Linear Algebra: Parametric Representations of Lines

Everything we've been doing in
linear algebra so far, you

might be thinking, it's kind of
a more painful way of doing

things that you already
knew how to do.

You've already dealt
with vectors.

I'm guessing that some of you
all have already dealt with

vectors in your calculus or
your pre-calculus or your

physics classes.

But in this video I hope to show
you something that you're

going to do in linear algebra
that you've never done before,

and that it would have been very
hard to do had you not

been exposed to these videos.

Well I'm going to start with,
once again, a different way of

doing something you already
know how to do.

So let me just define some
vector here, instead of making

them bold, I'll just draw it
with the arrow on top.

I'm going to define my vector
to be-- I can do with the

arrow on top or I can just
make it super bold.

I'm just going to define my
vectors, it's going to be a

vector in R2.

Let's just say that my vector
is the vector 2, 1.

If I were to draw it in
standard position,

it looks like this.

You go two to the right,
up one, like that.

That's my, right there,
that is my vector v.

Now, if I were to ask you, what
are all of the possible

vectors I can create?

So let me define a set.

Let me define a set, s, and
it's equal to-- all of the

vectors I can create, if I were
to multiply v times some

constant, so I multiply some
constant, some scalar, times

my vector v, and just to maybe
be a little bit formal, I'll

say such that c is a member
of the real numbers.

Now what would be a graphical
representation of this set?

Well, if we draw them all in
standard position, c could be

any real number.

So if I were to multiply,
c could be 2.

If c is 2, let me
do it this way.

If I do 2 times our vector,
I'm going to get

the vector 4, 2.

Let me draw that in standard
position, 4, 2.

It's right there.

It's this vector right there.

It's collinear with
this first vector.

It's along the same line, but
just goes out further 2.

Now I could've done another.

I could have done 1.5
times our vector v.

Let me do that in a
different color.

And maybe that would be,
that would be what?

That'd be 1.5 times 2,
which is 3, 1.5.

Where would that
vector get me?

I'd go one 1.5 and then I'd
go 3, and then 1.5,

I'd get right there.

And I can multiply
by anything.

I can multiply 1.4999
times vector v, and

get right over here.

I could do minus 0.0001
times vector v.

Let me write that down.

I could do 0.001 times
our vector v.

And where were that put me?

It would put me little super
small vector right there.

If I did minus 0.01, it would
make a super small vector

right there pointing
in that direction.

If I were to do minus 10, I
would get a vector going in

this direction that goes
way like that.

But you can imagine that if
I were to plot all of the

vectors in standard position,
all of them that could be

represented by any c in real
numbers, I'll essentially

get-- I'll end up drawing a
bunch of vectors where their

arrows are all lined up along
this line right there, and all

lined up in even in negative
direction-- let me make sure I

draw it properly-- along
that line, like that.

I think you get the idea.

So it's a set of collinear
vectors.

So let me write that down.

And if we view these vectors as
position vectors, that this

vector represents a point in
space in R2-- this R2 is just

our Cartesian coordinate plane
right here in every

direction-- if we view this
vector as a position vector--

let me write that down-- if
we view it as kind of a

coordinate in R2, then this set,
if we visually represent

it as a bunch of position
vectors, it'll be represented

by this whole line over here.

And I want to make that point
clear because it's essentially

a line, of slope 2.

Right?

Sorry, slope 1/2.

Your rise is 1.

Your rise is 1 for
going over 2.

But I don't want to go
back to our Algebra 1

notation too much.

But I want to make this point
that this line of slope 2 that

goes through the origin, this
is if we draw all of the

vectors in the set as in their
standard form, or if we draw

them all as position vectors.

If I didn't make that
clarification, or that

qualification, I could have
drawn these vectors anywhere.

Right?

Because this 4, 2 vector, I
could have drawn over here.

And then, to say that it's
collinear probably wouldn't

have made as much visual
sense to you.

But I think this collinearity of
it makes more sense to you

if you say, oh, let's draw them
all in standard form.

All of them start at the origin,
and then their tails

are at the origin, and their
heads go essentially to the

coordinate they represent.

That's what I mean by their
position vectors.

They don't necessarily have to
be position vectors, but for

the visualization in this video,
let's stick to that.

Now I was only able to represent
something that goes

through the origin
with this slope.

So you can almost view that
this vector kind of

represented its slope.

You almost want to view it as a
slope vector, if you wanted

to tie it in to what you
learned in Algebra 1.

What if we wanted to represent
other lines

that had that slope?

What if we wanted to represent
the the same line, or I guess

a parallel line-- that goes
through that point over there,

the point 2 comma 4?

Or if we're thinking in position
vectors, we could say

that point is represented
by the vector, and we

will call that x.

It's represented by
the vector x.

And the vector x is
equal to 2, 4.

That point right there.

What if I want to represent the
line that's parallel to

this that goes through
that point 2, 4?

So I want to represent
this line right here.

I'll draw it as parallel
to this as I can.

I think you get the idea, and it
just keeps going like that

in every direction.

These two lines are parallel.

How can I represent the set of
all of these vectors, drawn in

standard form, or all of the
vectors, that if I were to

draw them in standard form,
would show this line?

Well, you can think
about it this way.

If every one of the vectors that
represented this line, if

I start with any vector that was
on this line, and I add my

x vector to it, I'll show up
at a corresponding point on

this line that I
want to be at.

Right?

Let's say I do negative 2 times
my original, so minus 2

times my vector v, that
equaled what?

Minus 4, minus 2, so that's
that vector there.

But if I were to add x to it, if
I were to add my x vector.

So if I were to do minus 2 times
my vector v, but I were

to add x to it, so plus x.

I'm adding this vector 2 comma
4 to it, so from here I'd go

right 2 and up 4,
so I'd go here.

Or visually you could just
say, heads to tails, so I

would go right there.

So I would end up at a

corresponding point over there.

So when I define my set, s, as
the set of all points where I

just multiply v times the
scalar, I got this thing that

went through the origin.

But now let me define
another set.

Let me define a set l, maybe l
for line, that's equal to the

set of all of vectors where the
vector x, I could do it

bold or I'll just draw an
arrow on it, plus some

scalar-- I could use c, but
let me use t, because I'm

going to call this a
parametrization of the line--

so plus some scalar, t times my
vector v such that t could

be any member of the
real numbers.

So what is this going to be?

This is going to be
this blue line.

If I were to draw all of these
vectors in standard position,

I'm going to get my blue line.

For example, if I do minus 2,
this is minus 2, times my

vector v, I get here.

Then if I add x, I go there.

So this vector right here that
has its endpoint right there--

its endpoint sits
on that line.

I can do that with anything.

If I take this vector, this is
some scalar times my vector v,

and I add x to it, I end up
with this vector, whose

endpoint, if I view it as a
position vector, it's endpoint

dictates some coordinate
in the xy plane.

So it will [UNINTELLIGIBLE]

that point.

So I can get to any
of these vectors.

This is a set of vectors right
here, and all of these vectors

are going to point-- they're
essentially going to point to

something-- when I draw them
in standard form, if I draw

them in standard form-- they're
going to point to a

point on that blue line.

Now you might say, hey Sal, this
was a really obtuse way

of defining a line.

I mean we do it in Algebra 1,
where we just say, hey you

know, y is equal to mx plus b.

And we figure out the slope by
figuring out the difference of

two points, and then we do
a little substitution.

And this is stuff you learned
in seventh or eighth grade.

This was really straightforward.

Why am I defining this obtuse
set here and making you think

in terms of sets and vectors
and adding vectors?

And the reason is, is because
this is very general.

This worked well in R2.

So in R2, this was great.

I mean, we just have to worry
about x's and y's.

But what about the situation, I
mean notice, in your algebra

class, your teacher never really
told you much, at least

in the ones I took, about how do
you present lines in three

dimensions?

Maybe some classes go there,
but they definitely didn't

tell you how do you represent
lines in four dimensions, or a

hundred dimensions.

And that's what this is
going to do for us.

Right here, I defined x and
v as vectors in R2.

They're two-dimensional vectors,
but we can extend it

to an arbitrary number
of dimensions.

So just to kind of hit the point
home, let's do one more

example in R2, where, it's kind
of the classic algebra

problem where you need to find
the equation for the line.

But here, we're going
to call it the set

definition for the line.

Let's say we have two vectors.

Let's say we have the vector a,
which I'll define as-- let

me just says it's 2, 1.

So if I were draw it in standard
form, it's 2, 1.

That's my vector
a, right there.

And let's say I have vector
b, let me define vector b.

I'm going to define it as,
I don't know, let me

define it as 0, 3.

So my vector b, 0-- I don't
move to the right at

all and I go up.

So my vector b will
look like that.

Now I'm going to say that these
are position vectors,

that we draw them in
standard form.

When you draw them in standard
form, their endpoints

represent some position.

So you can almost view these
as coordinate points in R2.

This is R2.

All of these coordinate axes
I draw are going be R2.

Now what if I asked you, give
me a parametrization of the

line that goes through
these two points.

So essentially, I want the
equation-- if you're thinking

in Algebra 1 terms-- I want the
equation for the line that

goes through these two points.

So the classic way, you would
have figured out the slope and

all of that, and then
you would have

substituted back in.

But instead, what we can do is,
we can say, hey look, this

line that goes through both of
those points-- you could

almost say that both of those
vectors lie on-- I guess

that's a better-- Both of these

vectors lie on this line.

Now, what vector can be
represented by that line?

Or even better, what vector,
if I take any arbitrary

scalar-- can represent any other
vector on that line?

Now let me do it this way.

What if I were to take-- so this
is vector b here-- what

if I were to take b minus a?

We learned in, I think it was
the previous video, that b

minus a, you'll get this
vector right here.

You'll get the difference
in the two vectors.

This is the vector b
minus the vector a.

And you just think about it.

What do I have to add
to a to get to b?

I have to add b minus a.

So if I can get the vector b
minus a-- right, we know how

to do that.

We just subtract the vectors,
and then multiply it by any

scalar, then we're going to get
any point along that line.

We have to be careful.

So what happens if we take t,
so some scalar, times our

vector, times the vectors
b minus a?

What will we get then?

So b minus a looks like that.

But if we were to draw it in
standard form-- remember, in

standard form b minus a would
look something like this.

Right?

It would start at 0, it would be
parallel to this, and then

from 0 we would draw
its endpoint.

So if we just multiplied some
scalar times b minus a, we

would actually just get
points or vectors

that lie on this line.

Vectors that lie on that
line right there.

Now, that's not what
we set out to do.

We wanted to figure out an
equation, or parametrization,

if you will, of this
line, or this set.

Let's call this set l.

So we want to know what
that set is equal to.

So in order to get there, we
have to start with this, which

is this line here, and
we have to shift it.

And we could shift it either by
shifting it straight up, we

could add vector b to it.

So we could take this
line right here, and

add vector b to it.

And so any point on here
would have its

corresponding point there.

So when you add vector b, it
essentially shifts it up.

That would work.

So we could, say, we could
add vector b to it.

And now all of these points
for any arbitrary-- t is a

member of the real numbers, will
lie on this green line.

Or the other option we could
have done is we could have

added vector a.

Vector a would have taken any
arbitrary point here and

shifted it that way.

Right?

You would be adding
vector a to it.

But either way, you're going to
get to the green line that

we cared about, so you could
have also defined it as the

set of vector a plus this line,
essentially, t times

vector b minus a, where t is
a member of the reals.

So my definition of
my line could be

either of these things.

The definition of my line could
be this set, or it could

be this set.

And all of this seems all very
abstract, but when you

actually deal with the numbers,
it actually becomes

very simple.

It becomes arguably simpler than
what we did in Algebra 1.

So this l, for these particular
case of a and b,

let's figure it out.

My line is equal to-- let me
just use the first example.

It's vector b, so it's the
vector 0, 3 plus t, times the

vector b minus a.

Well what's b minus a?

0 minus 2 is minus 2, 3, minus
1 is 2, for t is a

member of the reals.

Now, if this still seems kind
of like a convoluted set

definition for you, I could
write it in terms that you

might recognize better.

If we want to plot points, if
we call this the y-axis, and

we call this the x-axis,
and if we call this the

x-coordinate, or maybe more
properly that the x-coordinate

and call this the y-coordinate,
then we can set

up an equation here.

This actually is the x-slope.

This is the x-coordinate,
that's the y-coordinate.

Or actually, even better,
whatever-- actually, let me be

very careful there.

This is always going to end up
becoming some vector, l1, l2.

Right?

This is a set of vectors, and
any member of this set is

going to look something
like this.

So this could be li.

So, this is the x-coordinate,
and this is the y-coordinate.

And just to get this in a form
that you recognize, so we're

saying that l is the set of
this vector x plus t times

this vector b minus a here.

If we wanted to write it in kind
of a parametric form, we

can say, since this is what
determines our x-coordinate,

we would say that x is equal to
0 plus t times minus 2, or

minus 2 times t.

And then we can say that y,
since this is what determines

our y-coordinate, y is equal to
3 plus t times 2 plus 2t.

So we could have rewritten that
first equation as just x

is equal to minus 2t, and
y is equal to 2t plus 3.

So if you watch the videos on
parametric equations, this is

just a traditional parametric
definition of

this line right there.

Now, you might have still viewed
this as, Sal, this was

a waste of time, this
was convoluted.

You have to define these
sets and all that.

But now I'm going to show you
something that you probably--

well, unless you have done
this before, but I guess

that's true of anything.

But you probably haven't
seen in your

traditional algebra class.

Let's say I have two points, and
now I'm going to deal in

three dimensions.

So let's say I have
one vector.

I'll just call it point
1, because these

are position vectors.

We'll just call it position 1.

This is in three dimensions.

Just make up some numbers,
negative 1, 2, 7.

Let's say I have Point 2.

Once again, this is in three
dimensions, so you have to

specify three coordinates.

This could be the x, the y,
and the z coordinate.

Point 2, I don't know.

Let's make it 0, 3, and 4.

Now, what if I wanted to find
the equation of the line that

passes through these
two points in R3?

So this is in R3.

Well, I just said that the
equation of this line-- so

I'll just call that, or the set
of this line, let me just

call this l.

It's going to be equal to-- we
could just pick one of these

guys, it could be P1, the
vector P1, these are all

vectors, be careful here.

The vector P1 plus some random
parameter, t, this t could be

time, like you learn when you
first learn parametric

equations, times the difference
of the two vectors,

times P1, and it doesn't matter
what order you take it.

So that's a nice thing too.

P1 minus P2.

It could be P2 minus P1--
because this can take on any

positive or negative value--
where t is a member of the

real numbers.

So let's apply it to
these numbers.

Let's apply it right here.

What is P1 minus P2?

P1 minus P2 is equal to-- let
me get some space here.

P1 minus P2 is equal, minus
1 minus 0 is minus 1.

2 minus 3 is minus 1.

7 minus 4 is 3.

So that thing is that vector.

And so, our line can be
described as a set of vectors,

that if you were to plot it in
standard position, it would be

this set of position vectors.

It would be P1, it would be--
let me do that in green-- it

would be minus 1, 2, 7.

I could've put P2 there, just as
easily-- plus t times minus

1, minus 1, 3, where, or such
that, t is a member of the

real numbers.

Now, this also might not
be satisfying for you.

You're like, gee, how do I plot
this in three dimensions?

Where's my x, y's, and z's?

And if you want to care about
x, y's, and z's, let's say

that this is the z-axis.

This is the x-axis, and
let's say the y-axis.

It kind of goes into our board
like this, so the y-axis comes

out like that.

So what you can do, and actually
I probably won't

graph, so the determinate for
the x-coordinate, just our

convention, is going to be
this term right here.

So we can write that x--
let me write that down.

So that term is going to
determine our x-coordinate.

So we can write that x is equal
to minus 1-- be careful

with the colors-- minus 1,
plus minus 1 times t.

That's our x-coordinate.

Now, our y-coordinate is going
to be determined by this part

of our vector addition because
these are the y-coordinates.

So we can say the y-coordinate
is equal to-- I'll just write

it like this-- 2 plus
minus 1 times t.

And then finally, our
z-coordinate is determined by

that there, the t shows up
because t times 3-- or I could

just put this t into
all of this.

So that the z-coordinate is
equal to 7 plus t times 3, or

I could say plus 3t.

And just like that, we have
three parametric equations.

And when we did it in R2, I did
a parametric equation, but

we learned in Algebra 1,
you can just have a

regular y in terms x.

You don't have to have a
parametric equation.

But when you're dealing in R3,
the only way to define a line

is to have a parametric
equation.

If you have just an equation
with x's, y's, and z's, if I

just have x plus y plus z is
equal to some number, this is

not a line.

And we'll talk more
about this in R3.

This is a plane.

The only way to define a line
or a curve in three

dimensions, if I wanted to
describe the path of a fly in

three dimensions, it has to
be a parametric equation.

Or if I shoot a bullet in three
dimensions and it goes

in a straight line, it has to
be a parametric equation.

So these-- I guess you could
call it-- these are the

equations of a line in
three dimensions.

So hopefully you found
that interesting.

And I think this will be the
first video where you have an

appreciation that linear algebra
can solve problems or

address issues that you
never saw before.

And there's no reason why we
have to just stop at three,

three coordinates, right here.

We could have done this
with fifty dimensions.

We could have defined a line in
fifty dimensions-- or the

set of vectors that define a
line, that two points sit on,

in fifty dimensions-- which is
very hard to visualize, but we

can actually deal with
it mathematically.