coordinate systems
in three-dimensional space.
And I really wanted to emphasize this, because this
is one of the most common points of confusion.
Are there any questions about this?

So, what do we do next?
Next, actually, we are ready to do some fun stuff in space.
We have, basically laid the groundwork for it.
We have developed the formulae we need.
By which I mean, finding vectors.
Defining vectors, and studying some of their properties.
And now, we can actually use these formulae to start
representing various geometric objects in space.
So, what kind of objects am I talking about?
Say again?
Conics we will also represent.
But there are simpler ones.
So, we will start with the simpler ones and we will
progress to the harder ones, which seems
like a good strategy.
Conics will be the next level of difficulty.
Lines and planes.
This is something that I already told you about earlier.
The simplest objects are linear objects.
Which means in dimension one, lines, and in
dimension two, planes.
I guess it's sort of a philosophical question as
to why these really are the simplest objects.
But mathematically it's quite clear.
Because the equations and formulae which you use to
represent them are the simplest possible ones.
As you will see.
So when we try to build a formula for representing
general curves and general surfaces in three-dimensional
space, it's natural to start with lines as the
simplest curves.
And planes as the simplest surfaces.
So let's start with lines.
And let's talk about ways of representing lines in
three dimensional space.

And here, the idea that will be very useful is the idea which
we learned when we studied parametric curves on the plane.
The idea we learnes, back then, is that to understand the
general curve, a good one way to understand the general curve
is to parameterize it by an auxiliary parameter.
So, here, you have a line.
And by this I mean the general line, not necessarily on this
blackboard, but it could be on one of those filtered

So, you have somewhere in the background a three dimensional
coordinate system, which you are going to use to
represent this line.
So we have to decide how best to parameterize it.
So, the first point is that what information do we need
to describe this line.
And here's one convenient way to describe it.
The To determine a line I have to tell you a particular point
on that line, and I have to tell you a direction
on this line.
And directions, we have now learned to describe
in terms of vectors.
So, what I'm saying is that in order to describe a line it
is sufficient to pick a point, let's call it l.
Pick a point of l.
And, second, pick a direction vector.
We are going to call this vector v.

And let's call this point c.
Now, neither of these objects are unique.
I could have chosen this point or this point or this point.
Anything from the infinitely many points which
belong to this line.
Likewise, the direction vector is not unique either.
I could take any multiple of this vector.
I could take various vectors, this vector.
Or, going in the opposite direction, like for example,
this one would work also.
So, this is not unique.
But I have to make some choices.
And once I make those choices, it will have determined
completely what the line is.
So, all we need to do now is to give a description, a
parameterization, of other points on this line.
We already have one of them, so we need to find other points.
And doing this is actually very simple.
Because we can now take advantage of the notion
of addition of vectors.
So, let me explain this.
Let me draw the position vector of this point.
I recall that a position vector of a point is a vector which
has the origin, the letter o, as the initial point.
And our point, p, as the final point.
So that is the position vector of this point.
So now I would like to find other points on this line.
Finding other points is the same as finding position
vectors of those other points.
So, for example, I know another point on this,
line, which is this point.
The endpoint of this vector.
So, the question is, how do I find this position.
How do I find that position vector of this point.
Well, for this, we can use additional vectors.
And I recall that if you have two vectors, you could use a
triangular rule to find the sum between them.
Here, they're perfectly arranged for that.
At first you move from the origin to this point.
And then I move from this point to this point.
So what's the net result?
The net result is that I move from the origin to this point.
Let's call this position vector by r0, and let's
call this vector r1.
And what I'm saying is that r1 is equal to r0 plus v.
So now, by using additional vectors, I have been able to
find a different point on the same line.
What about the rest of the line?
Well, to get to other points I can argue as follows If I take
another point -- say, this one -- the vector which connects my
initial point p to this one, let's call it p prime, can be
expressed as a multiple of the vector v.
Because parallel vectors are all proportional to each other.
Certainly the vector p prime, v p prime, is proportional to v.
So there is some number -- t1, say -- for which
we'll have this formula.
For some number.
This simply means the product of v and the scalar t.
We can write it like this, or we can write it in the reverse
order. t1 times v, it doesn't matter.
For some number t1.
But if so, then I can find the vector t prime as
r0 plus v times t1.
Just to show it's the t from the diagram.
Simply take this vector.
And you find the position vector for this point by taking
the sum of the initial one.
And this one which is v times t1.

But in fact, this is true for any point, t prime.
So for each point, t prime, there will be a particular
number t1, or maybe it's better if we call it t so that we
don't worry about the indices.
And so, we see that, this way, we can actually parameterize
all points on this line.
So, in other words, we can write the position vector, say,
r, for a general point on this line, as r0 plus v.
And that's going to be the crucial formula.
So you see, we have exploited the notion
of additional vectors.
This why it was important to talk about vectors and about
addition of these vectors.
Because otherwise it wouldn't be so clear how to get to
those other points which belong to the same line.
Now, let's write this formula in components.

Let's write it in components.
So we'll say r is xyz and r0 is x0 y0 z0 and v is a b c.
These are three vectors.
So now, we just write this formula out in components.
So we get the following: for the first one, we get x
is equal to x0 plus a t.

For the second one we get this.
For the third one, we get this.
And now we look at this, and we recognize that it looks very
similar to parametric representation of curves on the
plane, which we have talked about for two weeks,
or a week and a half.
And for very good reason.
Because, in fact, this is a parametric representation
of the line l.
Because I have now parameterized general points of
this line in terms of some of auxiliary coordinate t.
So, it certainly is very reminiscent of the general
formula for a parametric curve the plane.
The difference is that, when we talk about curves of the plane,
we only have two coordinates to describe, x and y.
That's why there are only two lines, instead
of three lines here.
Now we work in a three-dimensional space,
so we have to specify the dependence of each of the
three coordinates.
But, otherwise, it looks very similar because
this is some function.
We can call it f of t, if you want.
This is another function.
Call it g of t.
And this is a third function, which we can call h of t.
Are there any questions about this?
So I explained to you how to derive it, by using vectors.
But once you have the end result, you recognize that
this is simply a parametric representation of the line.
So, in practice on homework and on various tests, you're going
to be asked to find parametric representation for
a line, like this.
And this is really easy.
Because all you need to know is a particular point on this
line, and a particular vector on this line.
Or a particular vector which goes along this line.
Like this vector v, on this picture.
That's all you need to know.
And you can use the data which you are given, in a given
problem, to find the point and the vector.
And then you just combine this information in
a formula like this.
So here's an example.
It's a very typical question.
Find the line which passes through the points
3, 1, negative 1.
And 3, 2, negative 6.
So that's another way to describe a line.
Because the line is also determined if you specify
two points on it.
There is a unique line which passes through
two different points.
The points should be different.
Otherwise two points on a side.
There are many lines, which pass through one point.
But if points are different, then there is going to be a
unique line which passes through.
But we would like to convert this information into the
information that we need for this parametric representation.
And so, we need to know a point and the direction vector.

So, a point will represent by this division vector,
our 0, and the direction vector will be our v.
In this case, we actually have two points.
So you can choose either one.
It's your choice.
So pick one.
Let's say the first one, 3, 1, negative 1.
So your r0 is, so you see I'm being really rigorous
and pedantic here.
I'm distinguishing between the points and the position vector
of this point by using different brackets.
The round brackets for points, the angle brackets for vectors.
It's very important, as I already explained.
So, this is your 0, y0 and z0.
And now, to find v, we can simply take the difference
between the position vectors of these two points.
If you have two points on the line, you can surely find this
vector by taking the difference between this vector
and this vector.
So this is v, you can write it as o v prime.
So, I simply take the difference from 3 to negative
6, I subtract v1 negative 1.
What's the result?
The result is 0, 1, and negative 5.
So now I have found what I needed, which is, I have found
the point on this curve, on this line, and I have found
the direction vector.
And now I just put all this data into into this formula.

So the end result is x is equal to, these are my x0, y0 and z0.
So x is 3 plus, this is v.
Plus 0 times t, which I can just erase, like this, to 1
plus 1 times t, that's z is equal to negative 1 minus 5t.
So that's the answer.
It's not the answer, it's one possible answer.
Because someone else could instead choose, as a reference
point, not the first one but the second one.
Or take the vector v, the difference not from this
but from this to this, or even a multiple of that.
So in other words, you have to realize that there is
not a unique I parametric representation for a line.
There are many different ways.
So if you get a different answer from your friend,
it doesn't mean that one of you made a mistake.
It could be that both of you have the right answer.
Any questions about this?
Now, there is another way to write the equation of a line.
In general, we're going to have this formula, right?
So let's focus on this formula.
Suppose that a, b and c are all non-0.
Then we can express t from each of these formulae by
dividing by a, b, and c.
So we can find t is equal to x minus x0 over a. t
is y minus y0 over b.
And t is also z minus z0 over c.
But it's all the same t.
So, we could write the following formulas.
x minus x0 divided by a is equal y minus y0 divided by b
equals z minus z0 divided by t.
So that's another way to represent a line.
This is the first way.
And this is the second way.
I would like to explain what happens if one of those
numbers, a b c, is equal to 0.
Which is the case we have in this particular problem.
So if a is equal to 0, then we have the equation x equals x0
instead of x equals x0 plus a t.

So in this case, we cannot express t from the first
equation, because t is not in this equation.
But we can express t still from the second and third equations,
provided b and c are non-0.
So if a is 0 but b and c are non-0, we can write x equals
x0 and y minus y0 over b equals z minus z0 over c.
So we get two equations like this.
It's not surprising that we get two equations.
In fact, here it looks like one equation, but that's because of
a nice way in which I wrote it.
It's not really one equation, there are two equality signs.
This line actually represents two equations.
This is equal to this, and also this is equal to this.
There are two separate equations.
And someone can say, wait a minute, it's actually three
different equations.
Because it says this is equal to this.
This is equal to this.
And also this is equal to this.
But the third equation follows from the first two.
Because if this is equal to this and this is equal to
this, then the first and the third are equal.
That's the usual logical implication that we
have for equality.
So this really corresponds to two equations.

And we can see it more clearly in the case
when a is equal to 0.
There are indeed two equations in this case.
So, in both cases, in the general case when a,
b, and c are all on 0.
Or in the special case when a is equal to 0, but b and c are
non-0, we have two equations describing a line.
There is also a special case when both and b are equal to
0, but c is not equal to 0.

I'm not going to tell you what the answer is.
You'll have to figure it out on your own.
But you'll see that this is also two equations.
And a and b and c can not all be equal to 0.
Because that would mean that this vector v is equal to 0.
But I insisted that this is a non-0 vector.
Pick a direction vector.
I have to make this non-0.
Or otherwise we won't be able to argue in this way.

What's your question
Are you explaining?
Oh great.
If a is equal to 0, b is equal to 0, and c is equal to 0.
Then v is not defined.
Well, v is defined, but v cannot be used in this case.
v has to be a non-0 vector.

Now, so you see what happened is that we actually found
two different ways to represent the line.
This is the first way.
And then an important feature of this first way is that would
have one auxiliary parameter we use to parameterize it.
One parameter.
Or, we can write it in this way.
Which really means that we write two equations
for this line.
And now I want to remind you that when we talked about
curves on this plane, I explained that there are
two different ways to represent the curve.
This is the first way.
When we have one parameter.
But then there is also a second way in which
you have one equation.
Like y equals f of x,
or, the favorite equation for the circle. x squared
plus y squared equals 1.
So, remember, it was exactly the same picture.
In the case of curves on the plane.
We learned two different ways to represent such a curve.
One is a parametric way, where we use one parameter.
And another one is by means of an equation.
And then we have one equation.
So what I want to emphasize now is the dimension count.
Here, we talk about curves.
Which means dimension one.
Curve in the plane.
The dimension of the curve is 1.
And the dimension of the ambient space is 2.
In this case.
Because we talk about curves.
This is about curves on the plane.
I'm not going to write it down.
Yes, I did.
Curve on the plane.
So the object itself is dimension one.
And this ambience is put in the ambient space, which
has dimension two.
So, when you parameterize it, the number of parameters is
going to be equal to the dimension of your object.
In this case, the dimension is one, so there is one parameter.
But, the dimension of your object, dimension of your
object, as I explained earlier, is equal also to the dimension
of the ambient space.
Ambient space, minus the number of equations.

In this case, we have dimension one, which is equal to 2, which
is the number of variables, x and y, the dimension of the
ambient space, the plane.
So, we have to subtract 1 to get down from 2 to 1.
That means that the number of equations for a curve
on the plane has to be 1.
That's why a circle is given by one equation, x squared
plus y squared equals 1.
That's why the graph of a curve is given by one equation,
y equals f of x.

What happens now is that, instead of working on the
plane, we work in space.
And in space, the dimension is three and not two.
So the dimension of the ambient space now becomes three.
But we are still talking about a curve.
More specifically, we're talking about lines.
But lines are special examples of curves.
So, our object is a curve.

But now, so, the dimension of the curve is still 1.
But the dimension of an ambient space is three.
If you want to parameterize it, you have to still
use one parameter.
Just like the curves on the plane.
That's why it's not surprising that we end up with a
parametric representation, with one parameter.
But if you want to describe your line by an equation, you
have to use two equations.
Because you have to be able to come down from three to one.
You have to drop a dimension from three to one, so
you have to impose two independent equations.
So, 1 is equal to 3 minus 2.
That's the number of equations.
That's why it's not surprising that we really have two
equations describing this line.
Well, in this case, I arranged them neatly in one equation.
But as I just explained to you it's not one equation
but it's two equations.
Or, in that case, also clearly two equations.
So, the same principle that I was talking about earlier of
dimension count, dimension of an object being the dimension
of the ambient space minus the number of equations
still works here.

And it'll be interesting to see now how it works for planes.
Before I get to planes, I just want to mention one thing.
Which is that you can be asked various questions about lines.
Once you learn how to represent them, you can be asked about
relative positions of lines.
And what can happen with lines is that lines
can intersect or not.
That's one possible question.
Do these two lines intersect or not.
Another question is, are they parallel or not.
So, in fact, there are relative positions that can happen.
First of all, two lines can just intersect.
That's a possibility.
And you can easily find out whether it is the case.
You simply write the equation for the first line.
We're using a parameter for t, and then you write the equation
for the second line, using a different parameter for s.
Because this has two different lines.
A priori, don't talk to each other.
So, they have two different parameters.
What you need to know is, whether, for a special value,
let's say, this one is parameterized by a parameter
for t, but this one is parameterized by a
paramater for s.
So you want to know whether, for some t equals some special
value t0, there exists some special value s, which is equal
to s0, so that the resulting point is the same.
And if you write this down, you will get a system of equations.
Which may or may not have solutions.
And that's how you know whether they intersect or not.
The second issue which arises here is whether these two
lines are parallel to each other or not.
And that's real easy.
Because part of a parameterization of the line is
a choice of a direction vector.
Two lines are parallel if and only if the direction
vectors are proportional.
How do you know if they are proportional?
Well, you just see if one of them can be obtained by
multiplying the other one by a scalar.
That scalar will be immediately determined
by the first coordinate.
And then you check whether it works for the second
and third coordinates.
And that's how you see if they are parallel or not.
So, the question's like this.
But once you have this parameterization you
can handle them.
Finally, it's possible that the lines do not intersect and are
not parallel to each other.
It's also possible.
And here's an example.
So, let me keep this one This line.
So, let's suppose I have this line.
If I have also this line, first of all, this line
can intersect this one.
Even if it's not in this plane, it could be in another plane.
It could be a more general line.
It can also be parallel.
But these are not the only two choices.
Because it can be parallel to this plane, but not
parallel to the lines.
It can be like this.
So, you kind of look like in this way, you see that
they kind of intersect.
But actually, in three space they don't intersect because
there's some distance between this line and the plane.
So, in this case they're called skew.

And, again, it's very easy to find out whether they're
skewed or not by using parameterization.

So, in the remaining minutes I would like
to talk about planes.
So, I talked extensively about lines.
But now I want to talk about planes.
Which means that now these are objects of a different kind.
Because now these are objects which are two-dimensional.

In the case of curves, we had analogous objects on the plane.
Because the plane was two-dimensional.
So the plane fits one-dimensional objects
like curves in a nice way.
But a plane cannot fit any two-dimensional
object except itself.
Or maybe some, you can cut some piece of a plane.
It's also two-dimensional.
Because it's already two-dimensional itself.
But in three space we can now have all kinds of surfaces.
And, again, the simplest surfaces are the linear
surfaces, the plane.
The question is, what is the best way to represent the plane
in a three-dimensional space?
So, we look again at these two options.
One is to parameterize.
The other one is to write equations.
If we were to parameterize a plane, we would
need two parameters.
Two parameters,
not just one, because it's two-dimensional.
But, if we were to write down equations, we only
need one equation.
Because, for a plane, the dimension of a planes is two.
Which is 3 minus 1.
So that's the number of equations.
So we need to find a nice way to write an equation.
Clearly, it is more economical to represent a plane by an
equation than to try to represent that in a
parametric form.
Because in parametric form, we would need two parameters.
Foremost would be rather complicated.
But we only need one equation.
So let's just find a good way to right down
equations for this plane.
And the point is that there is a very nice way to
get to this equation.

Because a plane can also be determined by a vector.
And a point.
So, remember, we discussed that a line can be determined by a
point on this line and the direction vector.
In the case of a plane, there is no direction vector.
If there were, it would be a curve.
It would be one-dimensional.
So instead of talking about the direction vector of a plane, it
is much wiser to talk about perpendicular
vectors to a plane.
So, a plane, unlike a curve, is determined by
a point on the plane.
An the vector which is perpendicular to it.
So, instead of vector v, which is sort of like a direction
vector in the case of lines, we're going to have a
perpendicular vector.
Determined by, and I will just draw it as a picture.
By a point, p, and a vector, n.

Vector n is perpendicular to the plane.
So, now, how do we write down an equation?
Here we use a cross-product.
So that's where the cross-product really
comes into play.
Because, if you have a different point p prime, and
you connect those two points, this vector n is going to be
perpendicular to this vector.
You can easily see from that board.

So this is your vector from e to t prime.
This is your plane.
This vector is perpendicular to any vector that you could
have inside this plane.
But, the condition of being perpendicular can be expressed
by using the dot product.

So, just to write down this equation.
The equation is that n dot v p prime is equal to 0.
This is a dot product.
Because, remember, we decided, and we discussed and we agreed
that two vectors are perpendicular if and only
if the dot product between them is 0.
That's what I'm writing.
Now, if I write n as a b c, and I write v p prime as before as
x minus x0, y minus y0 and z minus z0 and use the rule of
dot product, I will find the following equation.

So, here you have one equation in which a, b, c x0,
y0 and z0 are given.
And x, y, z are the three variables.
And all x y z that would satisfy this equation
correspond to the totality of all points on this plane.
So, as promised, a plane is given by one equation and
here's a way to derive this equation.
So once you have this equation, there are all kinds of
questions that can be asked.
For example, if two planes, are they parallel to
each other or not.
What is the angle between these planes.
But all of this can be immediately learned from this
equation by looknig at it and interpreting the coefficients
a b c and x0, y0, and z0 as this data.
So, I'll stop here and I want to say one thing.
Which is that I have both of the solutions for the first
homework set the b space.
And I will be posting them regularly every Tuesday
night when the last of the section is over.

So I'll see you next Tuesday.