For more information, please see full course syllabus of Pre Calculus

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

### Dot Product & Cross Product

- Given two n-dimensional vectors →u = 〈u
_{1}, u_{2},…,u_{n}〉 and →v = 〈v_{1}, v_{2}, …, v_{n}〉, the*dot product*(symbolized by a · between the vectors) is the sum of the products of each pair of components.

Notice the dot product results in a→u

· →v

= 〈u _{1}, u_{2},…,u_{n}〉·〈v_{1}, v_{2}, …, v_{n}〉= u _{1}v_{1}+ u_{2}v_{2}+ …+ u_{n}v_{n}__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

[There are many equivalent words for `perpendicular': `normal', `orthogonal', `at right angles', but these all mean the same thing: θ = 90→u

· →v

=0 ⇔ →u

is perpendicular to →v

. ^{°}. (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 = 〈u_{1}, u_{2}, u_{3}〉 and →v = 〈v_{1}, v_{2}, v_{3}〉, the cross product is

[We say →u ×→v as `u cross v' when speaking.]→u

× →v

= 〈 u _{2}v_{3}− u_{3}v_{2}, u_{3}v_{1}−u_{1}v_{3}, u_{1}v_{2}− u_{2}v_{1}〉. - 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**u _{1}u _{2}u _{3}v _{1}v _{2}v _{3}← 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
: point the fingers of your*right-hand rule*__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

- →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

- →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

- →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.

We take each pair of components, then multiply them together. From there, it's just a matter of simplifying:→u· →v= 〈3, 9, −5 〉·〈6, 0, 4 〉 = 3 ·6 + 9 ·0 + (−5) ·4 3 ·6 + 9 ·0 + (−5) ·4 = 18 + 0 −20 = −2

- 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:

For this problem specifically, we're focused on finding the angle between the vectors, so we can formulate the above as→u· →v= | →u| | →v| cosθ cosθ = →u· →v| →u| | →v| - Therefore, to solve for θ, we need to know what |→u| and |→v| are. Let's find those:
| →u| =

√4^{2}+ 8^{2}=

√80= 4 √5 | →v| =

√6^{2}+ (−1)^{2}=

√37 - Plugging in to the formula:

Work out the dot product and simplify:cosθ = →u· →v| →u| | →v| = cosθ = 〈4, 8 〉·〈6, −1 〉 4 √5 ·

√37

Finally, solve for θ:cosθ = 4 ·6 + 8 ·(−1) 4

√185= 16 4

√185= 4

√185

Using a calculator, we get θ = 72.897cosθ = 4

√185⇒ θ = cos ^{−1}⎛

⎜

⎝4

√185⎞

⎟

⎠^{°}.

^{°}

- 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

Therefore, if θ = 90→u· →v= | →u| | →v| cosθ. ^{°}, 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

then work to solve for k:5·8 + (−4)·k = 0, 40 −4k = 0 ⇒ 4k = 40 ⇒ k = 10

- 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.

For this problem specifically, we're focused on finding the angle between the vectors, so we can formulate the above as→u· →v= | →u| | →v| cosθ cosθ = →u· →v| →u| | →v| - Therefore, to solve for θ, we need to know what |→u| and |→v| are. Let's find those:
| →u| =

√3^{2}+(−7)^{2}+ (−8)^{2}=

√122| →v| =

√2^{2}+ (−5)^{2}+ 10^{2}=

√129 - Plugging in to the formula:

Work out the dot product and simplify:cosθ = →u· →v| →u| | →v| = cosθ = 〈3, −7, −8 〉·〈2, −5, 10 〉

√122·

√129

Finally, solve for θ:cosθ = 3·2 + (−7)·(−5) + (−8) ·10

√122·

√129= −39

√122·

√129

Using a calculator, we get θ = 108.112cosθ = −39

√122·

√129⇒ θ = cos ^{−1}⎛

⎜

⎝−39

√122·

√129⎞

⎟

⎠^{°}.

^{°}

- 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:
〈u _{1}, u_{2}, u_{3}〉·〈v_{1}, v_{2}, v_{3}〉 = 〈u_{2}v_{3}− u_{3}v_{2}, u_{3}v_{1}−u_{1}v_{3}, u_{1}v_{2}− u_{2}v_{1}〉 - 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

- 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:

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:〈u _{1}, u_{2}, u_{3}〉·〈v_{1}, v_{2}, v_{3}〉 = 〈u_{2}v_{3}− u_{3}v_{2}, u_{3}v_{1}−u_{1}v_{3}, u_{1}v_{2}− u_{2}v_{1}〉〈2, 7〉 ⇒ 〈2, 7, 0〉 ⎢

⎢〈−8, 1〉 ⇒ 〈−8, 1, 0〉 - Now find the cross product for those vectors:

Simplify:〈2, 7, 0〉×〈−8, 1, 0〉 = 〈7 ·0 − 0 ·1, 0 ·(−8) − 2 ·0, 2 ·1 − 7 ·(−8)

Then take the magnitude of the cross product:〈2, 7, 0〉×〈−8, 1, 0〉 = 〈0, 0, 58 〉

√0^{2}+ 0^{2}+ 58^{2}= 58

*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

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

Using a calculator, we get y=68.40. Now we can put these together to find →F.sin(20 ^{°}) =y 200⇒ 200 sin(20 ^{°}) = y**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

^{°}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:

Thus, to find the work, we need to know what the force (→F) and distance (→d) are in component form.W = →F· →d - 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

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

Using a calculator, we get y=70. Now we can put these together to find →F.sin(30 ^{°}) =y 140⇒ 140 sin(30 ^{°}) = y**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

*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
- Introduction
- Dot Product - Definition
- Angle and the Dot Product
- Proof of Dot Product Formula
- Dot Product - Geometric Interpretation
- Dot Product - Perpendicular Vectors
- Cross Product - Definition
- Cross Product - A Mnemonic
- Cross Product - Geometric Interpretations
- Example 1
- Example 2
- Example 3
- Example 4
- Bonus Round
- Proof: Dot Product Formula

- 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.

### Precalculus with Limits Online Course

### 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 sense*0028

*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, u _{1}; our second component,*0042

*u _{2}; up until the n^{th} component, u_{n}...so u_{1}, u_{2}, u_{3},*0048

*all the way up until we get to u _{n}; and v, which is the same thing: v_{1}, v_{2},*0053

*up until v _{n}--so the first component of v, the second component of v...up until the n^{th} 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 u _{1}, u_{2}, up until u_{n},*0075

*dotted with v _{1}, v_{2}, up until v_{n}, would produce u_{1}v_{1}*0080

*(the u _{1}, 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 u _{n},*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 remember*0243

*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 v*0313

*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 u _{1}, u_{2}, u_{3}, and v is equal to v_{1}, v_{2}, v_{3},*0690

*it is this monster: u cross v equals u _{2}v_{3} - u_{3} v_{2},*0694

*u _{3}v_{1} - u_{1}v_{3}, u_{1}v_{2} - u_{2}v_{1},*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 product*0743

*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, u _{1}u_{2}u_{3}, on the second one, and then v_{1}v_{2}v_{3}.*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 u _{1}v_{1} 0775.8 and the jk; so we have i along with the determinant, u_{2}u_{3}/v_{2}v_{3},*0766

*times the unit vector i, minus...next we do j; j will knock out u _{2}v_{2}; i knocked u_{1}v_{1},*0782

*and j and k; j will knock out i and k, and will also knock out u _{2}v_{2},*0793

*leaving us with u _{1}u_{3}v_{1}v_{3}.*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 u _{3}v_{3}.*0805

*And so, we get u _{1}u_{2}v_{1}v_{2}k.*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 u _{2}v_{3} - u_{3}v_{2}i,*0820

*minus u _{1}v_{3}, minus u_{3}v_{1}j,*0827

*plus u _{1}v_{2} minus u_{2}v_{1}.*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 visualize*1005

*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, u _{2} (u_{2} would be, in this case, 4),*1158

*times the third component of the second vector, v _{3} (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, v _{2} (that is 1);*1171

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

*u _{3} is -5, times v_{1} is -3, minus u_{1} (u_{1}, in this case, is 2); v_{3} is 2,*1184

*as well; comma, u _{1}v_{2} (u_{1} is 2; v_{2} is 1),*1197

*minus u _{2} is 4...v_{1} 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 u _{1} (so we will make this u_{1});*1604

*and then the second component will be u _{2}, and then the third component would be u_{3},*1608

*and all the way up until some u _{n}, because every vector has to have some specific length.*1613

*It is not allowed to go on forever; so we will stop at u _{n}...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 n ^{th} 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 u ^{2}.*1634

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

*So, u dot u would be vector u _{1}, u_{2}, up until u_{n}, dotted with u_{1}, u_{2}, up until u_{n}.*1643

*All right, so u _{1} times u_{1}...well, u_{1} is just some number, so that is (u_{1})^{2},*1660

*plus...u _{2} times u_{2}...well, u_{2} is just some number, so that is (u_{2})^{2},*1665

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

*u _{n} times u_{n}...well, u_{n} is just some number, as well, so that is (u_{n})^{2}.*1673

*So, we have u _{1} squared, plus u_{2} squared, up until we get to u_{n} squared.*1677

*Now, there is not much we can do to simplify that there; so let's take a look at the magnitude of u ^{2}.*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 u _{1} squared plus u_{2} squared,*1717

*up until u _{n} 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 a ^{2},*1833

*the side a, squared, is equal to the other two sides squared (b ^{2} + c^{2}),*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

*a ^{2} = b^{2} + c^{2} - 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

1 answer

Last reply by: Professor Selhorst-Jones

Sun Sep 18, 2016 1:00 PM

Post by Khanh Nguyen on September 17, 2016

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.