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Lecture Comments (19)

1 answer

Last reply by: Professor Hovasapian
Wed Aug 3, 2016 6:09 PM

Post by Mohsin Alibrahim on August 3, 2016

Hello Professor Hovaspian,

First of all thanks for the wonderful lecture

In  Example 3 is it a coincidence the (x+1/2) = (x-x0/y0) and the same question goes for (y-2/6) ?

1 answer

Last reply by: Professor Hovasapian
Tue Sep 8, 2015 8:56 PM

Post by Jim Tang on September 6, 2015

Wait standard equations don't work in 3-D? Ax+By+Cz=D doesn't work? What I'm thinking is that you can just drop in another variable in the equation, or am I mistaken?

1 answer

Last reply by: Professor Hovasapian
Tue Sep 8, 2015 4:44 PM

Post by Jim Tang on September 6, 2015

Difference between |a dot b| vs. ||a dot b||? I'm assuming both just mean the norm.

In addition, in ||a dot b|| less than or equal to ||a|| ||b||, that ||a dot b|| is the LOWER LIMIT. In the video you said it was upper limit, but I think it's the opposite. Can you clarify this up?

Besides that, great videos!

2 answers

Last reply by: Hen McGibbons
Fri Aug 21, 2015 5:10 PM

Post by Hen McGibbons on August 19, 2015

@1:48 why does vector ab = vector b - vector a? i dont understand why vector ab can not equal vector a - vector b? is that just a definition? im having trouble understanding it intuitively but maybe i should ignore my intuition if it is just a definition.

1 answer

Last reply by: Professor Hovasapian
Sat Jun 20, 2015 5:09 PM

Post by Jeff Trondhjem on June 17, 2015

In you last example {3} regarding two dimensions you did not use the same parametric equation s(t) = a + t(b - a) you used s(t) = a + t(b)? Why the change in the generalized parametric equation?

1 answer

Last reply by: Professor Hovasapian
Sat Aug 2, 2014 3:08 AM

Post by Saad Athar on July 27, 2014

I had a question, in your eg 1, you asked find the coordinates 1/5th of the way from a to b. If the same question was asked but from b to a this time, so does the parametric equation change to s(t)= b + t(a-b)?

0 answers

Post by Juan Castro on April 19, 2014

Raffi- I am working on the practice exercises for the "Inequalities and parametric lines" lecture. The hint sections and answers show all messed up when using google chrome or firefox browsers. Is that a problem with my browsers? or do you guys need to fix it? if so, please do so as it is really inconvenient to try to figure all the programming syntax.

1 answer

Last reply by: Professor Hovasapian
Fri Feb 14, 2014 6:38 PM

Post by walid hamarneh on February 14, 2014

How could a point be a vector ?

2 answers

Last reply by: Professor Hovasapian
Fri Oct 25, 2013 3:59 PM

Post by Christian Fischer on October 25, 2013

Just a quick question for example 1: would it also be valid to say vector ab= a+b and vector ac = a+c instead of minus?

Inequalities & Parametric Lines

Let u = ( − 5,0,1), v = (0,4, − 3) and w = (7, − 10,0). Find the angle θ between \protectvu and \protectvw .
  • First we compute vu and vw by finding the difference between the vectors:vu = u − v = ( − 5,0,1) − (0,4, − 3) = ( − 5, − 4,4) andvw = w − v = (7, − 10,0) − (0,4, − 3) = (7, − 14,3).
  • The formula for finding the angle between two vectors is cos θ = [(a ×b)/(|| a |||| b ||)]. Let a = vu and b = vw .
  • Substitution yields cos θ = [(( − 5, − 4,4) ×(7, − 14,3))/(|| ( − 5, − 4,4) |||| (7, − 14,3) ||)] = [( − 35 + 56 + 12)/(√{( − 5)2 + ( − 4)2 + 42} ( √{72 + ( − 14)2 + 32} ))] = [33/(√{57} ( √{254} ))] = [33/(√{14,478} )].
Hence the angle bewteen the two vectors vu and vw is θ = cos − 1( [33/(√{14,478} )] ) ≈ 74.1o
Let a = ( [1/(√2 )], − [1/(√2 )] ), b = (3, − 2) and c = (1,1). Find the projection from \protectbc to a.
  • Note that a is a unit vector. Since we are projecting our vector into a, we use the formula Pbc a = (bc ×a)a.
  • Now, bc = (1,1) − (3, − 2) = ( − 2,3). Subsituting into our formula we obtain Pbc a = [ ( − 2,3) ×( [1/(√2 )], − [1/(√2 )] ) ]( [1/(√2 )], − [1/(√2 )] ).
Hence Pbc a = ( [( − 2)/(√2 )] + [( − 3)/(√2 )] )( [1/(√2 )], − [1/(√2 )] ) = [( − 5)/(√2 )]( [1/(√2 )], − [1/(√2 )] ) = ( − [5/2],[5/2] ).
Let p = (0,0) and a = (1,1). Use x(t) = p + ta to find:
i) x(0)
  • We have x(t) = p + ta = (0,0) + t(1,1).
Setting t = 0 yields x(0) = (0,0) + 0(1,1) = (0,0) + (0,0) = (0,0).
Let p = (0,0) and a = (1,1). Use x(t) = p + ta to find:
ii) x(1)
  • We have x(t) = p + ta = (0,0) + t(1,1).
Setting t = 1 yields x(0) = (0,0) + 1(1,1) = (0,0) + (1,1) = (1,1).
Let p = (0,0) and a = (1,1). Use x(t) = p + ta to find:
iii) x( [1/2] )
  • We have x(t) = p + ta = (0,0) + t(1,1).
Setting t = [1/2] yields x( [1/2] ) = (0,0) + [1/2](1,1) = (0,0) + ( [1/2],[1/2] ) = ( [1/2],[1/2] ).
Let p = ( \protect
[1/(√3 )]
[2/(√3 )]\protect
) and a = ( \protect
− 1
). Use x(t) = p + t a to find:
i) x(0)
  • We have x(t) = p + ta = (
    [1/(√3 )]
    [2/(√3 )]
    ) + t(
    − 1
  • Setting t = 0 yields x(0) = (
    [1/(√3 )]
    [2/(√3 )]
    ) + 0(
    − 1
  • (
    [1/(√3 )]
    [2/(√3 )]
    ) + (
( \protect
[1/(√3 )]
[2/(√3 )]\protect
Let p = ( \protect
[1/(√3 )]
[2/(√3 )]\protect
) and a = ( \protect
− 1
). Use x(t) = p + t a to find:
ii) x(1)
  • We have x(t) = p + ta = (
    [1/(√3 )]
    [2/(√3 )]
    ) + t(
    − 1
  • Setting t = [1/(√3 )] yields x(1) = (
    [1/(√3 )]
    [2/(√3 )]
    ) + [1/(√3 )](
    − 1
  • (
    [1/(√3 )]
    [2/(√3 )]
    ) + (
    [2/(√3 )]
    [( − 1)/(√3 )]
    [3/(√3 )]
( \protect
[3/(√3 )]
[( − 1)/(√3 )]
[5/(√3 )]\protect
Let p = ( \protect
[1/(√3 )]
[2/(√3 )]\protect
) and a = ( \protect
− 1
). Use x(t) = p + ta to find:
iii x ( [1/(√3 )] )
We have x(t) = p + ta = ( \protect
[1/(√3 )]
[2/(√3 )]\protect
) + t(
− 1
). Setting t = [1/(√3 )] yields x(1) = ( \protect
[1/(√3 )]
[2/(√3 )]\protect
) + [1/(√3 )]( \protect
− 1
) = (
[1/(√3 )]
[2/(√3 )]\protect
) + ( \protect
[2/(√3 )]
[( − 1)/(√3 )]
[3/(√3 )]\protect
) = ( \protect
[3/(√3 )]
[( − 1)/(√3 )]
[5/(√3 )]\protect
For x(t) = p + ta, x(0) = ( \protect
), find the standard form of the equation given that a = ( \protect
− 3
  • We use our given information to solve for p, substitution yields x(0) = (
    ) = p + 0(
    − 3
  • We have p = (
    ) and our parametric representation is (
    ) = (
    ) + t(
    − 3
  • We now have a system of two equations:
    x = 1 − 3t
    y = 2 + 7t
  • Solving for t in the first equations gives t = − [(x − 1)/3].
  • Substituting for t in our second equation gives y = 2 + 7( − [(x − 1)/3] ).
Simplifying yields 7x + 3y − 13 = 0.
Find a parametric representation of a line passing through (5, − 4) on the direction of ( − 1, − 1).
  • Using the equation x(t) = p + ta where p is a point on the line and a is the direction vector, we can find a parametric representation of a line.
Substitution yields x(t) = (5, − 4) + t( − 1, − 1).
Find a parmetric representation of a line segment passing through (2,2) and ( − 2, − 2).
  • Using the equation x(t) = a + t(b − a) where a and b are endpoints of the line segment, we can find a parametric representation the line segment with e0 ≥ t ≥ 1.
  • Let a = (2,2) and b = ( − 2, − 2). Computing b − a yields ( − 2, − 2) − (2,2) = ( − 4, − 4).
Substitution yields x(t) = (2,2) + t( − 4, − 4) with 0 ≥ t ≥ 1.
Find the coordinate point A at t = [1/2] given that a line segment passes through the origin and ( − [1/2],[(√3 )/2] ).
  • We can construct a parametric representation of the line segment using x(t) = a + t(b − a) for 0 ≥ t ≥ 1, letting a = (0,0) and b = ( − [1/2],[(√3 )/2] ).
  • Computing b − a yields ( − [1/2],[(√3 )/2] ), so that by substitution our parametric representation is ( \protect
    ) = t(
    − [1/2]
    [(√3 )/2]
Now, at t = [1/2] we have ( \protect
) = [1/2]( \protect
− [1/2]
[(√3 )/2]\protect
) = (
− [1/4]
[(√3 )/4]\protect
), so that our coordinate point A = ( − [1/4],[(√3 )/4] ).
Show that if two vectors a and b are seperated by [p/2] >θ > 0, then | a ×b | <|| a |||| b ||.
  • Recall that two vectors a and b are at an angle θ from each other by the equation cos θ = [(a ×b)/(|| a |||| b ||)].
  • Multiplying both sides by || a |||| b || yields || a |||| b ||cos θ = a ×b
  • Taking the absolute value of both sides gives || a |||| b ||| cos θ | = | a ×b |. Note that since [p/2] >θ > 0 then | cos θ | < 1.
Hence || a |||| b ||>|| a |||| b ||| cos θ | = | a ×b | or || a |||| b ||> | a ×b |.
Show that if a and b are orthoganal unit vectors, then || a + b || ≥ 2.
  • Recall that they pythagorean theorem for unit vectors is || a + b ||2 = || a ||2 + || b ||2.
  • Clearly || a + b || ≥ || a + b ||2. Also, since a and b are orthonganal we have || a ||2 = 1 and || b ||2 = 1.
Hence || a + b || ≥ || a + b ||2 = || a ||2 + || b ||2 = 1 + 1 = 2 or || a + b || ≥ 2.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Inequalities & Parametric Lines

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Inequalities and Parametric Lines 0:30
    • Starting Example
    • Theorem 1
    • Theorem 2
    • Definition 1: Parametric Equation of a Straight Line
    • Definition 2
    • Example 1
    • Example 2

Transcription: Inequalities & Parametric Lines

Hello and welcome back to educator.com0000

Welcome back to multivariable Calculus.0003

Last lesson, we introduced the notion of a projection.0005

We also introduced this notion of the angle between two vectors.0008

Remember a · b = the norm of a × norm of b × cos(θ).0012

We finished off an example and I would like to do one more example of that.0018

But this time with vectors that are not exactly located, that start from the origin.0022

Vectors that are just sort of random in space.0027

Let us go ahead and start with that example.0032

So, we have got... let us let a = the vector (1,2,3).0040

We will let b = (2,1,-4).0048

We will let c = (1,1,1).0055

I will go ahead and close that out.0061

We want to find the angle θ.0066

Between the vector ab, and the vector ac.0073

Again, a vector is just a point in space.0082

From this point to this point is a vector, and from this point to this point is another vector.0085

So we want to find the angle between those two vectors.0093

So, the vector ab.0097

Remember when we said we had two vectors, the vector that has one as a beginning point and the other as an ending point, that is equal to the ending point vector - the beginning point vector.0100

So it actually equals b - a.0113

If you look back a lesson ago, or two lessons, you will find it in there.0115

The vector b - a = (1,3,-7), just this - that.0121

b - a... oh, you know what... oh this is -2, that is why.0130

There you go, so (1,-2,3), (2,1,-4), and (1,1,1), there we go.0135

And -4 - 3 is -7.0143

And the vector ac = the vector c - a.0146

So that will equal (0,3,-2).0153

Let us check that real quickly.0156

1-1=0, 1--2=3 1 - 3 = -2, that is correct.0158

Now we can do our cos(θ) = ... now here our are vectors, these are the 2 vectors right here that we are finding the angle between.0166

So it is going to be the vector ab · ac/norm(ab) × norm(ac).0170

Okay, when we do ab · ac, we end up with 0+9+14.0188

When we run the norm of these vectors, we end up with sqrt(59) and sqrt(13), so it will = 23/... when we multiply this and take the radical, over 27.69480198

When I take arccosine of that, I get θ = 33.8 degrees.0217

There we go.0227

Let us see what this actually looks like in 3 space.0230

I am not going to do the actual vectors, I am just going to kind of do the random vectors.0235

This is the vector a.0239

Well, let us say this is the vector b, and let us say that this is the vector c.0242

Now, the vector ab, this is a and this b, and this is c, it is just these vectors here.0250

Those vectors are just 3 different points.0260

ab is that vector right there, and ac is that vector right there.0268

We are taking these vectors and we are creating other vectors.0277

This vector is b - a, and this vector is c - a.0282

So this is the vector ab, and this is the vector ac.0286

This θ is the angle between them, that is all we did, so we have to be very careful when we read these.0292

If they say a vector a and a vector b, and the angle between them, or if say they give us another point, we have to know which vector they are talking about.0303

Be very careful about how you read these, do not just jump into the problem.0307

Okay, let us go ahead and state a couple of theorems.0311

Very important theorems.0316

We are just going to state them again, we are not going to prove them, but we just want to be aware of them because they will come up from time to time.0318

They are very important actually.0325

They are important for theoretical reasons actually, but we do at least want to be aware of them.0328

So, let a and b be 2 vectors in RN.0334

Again, we are being as general as possible.0346

We will also use the notation instead of actually writing all of this out, we will often say a and b are members of RN.0350

We will often just do it that way.0359

Then, a · b, the absolute value of a · b, is less than or equal to norm(a) × norm(b).0366

This is called the Koshi Shwartz inequality.0381

It is a very important inequality in mathematics.0390

Not only does it hold in N space, any number of dimension of Euclidian space, it actually even holds more generally than that.0395

It actually even holds when you talk about a function space. or any other kind of space.0404

It is a very beautiful theorem.0409

Let us just talk about what it means.0412

It says if you take two vectors in RN, and you take their dot product, you know that you end up with a number.0414

It says that that number is going to be... that number has an upper limit... in other words, when you take the absolute value of that, the magnitude of that number is going to be less than the product of the norms of the two vectors involved in that scalar product.0420

Again, very important so we will just be aware of it.0436

Now, let us move on to the second theorem, which will probably make more sense geometrically.0444

Again, let a and b, ok let me draw these a little bit better here.0450

Let a and b be in RN, then the norm of a + b, in other words if I take a + b, and then I take the norm of that, it is going to be < or = the norm of a separately, and adding it to the norm of b.0461

This is called the triangle inequality.0485

We will go ahead and give you a nice geometric interpretation of this, so that you can see what it is.0490

It makes complete sense. You actually know this. It makes complete sense.0495

I mean you actually know this from... intuitively you know what this means from the time you are 5, 6, 7 years old.0503

You just looked at some triangular shape and you knew intuitively what this meant, but again, we just want to formalize things.0504

So this says that if I have 2 vectors and I add them, and I take their norm, the length of that vector, the magnitude, the length of that has an upper limit.0510

That limit is the sum of the two norms taken individually.0520

Let us just work in 2 space here.0526

If I have some vector a, and I have another vector b, well I know a + b is going to be this vector here.0530

It just means you go to a and then do b, so this vector is a + b, and this is b, and this is a.0538

I get a triangle right here, this little triangle right here.0548

This triangle inequality says that if I add the length of one side and the length of the other side, it is going ot be bigger than the length of the third side.0553

That has to be true, we do not know this just from geometry, but we know this from looking at it.0565

If the length of the third side were somehow equal to the length of this and this, you would not have a triangle.0568

What you would have is, one side, one side, and then on the third side, you would basically just have a triangle that collapses on itself.0575

If however for some odd reason the length of one side and then the length of the other side is somehow longer than the length of the third side, you do not even have a triangle at all.0582

You do not even have something that connects to any one side.0588

That is all this says.0594

It says that, for a triangle, we know that the sum of the two sides has to equal bigger than the third side, that is all this says.0597

This is just a generalization into any number of dimensions.0605

So again, geometry helps to see it, it is the algebra that is important.0608

This is true in any number of dimensions.0614

We are going to start talking about... we are going to give the definition of a parametric line.0620

Later on, when we start talking about functions of several variables, we are going to define a function parametrically.0627

This notion of a parametric definition is profoundly important.0633

We will talk about it a little bit now.0639

I will actually talk about it broadly and theoretically, but without proofs or anything, I will take some time to discuss what this means in the next lesson.0641

Right now, let us just get a sense of what is going on here.0650


A parametric equation, also a representation, in fact I prefer the word representation so a parametric representation of a straight line.0659

We are only talking about a straight line so far.0678

The straight line passing through a point p, which is a vector in the direction of a vector "a" is x = the point p + some parameter t × the vector a, where t can be any number. Any real number.0681

Okay, what this means is this x right here, it means that if I start at p, and if I go in the direction of a any length, I am going to end up at a different point, right?0728

Right, all those different points that I end up, that is what the x is.0739

All those points in this direction and that direction, they form the line.0745

I will draw a picture in just a minute.0748

Just to let you know, the vector x that you get, or the points in space that make up this line, they are functions of t.0752

So it equals p + t × vector a.0762

Now let us draw a picture and see what it is we mean, this is very, very, important.0768

You must absolutely understand what a parametric representation of an equation, in general of a function is.0775

You are going to need as we go along later in the course, you are going to need to actually write down... we are going to give you some equation in x,y, or maybe x,y,z, and you have to give the parametric representation.0785

It is very important that you understand what we mean by this thing called a parameter.0794

Like t, why it is we write things this way.0798

Let us see if we can develop some intuition.0800

This is our point p and we know that our line is going to be passing through this.0805

Well, we want to pass, we want the line to be in the direction a, so let us just pick some random vector a, that is a.0808

So we know the line is going to pass through here, and it is going to be in this direction.0817

In other words, the line is going to be that line right there, going this way and going that way.0823

We want to be able to write an equation for that.0830

Watch what happens.0833

This is the equation that I gave, right?0834

Let us just do a couple of examples for t.0837

Well, when t = 0, if I put t = 0 in here, t = 0 and 0 × a = 0, therefore our point is just our vector p, so the point is there, no problem.0840

Now let us take t = 1, well now, if I put 1 in here, that means it is going to be p, our new point x is going to be p + 1 × vector a.0855

I start here and I do just one length of vector a in that direction.0870

Now my new point is right there.0874

So it is going to be p + the vector a.0877

If I do t = 2, that is going to be p + 2 × the vector a.0881

That means I start at p, then I go one length of vector a, and then I go another length of vector a, that puts me at that point there.0891

Well, what if I do t = -6?0901

That means my new point along that line is going to be... start at p, and since it is - 6, that means -6a.0905

So start a p, then go in the direction of a, but opposite, -6.0916

So it is going to be 1,2,3,4,5,6 that puts me at that point.0921

Now, if I take all values of t, t = - infinity to positive infinity, .6 sqrt(13), -pi, I get all of these points.0929

I get all of these points along that line.0940

What I have done is now I have an equation for that line, with respect to one parameter.0943

It is no longer y as a function of x.0950

What I have done is I have freed myself from thinking of things in a coordinate system, y as a function of x, two variables.0953

Now I have created an equation in one variable t, a parameter.0961

I have picked a starting point, I have picked a direction, and now as t goes from - infinity to + infinity, it represents all of these points.0967

That is the power of a parametric representation.0975

It frees me from expressing one variable in terms of another variable.0980

I can recover it if I want, but it totally frees me from that.0984

Now I can treat each variable separately as a function of t, you will see that later on.0988

Let us talk a little bit more about this.0993

We just said, the vector x(t) is some starting point + numbers × some vector a which is the direction of the line.0998

Let us do this in component form, x is just a vector.1015

Let us just take components in 3 space.1017

In 3 space, this would just be x,y,z.1020

I am going to write these vertically, is equal to some p.1026

Let us say p1,p2,p3 + t × a1,a2,a3.1030

This is the vector form, and this is the component form.1043

When we do a specific problem, we work in components.1048

When we work theoretically, algebraically, we are not doing numbers, we work in vector form... that is the best way to do it.1050

Now, let us give another definition.1058

Let a and b be vectors in RN.1062

Then, the line segment between a and b is... we will call it segment t... and again, we are still talking about vectors here.1070

it is equal to a + t × b - a.1095

This is a parametric representation for the line segment between the vector a and b, between the point a and b.1100

Let us just... this is the definition we gave, let us just try a couple values and see if this makes sense.1110

If I let t = 0, then s(0) = a + 0, t = 0, so that is equal to a.1116

What happens when t = 1?1131

Well s(1) = a + 1 × b - a, which is distribute the 1 and we get + b - a, a's cancel... and I get b.1136

Oh, I forgot to say t > or = 0 and < or = to 1.1150

The line segment between two points, two vectors, is this equation when t takes on the values 0 and 1 and everything between.1156

At 0, it is the point a, at 1, it is the point b.1167

At every point in between, 0.1,0.2,0.5,0.6,0.7,2/3, things like that, you are going to get all the points in between.1173

So here is what it looks like.1181

This is the point a, the vector a.1185

Oops, we have our stray lines again, this is the point b, the vector b.1191

So this is a, and this is b, a, b.1196

We want to find a parametric representation for this line segment in between them, this is what we want.1202

Well, this is the equation as we set it up.1210

If I start here, that is my starting point, right... this is the same as the other parametric, the definition we gave was just some starting point + t × some direction vector.1214

Well, the direction vector is this direction.1226

This direction vector is equal to b - a, there is nothing different here.1228

Now, basically what happened is we set t=0, we get this point. t = 1, we get this point, well all the points in between is when t = everything between 0 and 1.1235

Literally, what you are doing is you are starting here and you are moving along this direction, you are just moving in this direction taking all the points.1248

Eventually, if you take all of the values between 0 and 1, you end up with all of the points in that line segment.1256

So again, this is the definition of the line segment between 2 points, a and b.1270

Okay, let us go ahead and do some examples.1279

Let us do example 1.1283

Actually, let me go back to black here.1286

Now, we will let a = (1,3,5), and we will let b = (1,-3,-5).1290

Our problem in this case is to find the coordinates of the point, 1/5 of the way from a to b.1302

So we want to find the coordinates of the point which is 1/5 of the way from a to b.1332

We have some... it is a line segment from a to b... we want to find the coordinate, the distance, we want to find the coordinate of that point.1337

So pictorially we are looking at something like this.1347

We said if this is a, and this is, that is the line segment, 1/5, we want to find that coordinate.1350

In other words, we want to find that vector, not this length, we want to find this vector.1357

Here is how we do it.1363

We said that the line segment t = a + t × b - a, when t > or = to 0, < or = to 1.1365

In this case t = 1/5, this is the starting point.1380

1/5 of the way here, so here we are going to take t = 1/5, this is what we want.1385

Well, let us just do the general version.1393

a = (1,3,5), I am going to write these as vertical vectors, + t × vector b - a.1395

Well, b - a, 1-1 is 0, -3 -3 is 6, -5 -5 is - 10, so this is our generic equation for this line segment in coordinate form.1406

That is the vector form, this sis the coordinate form, this is the equation, this is our parametric equation, it is a function of t.1419

Now, we know t = 1/5, so it is going to equal (1,3,5) + 1/5(0,-6,-10).1430

That is going to equal... I will do it down here.1443

It equals (1,3,5) + (1/5 × 0 is 0, 1/5 × -6, is -6/5, and 1/5 × -10 is -2).1448

Now, when I add those together I end up with 1+0 is 1, 3 + -6/5... when you do the math it end sup being 9/5, and 5 - 2 is 3.1465

There we go.1482

The coordinates of this point is (1, 9/5, 3).1484

They are the coordinates of the point that is 1/5 of the distance from the point (1,3,5) to the point (1,-3,-5).1490

Notice how when we are solving a specific problem, we work in coordinates.1500

This is the vector representation, when we actually do the problem, we work in coordinates.1506

We actually put the vectors in.1512

I wrote them vertically because it makes more sense for me when I put them in, you can do horizontally, it is whatever works for you.1514

Okay, so let us do another example here.1520

Example 2.1528

Find a parametric representation for the line passing through the vector a, which is (2,4,6), and the vector b... let us get some room here... and b equals (-1,-2,6).1530

Let us make sure we understand what it is this is saying.1575

This is saying find a parametric representation for the line passing through a and b.1580

We want to find a parametric line that passes through a and b, notice it does not say passing a in the direction of b, it says passing through a and through b, these are two different points.1585

Well, the direction from a to b, or form b to a, it does not matter which way you do it, let us call that vector c.1596

So, c = the direction, so it is going to pass through a and b, I can choose either one as my starting point.1605

I am going to go ahead and choose a as my starting point, so it is going to go in the direction b, ab.1615

So C... actually let me write it at, yea it is fine, we will go and say that... this is my direction vector c, it is going to = b - a.1620

When we have two points, the direction, this direction is the vector, that vector minus this vector.1630

The arrow, the head minus the tail.1635

Therefore, we are going to have x(t) = our starting point a, so again we just want to write out the generic form so we know what it is we are working with, + t × vector c.1640

That is the direction we are going in.1657

That equals a + t × b - a.1659

As you can see, this looks like the line segment definition, except now we want the entire line, not the segment, so we are not going to restrict our values of t between 0 and 1.1667

We are going to let t be any value at all.1677

So let us go ahead and put our coordinates in.1679

a = (2,4,6) + t × b -a, now let us take b - a.1680

-2 - 1 is -3, -2 - 4 is -6 , 6 - 6 is 0, and boom, there you go, it was as easy as that.1690

We just have to make sure we understand what it is asking.1705

It is passing through these two points, which means you pick one point to start from, and the direction is from a to b, or from b to a, it does no matter what direction you take or what parameterization you use as long as you know what it is asking.1708

As long as you have a direction vector, the proper direction vector.1718

Notice that b is not the direction, that is the direction vector, not b, it is passing through b and a.1727

Sorry if you think I am belaboring the point, I understand, forgive me if I do that but it is very important to get this right.1736

Example 3.1745

Let a = (-1,2)... this time we are working in 2 space... and we will let b = (2,6).1749

Wow, now these crazy lines are coming up all over the place.1764

A line passing through a in the direction of b is, parametrically... ok now that we are getting to the bottom of the page these lines are going to show up, so I am actually going to skip and go over one.1770

So we have x(t) = a + t × b -- which is going to equal a, which is (-1,2) -- + t × (2,6).1800

Let me write this in component form.1820

The vector x is just a vector, so it is some point (x,y) = -1 + 2t, 2 + 6t.1824

There you go, so now I have the component version of this, now let me separate these.1840

x = -1 + 2t, y... let me do this, y = 2 + 6t, let me solve for t.1845

This is x +1/2 = t.1860

This is y-2/6 = t, the t's are equal to each other, it is one parameter, this is the power of it, it is one parameter.1866

So let me set these equal to each other.1875

x+1/2 = y-2/6, let me cross multiply.1878

I get 6x+6 = 2y-4.1886

Let me bring the y over. I get 6 -2y = -10, look at that.1890

I have taken my parametric equation and now I have recovered the standard form of the equation that you are used to ever since middle school and high school1895

An expression of the line in terms of the two variables x and y.1903

We can do this in 2 space because we have just the 2 coordinates, two equations we can set equal to each other.1908

The problem is, when we try a parametric representation in lets say 3 space, we cannot solve for t and set these things equal to each other and come up with some, you know, equation like this.1913

Now that we are actually moving into higher dimensions, the parametric representation of a line is the only representation of a line.1925

It is the best, this whole idea of working with x's and y's, it is good, and it is fine for what we did, but now we need to move on to a more powerful tool.1936

We need to move on to parametric representations of lines.1948

The parametric representation of a line is valid in any number of dimensions.1954

I no longer need to work with x and y, we are moving up, we are becoming more sophisticated with our tools.1955

That is why we introduced this notion of a parametric equation.1960

That is it, so again, if we need to recover it for dimension 2, that is fine.1968

This is certainly doable, you solve for t and you do this.1973

But we are going to be working strictly with parametric equations because it is a more powerful tool and it works in any number of dimensions.1980

In fact, parametric representation is the only representation that works in any number of dimensions greater than 2.1985

Okay, thank you for joining us here at educator.com1992

We will continue our discussion of parametric representations and lines next time. Bye-bye.1995