Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Pre Calculus
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (4)

1 answer

Last reply by: Professor Selhorst-Jones
Sun Sep 18, 2016 1:00 PM

Post by Khanh Nguyen on September 17 at 11:57:17 PM

I love the bonus round.

1 answer

Last reply by: Professor Selhorst-Jones
Tue Jan 6, 2015 12:20 PM

Post by Jamal Tischler on December 29, 2014

I thought the cross product was UxV=u*v*sin(thetha) and than we use the right-hand-rule to see it's direction.

Dot Product & Cross Product

  • Given two n-dimensional vectors u = 〈u1, u2,…,un〉 and v = 〈v1, v2, …, vn 〉, the dot product (symbolized by a · between the vectors) is the sum of the products of each pair of components.

    u
     

     
    ·

    v
     

     
    =
    〈u1, u2,…,un〉·〈v1, v2, …, vn
           = u1v1 + u2v2 + …+ un vn
    Notice the dot product results in a scalar, not a vector. [We say u ·v as  `u dot v' when speaking.]
  • The dot product of two vectors is deeply related to the angle between the two vectors (if put tail to tail) and their magnitudes.

    u
     

     
    ·

    v
     

     
      =  |

    u
     

     
    | |

    v
     

     
    | cosθ
  • We can interpret the dot product geometrically as projecting one vector on to the other, then multiplying the length of the resulting projection by the length of the vector being projected onto. In general, we can interpret the dot product as a measure of how long and how parallel two vectors are.
  • From the formula above, we see

    u
     

     
    ·

    v
     

     
    =0     ⇔    

    u
     

     
    is perpendicular to

    v
     

     
    .
    [There are many equivalent words for `perpendicular': `normal', `orthogonal', `at right angles', but these all mean the same thing: θ = 90°. (It's a very important idea in math, so that's why there are so many synonyms.)]
  • Unlike the dot product, the cross product only works in three dimensions. It takes two vectors and produces a third vector that is perpendicular to both. If u = 〈u1, u2, u3〉 and v = 〈v1, v2, v3〉, the cross product is

    u
     

     
    ×

    v
     

     
      =  〈 u2 v3 − u3 v2,    u3 v1 −u1v3,    u1 v2 − u2 v1 〉.
    [We say u ×v as `u cross v' when speaking.]
  • While we haven't learned about matrices or their determinants (yet!), you may have learned about them in a previous math course. A great mnemonic for remembering the cross product is with the determinant of a 3×3 matrix and standard unit vectors:

    u
     

     
    ×

    v
     

     
    =
    i
    j
    k
    u1
    u2
    u3
    v1
    v2
    v3
    unit vectors
    first vector
    second vector
  • Given two vectors u and v, the cross product produces a third vector u ×v that is perpendicular to both. However, notice that there are two possible directions such a perpendicular vector could point in. We orient the cross product based on the right-hand rule: point the fingers of your right hand in the direction of the first vector, with your palm in the direction of the second. The cross product's direction is the direction your thumb points.
  • The magnitude (length) of u ×v is equal to the area of the parallelogram enclosed by u and v. In general, we can interpret the length of u ×v as a measure of how long and how perpendicular the two vectors are.

Dot Product & Cross Product

Let u = 〈7,  12 〉 and v = 〈−3,  5 〉. What is u ·v?
  • u ·v is the dot product of u and v. The dot product of two vectors is the sum of the products of each pair of components.


  • u
     
    ·

    v
     
        =     〈7,  12 〉·〈−3,  5 〉    =     7 ·(−3)   +   12 ·5     =     −21 + 60     =     39
u ·v = 39
Let a = 〈−5,  8 〉 and b = 〈−7,  −11 〉. What is a ·b?
  • a ·b is the dot product of a and b. The dot product of two vectors is the sum of the products of each pair of components.


  • a
     
    ·

    b
     
        =     〈−5,  8 〉·〈−7,  −11 〉    =     (−5) ·(−7)   +   8 ·(−11)     =     35 −88     =     −53
a ·b = −53
Let u = 〈3,  9,  −5 〉 and v = 〈6,  0,  4 〉. What is u ·v?
  • u ·v is the dot product of u and v. The dot product of two vectors is the sum of the products of each pair of components. It doesn't matter how many components the vectors have, as long as they have the same number of components. Just take each pair of components, multiply them together, then sum it all up.


  • u
     
    ·

    v
     
        =     〈3,  9,  −5 〉·〈6,  0,  4 〉    =     3 ·6  +  9 ·0  +  (−5) ·4
    We take each pair of components, then multiply them together. From there, it's just a matter of simplifying:
    3 ·6  +  9 ·0  +  (−5) ·4     =     18 + 0 −20     =     −2
u ·v = −2
Let u = 〈4,  8 〉 and v = 〈6,  −1 〉. What is the angle θ between u and v?
  • The dot product of two vectors is connected to the angle between those vectors and their magnitudes. The following formula is very useful, and what we will need for this problem:

    u
     
    ·

    v
     
      =  |

    u
     
    | |

    v
     
    | cosθ
    For this problem specifically, we're focused on finding the angle between the vectors, so we can formulate the above as
    cosθ = 

    u
     
    ·

    v
     

    |

    u
     
    | |

    v
     
    |
  • Therefore, to solve for θ, we need to know what |u| and |v| are. Let's find those:
    |

    u
     
    |     =    

     

    42 + 82
     
        =    

     

    80
     
        =     4 √5

    |

    v
     
    |     =    

     

    62 + (−1)2
     
        =    

     

    37
     
  • Plugging in to the formula:
    cosθ = 

    u
     
    ·

    v
     

    |

    u
     
    | |

    v
     
    |
        =     cosθ = 〈4,  8 〉·〈6,  −1 〉

    4 √5 ·


    37
    Work out the dot product and simplify:
    cosθ    =     4 ·6 + 8 ·(−1)

    4


    185
        =     16

    4


    185
        =     4




    185
    Finally, solve for θ:
    cosθ = 4




    185
        ⇒     θ = cos−1

    4




    185


    Using a calculator, we get θ = 72.897°.
θ = 72.897°
Let a = 〈5,  −4 〉 and b = 〈8,  k 〉. What value must k be for a and b to be perpendicular?
  • If a and b are perpendicular, then we know the angle between them must be θ = 90°. From the dot product, we know for any two vectors

    u
     
    ·

    v
     
      =  |

    u
     
    | |

    v
     
    | cosθ.
    Therefore, if θ = 90°, since cos(90°) = 0, it must be that the dot product of a and b comes out to 0:

    a
     
    ·

    b
     
    = 0
  • Since we know a ·b = 0, we can set up the following:
    〈5,  −4 〉·〈8,  k 〉 = 0
  • From there, we can do the dot product
    5·8 + (−4)·k = 0,
    then work to solve for k:
    40 −4k = 0     ⇒     4k = 40     ⇒     k = 10
k=10
Let u = 〈3,  −7,  −8 〉 and v = 〈2,  −5,  10 〉. What is the angle θ between u and v?
  • Previously, we talked about an equation that connects the dot product of two vectors with the angle between them and the lengths of the vectors. This equation is true no matter how many components the vector has. While we're used to working with two-dimensional vectors, it makes no difference that u and v are three-dimensional this time: the equation is still just as true.

    u
     
    ·

    v
     
      =  |

    u
     
    | |

    v
     
    | cosθ
    For this problem specifically, we're focused on finding the angle between the vectors, so we can formulate the above as
    cosθ = 

    u
     
    ·

    v
     

    |

    u
     
    | |

    v
     
    |
  • Therefore, to solve for θ, we need to know what |u| and |v| are. Let's find those:
    |

    u
     
    |     =    

     

    32 +(−7)2 + (−8)2
     
        =    

     

    122
     

    |

    v
     
    |     =    

     

    22 + (−5)2 + 102
     
        =    

     

    129
     
  • Plugging in to the formula:
    cosθ = 

    u
     
    ·

    v
     

    |

    u
     
    | |

    v
     
    |
        =     cosθ = 〈3,  −7,  −8 〉·〈2,  −5,  10 〉




    122
    ·


    129
    Work out the dot product and simplify:
    cosθ    =     3·2 + (−7)·(−5) + (−8) ·10




    122
    ·


    129
        =     −39




    122
    ·


    129
    Finally, solve for θ:
    cosθ = −39




    122
    ·


    129
        ⇒     θ = cos−1

    −39




    122
    ·


    129


    Using a calculator, we get θ = 108.112°.
θ = 108.112°
Let p = 〈4,  0,  −3 〉 and q = 〈−2, 5,  7 〉. Find p ×q. Verify (by the dot product) that p ×q is actually perpendicular to p and q.
  • The cross product of two vectors produces a new vector that is perpendicular to the two vectors used in the cross product. It only works in three dimensions, and is formulated as below:
    〈u1, u2, u3〉·〈v1, v2, v3〉 = 〈u2 v3 − u3 v2,   u3 v1 −u1v3,   u1 v2 − u2 v1
  • This means we need to use the components of p and q based on their locations within the vector-whether they are first, second, or third. Based on the above formula for the cross product:

    p
     
    ×

    q
     
        =     〈0 ·7 − (−3)·5,    (−3)·(−2) − 4·7,    4 ·5 − 0 ·(−2) 〉
  • Simplify:

    p
     
    ×

    q
     
        =     〈15,   −22,   20 〉
  • To verify that p ×q is actually perpendicular to p and q, we can use the dot product. Remember, if the dot product of two vectors is 0, then those two vectors must be perpendicular. Therefore we can check for perpendicularity by making sure the dot product indeed comes out to be 0.

    (p ×q) ·p:
    〈15,  −22,  20 〉·〈4,  0,  −3 〉    =     15 ·4 + (−22) ·0 + 20 ·(−3)     =     0    


    (p ×q) ·q:
    〈15,  −22,  20 〉·〈−2, 5,  7 〉    =     15 ·(−2) + (−22) ·5 + 20 ·7     =     0    
p ×q = 〈15,  −22,  20 〉 To verify that p ×q is perpendicular to p and q, compute the dot products with each and show that they are equal to 0.
Find the area enclosed by the parallelogram where the vectors 〈2,  7〉 and 〈−8,  1〉 make up two of the sides.
  • We could approach this problem by finding a formula that gives the area of a parallelogram based on some of its measurements, then use a combination of trig and geometry to find those elements. However, using what we know about vectors, there's an easier way! For any two vectors u and v, the magnitude of their cross product u ×v is the area of the parallelogram they enclose.
  • First, we check how to find the cross product:
    〈u1, u2, u3〉·〈v1, v2, v3〉 = 〈u2 v3 − u3 v2,   u3 v1 −u1v3,   u1 v2 − u2 v1
    Ack! There's a problem! The above uses three-dimensional vectors, but the vectors we are working with are only two-dimensional! Luckily, there's an easy fix: we can give them an extra dimension, but just say they have 0 in that component. After all, in a way, everything in two-dimensional space is inside a three-dimensional space: it just doesn't poke off the plane into that third dimension. Thus, we can use the below vectors:
    〈2,  7〉    ⇒     〈2,  7,  0〉          
               〈−8,  1〉    ⇒     〈−8,  1,  0〉
  • Now find the cross product for those vectors:
    〈2,  7,  0〉×〈−8,  1,  0〉    =     〈7 ·0 − 0 ·1,     0 ·(−8) − 2 ·0,     2 ·1 − 7 ·(−8)
    Simplify:
    〈2,  7,  0〉×〈−8,  1,  0〉    =     〈0,  0,  58 〉
    Then take the magnitude of the cross product:


     

    02 + 02 + 582
     
        =     58
The area of the parallelogram is 58 square units.
In physics, the work done by a force F over a distance d is defined as W = F ·d. You push a crate along the floor for a distance of 15 m. Your push has a total force of 200 N (newtons) and is directed at an angle of 20° below horizontal. What is the total work you have done on the box by the end? [The unit for work/energy in the metric system is the joule→ J.]
  • Since work is defined as W = F ·d, we need to find F and d.
  • Finding d is quite simple. We know the box is pushed along the floor for a distance of 15 m. It doesn't move off the floor, so its distance vector is entirely horizontal:

    d
     
    = 〈15,  0 〉
  • Finding F is a little more complicated, but we can find it using some simple trigonometry. The length of its horizontal component (x) can be found as
    cos(20°) = x

    200
        ⇒     200 cos(20°) = x
    Using a calculator, we get x = 187.94. We find the length of the vertical component (y) similarly:
    sin(20°) = y

    200
        ⇒     200 sin(20°) = y
    Using a calculator, we get y=68.40. Now we can put these together to find F. However!, it is important to note that we only found the lengths of the components. We have to pay attention to the diagram to figure out their signs (+ or −). The box is being pushed to the right, so its x component will be positive. However, we see that the box is being pushed down (although it's against the floor, so it won't actually move down), so its y component is negative. Thus,

    F
     
    = 〈187.94,   −68.40 〉
  • Now that we know the vectors in component form for F and d, we only have to take the dot product to find the work:
    W = 〈187.94,   −68.40 〉·〈15,  0 〉    =     187.94 ·15 + (−68.40) ·0     =     2819.1
2819.1 j
Taking a nice stroll in the middle of a snowy, winter landscape, you see two children coming down a hill on a sled... but something looks wrong. Oh no! The sled is out of control! It's coming down way too fast, and the children don't know what to do! You are currently standing on a flat section of ground, and you see that the sled will come by near to where you are. The flat stretch of ground leads right to a clump of trees, and the sled will soon slam into them if you don't do something. Ever the hero, you decide to act. As the sled approaches, you notice a coil of rope on the back of the sled. At the moment the sled passes you, you snatch up the rope, and run along behind the sled, pulling backwards on the rope to provide some braking force on the sled. You pull backwards on the rope with a force of 140 N, and the rope makes an angle of 30° with the ground. You finally manage to bring the sled to a stop after it has slid a further 11 m. Much farther and it would have hit the trees, but thanks to your intervention, the children are safe! Congratulations! You're a hero! How much work did you do on the sled as it was brought to a stop? [Remember, work is defined as W = F ·d and the metric unit of work/energy is the joule→ J.]
  • First off-good on you! You really did an amazing thing, saving those kids like that. What a selfless act! Alright, now let's get down to analyzing what happened. To figure out the work involved, we can use the definition of work:
    W =

    F
     
    ·

    d
     
    Thus, to find the work, we need to know what the force (F) and distance (d) are in component form.
  • Finding d is pretty simple. We know the sled moved a further 11 m after you began pulling back on it. It didn't move into the air, so it only moved horizontally. Thus, we have

    d
     
    = 〈11,  0 〉.
  • Finding F is a little more complicated, but we can find it using some simple trigonometry. We know the magnitude of the force is 140 N and the angle is θ = 30° because the rope is at that angle. Using that, the length of its horizontal component (x) can be found as
    cos(30°) = x

    140
        ⇒     140 cos(30°) = x
    Using a calculator, we get x = 121.24. We find the length of the vertical component (y) similarly:
    sin(30°) = y

    140
        ⇒     140 sin(30°) = y
    Using a calculator, we get y=70. Now we can put these together to find F. However!, it is important to note that we only found the lengths of the components. We have to pay attention to the diagram to figure out their signs (+ or −). You are pulling backwards on the sled, so because the horizontal goes to the left, the x component must be negative. You pull up on the sled, so the y component is positive

    F
     
    = 〈−121.24,   70 〉
  • Now that we know the vectors in component form for F and d, we only have to take the dot product to find the work:
    W = 〈−121.24,  70 〉·〈11,  0 〉    =     (−121.24) ·11 + 70 ·0     =     −1333.64
−1333.64 j [If you're confused why the answer comes out to be negative, pay attention to the direction of the force compared to the direction of the sled's motion. Also, it makes sense that the work done on the sled is negative: you took energy out of the sled to slow it to a stop.]

*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.

Answer

Dot Product & Cross Product

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
  • Introduction 0:08
  • Dot Product - Definition 0:42
    • Dot Product Results in a Scalar, Not a Vector
    • Example in Two Dimensions
  • Angle and the Dot Product 2:58
    • The Dot Product of Two Vectors is Deeply Related to the Angle Between the Two Vectors
  • Proof of Dot Product Formula 4:14
    • Won't Directly Help Us Better Understand Vectors
  • Dot Product - Geometric Interpretation 4:58
    • We Can Interpret the Dot Product as a Measure of How Long and How Parallel Two Vectors Are
  • Dot Product - Perpendicular Vectors 8:24
    • If the Dot Product of Two Vectors is 0, We Know They are Perpendicular to Each Other
  • Cross Product - Definition 11:08
    • Cross Product Only Works in Three Dimensions
  • Cross Product - A Mnemonic 12:16
    • The Determinant of a 3 x 3 Matrix and Standard Unit Vectors
  • Cross Product - Geometric Interpretations 14:30
    • The Right-Hand Rule
    • Cross Product - Geometric Interpretations Cont.
  • Example 1 18:40
  • Example 2 22:50
  • Example 3 24:04
  • Example 4 26:20
  • Bonus Round 29:18
  • Proof: Dot Product Formula 29:24
    • Proof: Dot Product Formula, cont.

Transcription: Dot Product & Cross Product

Hi--welcome back to Educator.com.0000

Today, we are going to talk about the dot product and the cross product.0002

Now, we have some idea of vectors learned, so we can move on to looking at two new ways that vectors can interact: the dot product and the cross product.0005

These ideas are very important and useful in advanced math, science (especially in physics), and engineering.0014

If you have any interest in those fields, you definitely want to pay extra attention here.0019

But you are also going to need it just for this course.0022

All right, with each of these, we will start by looking at an algebraic definition,0025

and then exploring what that means in a geometric interpretation, so we can get a sense0028

of how this would look, as opposed to just a bunch of numbers.0031

All right, let's go!0035

First, the dot product: Given two n-dimensional vectors (that is, vectors that have n components,0036

where n will just be some number), u, which will be our first component, u1; our second component,0042

u2; up until the nth component, un...so u1, u2, u3,0048

all the way up until we get to un; and v, which is the same thing: v1, v2,0053

up until vn--so the first component of v, the second component of v...up until the nth component,0058

the last component, since it is just n-dimensional; then the dot product, which is just symbolized exactly as you would guess,0063

a dot between the vectors, is the sum of the products of each pair of components.0070

So, u dot v, which would be the same thing as u1, u2, up until un,0075

dotted with v1, v2, up until vn, would produce u1v10080

(the u1, the first component here, and the first component here multiply together),0086

and then we add that to the second component here and the second component here;0091

and then we add that to all of the ones in between, and then we add that to un,0096

the last one, multiplied by the last one of v, as well.0104

So, we have the first two components multiplied together, the second two components multiplied together...0108

we do this for every one of them, until we get to the last two components, multiplied together.0113

So, we add each of the possible component pairs, each of the component pairs between u and v, for everything that is in the same component location.0119

We multiply each of those together, and then we sum it all up.0126

Notice that the dot product will result in a scalar, not a vector.0129

What we get out of this in the end isn't another vector; it is just a real number.0132

And if we are talking about this, we say u, dot, v: so the symbols are vector u, dot product with, vector v.0137

We just say it as "u dot v," as you have heard me repeatedly say.0144

OK, so let's see an example in two dimensions.0149

If we have (5,2) dotted with (-7,12), then what we do is multiply the first two components, 5 and -7.0150

We have 5 times -7, and then we add that with 2 and 12, multiplied together.0159

5 and -7 gets us -35, plus 24; and finally, we get -11 as our answer when we add it together.0167

All right, it turns out that the dot product of two vectors is deeply related to the angle between the vectors and their magnitude.0175

Now, for us to talk about the angle between the vectors, we can't have them in totally separate places.0182

We have to put them together; so let's say we have two vectors like this and this.0186

Before we can talk about the angle and dot product, we have to put them tail-to-tail.0190

Now, once they are tail-to-tail, we can talk about the angle between them; so right from here to here, where my face is, would be the angle.0193

So, u with θ in between, and then v, is the two angles.0200

Now, it turns out that u dot v is equal to the length of u (remember, if we have vertical bars on either side,0203

that says the length which we get by that magnitude formula, the square root of each component squared and added together),0209

times the length of v, times cosine θ.0217

If we want, we can shift this around to have the formula that makes the angle a little more accessible.0222

Right now, we have cos(θ), but there is other stuff multiplied by it.0226

So, we could solve for cos(θ); we would have cos(θ) = u ·v divided by the length of u, times the length of v.0229

And then, at that point, if we want to find out the angle, we could just take the arc cosine of both sides, the cosine inverse of both sides.0236

And we would figure out what our angle is.0241

For the most part, though, I prefer this formula, because I think it is a little bit easier to remember0243

that u · v is equal to length of u, length of v, times the cosine of the angle between them.0247

All right, it turns out that we can actually prove this formula pretty easily.0252

It is not going to be that hard for us to prove it.0257

But it won't really help us directly understand vectors any better.0259

As such, the proof is still going to be in this lesson, but it has been put at the end, after the examples.0263

So, if you wait until we get through all of the examples, you will get to see the proof, if you are curious about it.0268

If you have a few extra minutes, and you are interested, awesome--I would love to have you check it out.0272

It is really cool to just get the sense of more things in math--all of the proofs.0276

But if you are busy, and you don't have time for it, that is OK, too.0279

It is really...if you had to miss one proof, this is probably the best one to just skip and take in faith.0281

So, don't worry about it if you end up not being able to watch this proof.0286

But if you have the time, and you are interested in proofs, it is really cool, and it is totally something we can actually manage pretty easily.0289

OK, let's interpret this geometrically: we can break up our new formula, u · v = length u times length v times cos(θ),0294

into length u times length v times cos(θ); so we think of it as two different pieces:0302

the length of one vector, times the length of the other vector, multiplied by cosine θ.0308

So, what that gives us: if we look at this, this length v times cosine θ is a way of thinking of the length of v0313

if we projected it onto u--v as a projection onto the vector u.0320

What does that mean? Let's look at these pictures here.0325

We have v here; v is this vector here, so if v is shorter than u, we project it by dropping a perpendicular down to u;0328

and then, what we have is: wherever u went, up until that perpendicular that we just dropped down,0338

that is our length, length v times cosine θ.0344

So, that gives us the length of the projection.0349

We can think about this idea as how far we drop down; once we drop down this perpendicular onto u, we have created a projection of v onto u.0351

It is like you take a flashlight and shine it directly down onto it.0358

It is the shadow that v would cast on u.0361

So, length of v, times cosine θ--well, remember: this is, after all...0365

since it is a perpendicular that we dropped down, we have a right angle in there.0371

So, since we have a right angle in there, and we have θ, it is just basic trigonometry.0374

Cosine of θ times the length of the hypotenuse, length v, will give us the length that it is for that projection--the shadow that we just created.0377

And we can also do this if our vector v ends up being longer than u.0386

Instead of worrying about how long we are on u, it is if u had continued (this dotted line right here),0390

and then we had dropped a perpendicular onto if-it-had-continued, where would we end up being on that continuation?0399

So, we think of u, and then we think of u continuing off forever, and then v drops down onto that, and that gives us our projection of v.0405

So, this gives us another way to think about where this is coming from.0411

It is the length of the projection, multiplied by the length of the vector it is being projected onto.0415

So, we see that u · v is the length of one vector (u in this case), multiplied by the length of the other vector's projection.0421

Our other vector would be v, and its projection was this part right here; that is length v cosine θ.0432

In general, we can interpret the dot product as a measure of how long and how parallel two vectors are.0442

So, for example, if we have two vectors like this, we are going to get a larger dot product the smaller our θ is,0448

because cosine of numbers close to 0 gets us numbers that will be close to 1.0456

Cosine of 0 is 1; that is the absolute maximum.0461

When they are perfectly parallel, that is going to be the largest possible projection that v can make onto u.0464

As it gets less and less parallel, more and more perpendicular, though, cos(θ) is going to become smaller and smaller.0469

This projection that is dropping down is going to become smaller and smaller, until finally, we eventually hit perpendicular,0476

and it drops down, and there is no projection whatsoever.0481

It is like shining a shadow on something pointing straight up; it doesn't cast a shadow at all--there is no projection that comes out of it.0484

So, once they are perfectly perpendicular, we are going to get nothing out of it.0490

But if they are parallel, the more parallel they are, the longer they are, the larger the value that we will get out of the dot product.0493

This brings up a good point--this idea that, when they are perpendicular, when we have cosine of 90 degrees,0501

or cosine of π/2 (degrees versus radians), we end up getting a cosine that comes out as a 0--it goes to 0.0508

So, we will have nothing coming from the dot product, because there will be no projected shadow onto it.0515

From our formula, u · v = length u(length v)(cos(θ)), we see that, if u · v = 0,0521

then it must be the case that cos(θ) = 0, and if cos(θ) = 0, then θ = 90 degrees.0527

So, if we have the dot product of two vectors as 0, we know we have perpendicularity; they have to perpendicular to each other.0535

If u · v = 0, then u is perpendicular to v; and if u is perpendicular to v, then u · v = 0.0545

I also want to point out that there are many equivalent words for "perpendicular."0552

Perpendicular is...sometimes you will hear "normal," or maybe even "orthogonal,"0555

although you normally don't hear that until later math classes, or "at right angles."0560

But all of these might end up meaning the same thing: θ = 90 degrees, or alternatively, θ = π/2, if we are in radians.0564

So, perpendicular, normal, orthogonal, right angles, θ = 90 degrees, θ = π/2--0573

they all mean the same thing, this idea of "perpendicular," something that we are used to from geometry.0579

Don't get confused if you end up hearing one of these synonyms; they just all mean the same thing.0584

You are probably wondering why there are so many synonyms, and it is because it is a really important idea,0588

and it shows up in a bunch of different places; so at different times, different people used different words.0592

And so, we ended up having something like four different ways of talking about this; so, that is why we see so many.0597

Also, I want to point out one other thing: technically, up here, if we have u · v = 0, well, u or v could be the zero vector, as well.0602

If u or v was the zero vector, all 0's in all of the components, then u · v is going to come out as 0, also.0612

We would get u · v = 0 from that, as well.0619

But is our θ really going to be 90 degrees?0621

Well, at that point, if we are just a dot, we are going to say that a dot, that 0 vector, is just going to be perpendicular to everything,0624

because at that point, it is not really sticking out in any direction.0631

So, it seems reasonable to say that it is perpendicular.0634

And it works well with our new idea of "perpendicular" meaning that the dot product equals 0.0637

So, we get around this by saying that the zero vector is going to be perpendicular to all vectors.0642

So, if something is the zero vector, it automatically is going to be perpendicular.0648

But mostly, when we think of this, it is going to work perfectly well.0651

And because of this new thing about saying that the zero vector is always going to be perpendicular, it works out all the time.0656

So, that is just always true; we can think of perpendicular as meaning that the dot product comes out to be 0.0661

All right, let's talk about the cross product now.0667

Unlike the dot product, the cross product will only work in three dimensions.0670

When you are in space, like three-dimensional space, that is when you can use the cross product.0674

It takes two vectors, and it produces a third vector that is perpendicular to both.0678

What comes out of the cross product is something that will be perpendicular to our first vector and our second vector.0684

If u is equal to u1, u2, u3, and v is equal to v1, v2, v3,0690

it is this monster: u cross v equals u2v3 - u3 v2,0694

u3v1 - u1v3, u1v2 - u2v1,0700

where each one of those is multiplied and then subtracted from the other.0704

So, those are our three components.0707

We say u cross v as "u cross v," this vector u cross product with vector v; we just say it as "u cross v" to make it easy.0708

We will calculate a product in Example 1; and basically, this formula is tough to remember, because it is just so many symbols at once.0717

So, there is a pretty good mnemonic for this; but you might not have seen what we are about to use for it.0725

So, we haven't learned about matrices or their determinants yet; but there is a great mnemonic, if you are familiar with it.0731

If you learned about this in a previous math course, or if you go on and watch the lesson on determinants in a little bit--0738

and matrices--there is a great mnemonic for remembering the cross product0743

by using the determinant of a 3x3 matrix and the standard unit vectors.0747

So, if we have u cross v, we can also write that as the determinant of the 3x3 matrix, ijk on the top,0751

then the first vector, u1u2u3, on the second one, and then v1v2v3.0759

At this point, we take ijk; we do a co-factor expansion on ijk, so when we look at i, that will knock out the u1v1 0775.8 and the jk; so we have i along with the determinant, u2u3/v2v3, 0766

times the unit vector i, minus...next we do j; j will knock out u2v2; i knocked u1v1,0782

and j and k; j will knock out i and k, and will also knock out u2v2,0793

leaving us with u1u3v1v3.0799

So, we take the determinant of that and multiply that by our j.0802

And then finally, we get to k; k and its cofactor will knock out ij and u3v3.0805

And so, we get u1u2v1v2k.0812

And then, if you take the determinants of each of those 2x2 matrices, multiplying on the diagonal going down,0814

that is the positive; and then subtract on the diagonal going up...so u2v3 - u3v2i,0820

minus u1v3, minus u3v1j,0827

plus u1v2 minus u2v1.0830

That is another way of doing it; it is a pretty good mnemonic; if you are familiar with determinants, it works great.0832

If you are not familiar with this, that all probably didn't make very much sense.0838

So, you can go ahead and watch the later lesson on determinants, or you can just go back to that previous slide and just end up memorizing that formula.0841

There is not really a very easy way to memorize it, other than this mnemonic, which works great.0847

But if you don't know the mnemonic, it is a little bit difficult.0851

Yes, it is really great if you learned determinants, though.0855

And you will learn that later on, so it might be worth just learning it now,0857

so you can have this stuck in your head if you end up having to do a lot of work with cross products.0860

All right, what does this mean geometrically?0864

If you are given two vectors, u and v, the cross product produces a third vector, u cross v, that is perpendicular to both.0868

Say we have some u going off like this, and we have v going off like this.0874

Now, what we have, u cross v, is this third vector that comes out of both that will be perpendicular to both.0881

I think you can see this: so, u cross v would end up being perpendicular to both.0889

It is perpendicular to u, but it is also perpendicular to v.0893

And so, it comes out like that, and we have u cross v.0897

However, how do we tell which way u cross v is going to point?0900

The perpendicular vector can come out like this, but the perpendicular vector could also come out like this.0905

There is nothing wrong with being perpendicular on the underside, as well.0910

So, how do we tell which way you end up going?0913

The trick to this is the right-hand rule: you point the fingers of your right hand in the direction of the first vector.0916

And then, your palm goes in the direction of the second vector; you can also think of it as curling your fingers toward it.0923

And the cross-product's direction is the direction your thumb points.0928

In this one, if this is u and this is v, then you put u...my fingers are along u, and then my palm goes towards v.0931

So, u, v...and u cross v comes out like this, which is exactly what you get from this picture right here.0940

If you do u with your fingers, and then v with your palm, and then bring your thumb out, you will be able to see u cross v coming out in purple there.0947

I really recommend trying this out right in front of you right now, because there is really no way to get this done geometrically,0956

in your head, without actually seeing it visually in front of you.0961

And then, if we wanted to see what v cross u would be, right-hand rule: this is our v; this is our u.0964

So, v goes first; the fingers go in the v direction.0970

And then, u: the palm goes in the u direction, so now our thumb is pointing down.0975

v, u, down; v cross u will go down like that; that is the right-hand rule.0980

Fingers go in the first vector's direction; palm (or fingers curling--either way you want to think about it) goes in the direction of the second vector.0987

And whatever you have with your thumb, that is the direction that your cross-product is going to come out of it.0995

It is definitely worth trying that; try it right now--make sure you end up seeing that you are getting the same thing,1000

that you can do this with your hands, that you can see this, because it is really difficult to visualize1005

purely with your mind; but if you use it with your hands, you will be able to see it very well.1010

This stuff comes up all of the time in physics and engineering--it is really important stuff there.1014

All right, cross-product...now, another way to interpret this geometrically...1019

We talked about the direction, but we haven't talked about how long u cross v is going to be.1023

The magnitude--how long u cross v will be--is equal to the area of the parallelogram that is enclosed by u and v.1027

So, we have u and v; if we continue those out, u is down here, and so we also do a parallel one here.1035

This is parallel to this, and then our v here is parallel to this.1042

So, that makes a parallelogram; the area inside of that parallelogram is how long u cross v will be.1048

So, notice that the more parallel u is to be, the more squished it becomes.1055

If this is our u and v, the more parallel they are, the less area that there is going to be inside of them.1059

They get squished more and more; as we open it up more and more, though, we have this larger area inside of it.1064

So, as we squish it down, as u becomes more parallel to be, the less area it has.1070

The more perpendicular, the more we open it up...we have this wide area.1076

When we are perpendicular, we are going to have the maximum amount of area, because we will be a perfect square.1080

So, the more it opens up, the more perpendicular it is; the more it opens up, the larger the area becomes.1085

With this in mind, we can interpret the length of u cross v as a measure of how long and how perpendicular the two vectors are.1091

If the vectors aren't very perpendicular at all, then we are not going to get much out of the cross product, in terms of its length.1099

If they are really perpendicular, we are going to get a lot more out of it.1106

And of course, we can just make the vectors longer in the first place to increase this area.1110

All right, we are ready for some examples.1114

The first one: A vector, a, equals (2,4,-5); vector b = (-3,1,2); the first thing to do is give the cross product of a and b, a cross b.1116

Then, we want to show, by the dot product, that a is perpendicular to a cross b and that b is also perpendicular to a cross b.1125

So, our first thing to do is just to figure out what a cross b is.1134

We have (2,4,-5), (-3,1,2); so (2,4,-5) is crossed with (-3,1,2).1137

We have this formula here: here is our formula; so for the first coordinate of our outcoming cross-product,1150

it is going to be the second component of the first vector, u2 (u2 would be, in this case, 4),1158

times the third component of the second vector, v3 (so that would be 2 here), so 4 times 2;1165

minus the third component of the first vector (that is a -5), times the second component of the second vector, v2 (that is 1);1171

the same thing is going on; see if you can follow along here...1182

u3 is -5, times v1 is -3, minus u1 (u1, in this case, is 2); v3 is 2,1184

as well; comma, u1v2 (u1 is 2; v2 is 1),1197

minus u2 is 4...v1 is -3; great.1204

We start simplifying this out; we have 8 minus a negative--that cancels out, so we have 8 + 5.1211

-5 times -3 becomes positive 15, minus 2 times 2...minus 4...2 times 1 is 2; minus 4 times -3...they cancel out, and we have addition there,1217

as well; plus 12; simplify that, and we get (13,11,14), so that is a cross b.1229

So, there it equals a cross b; there is our cross-product vector.1243

Now, we want to verify this; we want to show that it is, indeed, going to be perpendicular to both a and b,1249

because we know that the cross-product has to be perpendicular to both of them, so that had better come out.1254

So, if a is perpendicular to a cross b, then that will be true if a dot a cross b comes out to be 0.1258

So, if a dot a cross b comes out to be 0, we know that that is perpendicular by how the dot product works.1267

Remember: that was one of our big realizations about the dot product--that if θ is equal to 90 degrees,1271

if the two vectors are perpendicular to each other, then the dot product of the two vectors always comes out to be 0.1276

a dot a cross b: our a is (2,4,-5); our b...sorry, not our b; we are dotting that with a cross b; a cross b is (13,11,14).1281

2 times 13 is 26, plus 4 times 11 is 44, plus -5 times 14 is -70.1295

26 + 44 becomes 70, so we have 70 - 70; that comes out to be 0, so that checks out.1307

Our dot product came out to be 0, so we know that they must be perpendicular; great.1313

The next one: Show that b is perpendicular with a cross b: so b, dotted with a cross b:1318

b is (-3,1,2); our a cross b that we are dotting with is (13,11,14) (I am running a little bit out of room there);1332

-3 times 13 becomes -39; plus 1 times 11; plus 2 times 14 is 28; -39 + 39...11 + 28...-39 + 39 becomes 0, so that checks out, as well.1344

The dot product of b with a cross b comes out to be 0, so we know that those two vectors must be perpendicular.1361

So, there we are; we have finished that one.1367

All right, the second example: we have u = (5,2,8,k); v = (3,-4,1,3).1369

What is k if u is perpendicular to v? If u is perpendicular to v, then that tells us that u dot v equals 0.1377

Great; so if u dot v equals 0, then we have that (5,2,8,k), dotted with (3,-4,1,3), equals 0.1389

So, we work this out: 5 times 3 is 15; plus 2 times -4 is -8; plus 8 times 1 is 8; plus k times 3 is 3k; equals 0.1406

-8 + 8...they cancel each other out, so we have 15 + 3k = 0; 3k = -15, which gives us k must equal -5.1420

So, there is nothing really difficult there; as soon as we realized that if they are perpendicular,1433

then it must be that their dot product is 0, at that point, we can set something up that we can just solve through simple algebra.1437

The third example: In physics, the work done by a force f over a distance d is defined as force dotted with distance.1443

So, the work, W, is equal to the force vector, dotted with the distance vector.1452

If you push a box with a mass of 20 kilograms with a force of 100 Newtons at an angle of 15 degrees above the horizontal,1457

for 10 meters, how much work have you done on the box?1463

So, the first thing we can do is figure out: All right, what is our force vector in terms of its components?1467

So, we could get the force vector...force equals component stuff; and then we will figure out d = component stuff...1473

well, that will be easy, because it is entirely horizontal; so we will have to use trigonometry to figure out what the force vector is,1481

in its component form, and then we can dot the two together.1486

But that is actually more work than we have to do.1488

All we have to do is remember that we don't need component form at all, because we know that,1490

if work equals the force vector dotted with the distance vector, then, well, u dotted with v is the same thing as length u times length v,1495

times cosine θ, so this is the length of our force vector,1505

times the length of our distance vector, times the cosine of the angle between them.1509

We know what our force vector is--it came out to be 100 Newtons; the force was 100 Newtons.1514

We know what our distance is: we go for a distance of 10 meters.1519

And we know what our angle θ is; do we need the 20 kilograms of mass?1523

20 kilograms of mass actually never shows up for figuring out the work; the mass of the object has no effect on the amount of work that goes in.1526

It is all about the force, the distance that happens, and the angle between those two--how the two interrelate.1533

So, we actually don't need to know the mass of the box at all to figure this one out; it is just a "red herring."1540

So, force is 100 Newtons, times distance (is 10 meters), times cosine of the angle of 15 degrees.1545

We work that out with a calculator: we have 1000 times the cosine of 15 degrees.1556

And that comes out to be 965.93; now, what are the units of work?1560

They told us in the problem that the unit is the Joule, or joules, which is signified with a J.1566

So, we use the unit of J at the end of it.1572

And there we are; there is our work; great.1576

All right, the final example: Prove that u dot u equals the magnitude of u, squared.1579

All right, the first thing to do: we have to have a way of talking about just some general vector u,1585

because they didn't tell us much about u at all.1590

They told us just "vector u"; so we need to be able to talk about what vector u is, in a way that we can actually work with it.1592

Let's just give the components names; we will do it in the same way that has happened in all of the previous stuff,1598

where we have just said that the first component is u1 (so we will make this u1);1604

and then the second component will be u2, and then the third component would be u3,1608

and all the way up until some un, because every vector has to have some specific length.1613

It is not allowed to go on forever; so we will stop at un...n will be just the length of our vector.1619

This is going to be the case for any vector at all; we could put it in this form of first component, second component,1625

up until its last component, which we will say will be its nth location in the thing.1630

All right, so now we have a way of doing this; let's just look at what u dot u is, and then what the magnitude of u2.1634

If they end up being equal, we have proved this thing.1640

So, u dot u would be vector u1, u2, up until un, dotted with u1, u2, up until un.1643

All right, so u1 times u1...well, u1 is just some number, so that is (u1)2,1660

plus...u2 times u2...well, u2 is just some number, so that is (u2)2,1665

plus...this is just going to keep happening, until we get to our final component.1670

un times un...well, un is just some number, as well, so that is (un)2.1673

So, we have u1 squared, plus u2 squared, up until we get to un squared.1677

Now, there is not much we can do to simplify that there; so let's take a look at the magnitude of u2.1682

So, first, what is the magnitude of u? Well, the magnitude of any vector, remember,1688

is the square root of each component squared added together underneath the square root.1693

So, it would be the first component squared, plus the second component squared,1699

up until the final component squared, and all of that underneath the square root.1703

Well, if we square this, then we have the magnitude of u, squared, equals...well, if we square both sides of that equation above,1708

the square root times the square root cancels out, and we just have u1 squared plus u2 squared,1717

up until un squared; well, look: this and this are the same thing.1724

So, since they are the same thing, we have just shown that u dot u is equal to the magnitude of u, squared.1733

Great; the proof is finished.1742

All right, that finishes for the examples; thanks for coming to Educator.com.1744

And we will see you in the next lesson, when we start talking about matrices--all right, goodbye!1748

All right, I think they are gone--everybody who is not actually interested in the proof.1754

So, are you ready for this: Bonus round--here we are, ready for the proof!1757

Dot product formula: let's prove that u · v is equal to the length of u, times the length of v, times cosine θ.1764

The very first thing that you want to do any time you are really trying to think about anything analytically is draw a picture.1770

A picture is always a useful way to think about things.1775

So, we start by drawing a picture: we have u and v, with θ in between.1778

Now, we look at this, and we want to see: is there any way to connect the length of u, the length of v, and θ together?1783

Is there some way to get these things to talk to each other?1788

Do we know any way to say, "Yes, I know these are related"?1791

Well, we look at this for a while, and we say, "Well, I don't see anything yet."1794

But that looks kind of like a triangle; and I know a lot of things about triangles from trigonometry.1798

So, let's say it looks like a triangle without a top; let's give that triangle a top!1803

We draw in the top in this purple color; and we might realize that there are three sides to a triangle;1808

I know an angle...sort of...oh, I can connect the length of u, the length of b, that angle θ, and the length of the top,1816

together with something that we learned in trigonometry: the law of cosines.1823

We might remember this; and if we remember this, we just go back and look up the law of cosines.1827

There is no reason not to just go and look it up.1831

We look up the law of cosines in a book; we refresh ourselves, and we get that the length a2,1833

the side a, squared, is equal to the other two sides squared (b2 + c2),1839

minus 2 times b times c times cosine of capital A, where little a and capital A are the side and then the angle opposite;1846

so little a is the length of one side, and then capital A is the angle opposite that side; so we have:1857

a2 = b2 + c2 - 2bc times cos(A).1863

That is the law of cosines; now, we bring up our vectors picture, and we look at this.1868

Now, one thing before we keep going: how are we going to have this top as u - v?1872

We can see the top as the vector u - v, since it runs from the head of v to the head of u.1877

And think about this: if we take v, and we add that to u - v, well, the -v and the +v would cancel out, and we would be left with just u.1883

So, it must be the case: we can see graphically, through this algebra, that we can get from v to u by using u - v,1891

because it will take away the v and give us the u, and we will manage to get from the head of v to the head of u.1899

Cool; so that is how we have that vector u minus vector v is the way to be able to talk about the top as a vector.1905

All right, using the law of cosines, then we have that the length of u - v, our a, squared, is equal to the length of u, our b,1911

this part right here, squared, plus the length of v, this part right here, c, squared, minus 2 times the length of u,1926

times the length of c, times cosine of the angle between them (cosine of θ).1948

So, we have the length of vector u - v, squared, equals the length of u, squared, plus the length of v, squared,1957

minus 2 times the length of u times the length of v, times cos(θ).1965

All right, it looks like we are getting somewhere: we have some relationships going on.1969

We even have that cos(θ), if we are trying to prove that thing.1972

So, it looks like we are getting there; but we still have this problem, where we don't have any dot products there.1975

So, how can we get u · v to show up in there? We want u · v in there.1980

Well, remember: in Example 4, we just proved, for any vector a, a dotted with itself (a vector dotted with itself)1985

is equal to the magnitude of that vector, squared; so a dot a equals the magnitude of a, squared.1992

Thus, we can swap out each of these magnitude-squareds for a dot a.1998

So, we have that relationship, whatever they are.2004

So, u - v: the length of u - v squared will become the vector u - v, dotted with the vector u - v.2006

u squared: the magnitude of u squared will become the vector u, dotted with the vector u.2014

The magnitude of v squared will become v dot v.2018

So, we have all of these swapping out right here.2022

OK, at this point...we didn't show this technically, but you can prove it to yourself--it is not that difficult:2026

the dot product is distributive, so we can actually distribute using this dot product.2032

u - v dot u - v: well, then, we have u dot u minus u dot v, minus v dot u plus v dot v.2038

u dot u minus u dot v, minus v dot u minus...times a negative again, so plus v dot v.2048

Great; the stuff on the right just stays the same.2060

At this point, we see that we have certain things on the right and the left.2062

So, if we have v · v on both sides, let's just subtract it on both sides.2065

We have u · u on both sides; let's just subtract it on both sides.2068

So, we have - u · v - v · u.2071

Well, notice: u · v is just the same thing as v · u, so we can combine them together.2073

If we have u · v + v · u, then that is the same thing as 2 times u · v.2080

So, if we have - u · v - v · u, then that is the same thing as -2(u · v).2085

So, we have -2(u · v) there on the left; it equals -2 length of u, times length of v, times cosine θ.2090

We have -2 on both sides; divide by -2 on both sides; those cancel out; we have u · v equals the length of u, times the length of v, times cosine θ.2097

Cool; and our proof is finished--that was not too tough.2104

All right, thanks for staying around; I think proofs are really cool.2107

They are really, in my mind, the heart of mathematics--being able to show that this stuff is definitely always true.2110

I think it is awesome; thanks for staying around--I was glad to share it with you.2116

We will see you at Educator.com later--goodbye!2118