# Dot Product & Angle Between Vectors

BAM!!! Mr. Tarrou. In this video we are going

to take a look at the Dot Product and see how it helps us find the angle between two

vectors. Now the Dot Product is the process, of well let’s see, first you have to have

two vectors in their horizontal and vertical components. Here I have vector v which is

a sub 1 times vector i plus b sub 1 times vector j and vector w which is a sub 2 times

vector i plus b sub 2 times vector j. I am using these i and j notation, this horizontal

component vector and the vertical component vector which each has the length of one because

my textbook stresses that a lot. You might see this in a bit of a short hand version

where it is just a,b in a set of pointed parenthesis if you will. The Dot Product itself is found

by taking the product, or multiplying the a values a sub 1 times a sub 2 plus b sub

1 times b sub 2. You just this scaler multiple, this numerical value. It alone does not necessarily

mean a tremendous amount as it stands alone, except for as you will see later and I will

remind you now… That when the dot product is equal to zero those two vectors are perpendicular.

Now all this math operations, all of this arithmetic is just dealing with addition and

multiplication. Addition and multiplication are commutative and associative so a lot of

the properties that you learned way bay in say algebra 1, a lot of those properties apply

to Dot Products as well. The dot product is associative…excuse me not associate but

commutative. So it does not matter what order you do the dot product in. If you are taking

the dot product of a vector and the resultant of two other vectors being added together,

you can basically distribute that dot product through the parenthesis and get vector u dot

product of vector v plus vector u dot product vector w. If you do the dot product between

the zero vector and any other vector you will get a value of zero. If you take the dot product

between a vector and its self as you will see in this example, that is the same as squaring

the magnitude of that same vector and do actually have to write that out. The square root that

would see in the magnitude which is the distance formula or the Pythagorean Theorem will cancel

away with the power of 2. Why don’t I just show you that? So the magnitude of a vector

if it is in component form is equal to the square root of a squared plus b squared. Then

it is the magnitude squared, so that is going to cancel out that square root symbol giving

you the a^2+b^2 I had a second ago. And we have kind of like a combination of the associate

and the commutative property a little bit with the scalar multiple and the dot product.

Let’s go through an example. We have vector v and vector w, and we want to find the dot

product. We are going to take the two values of ‘a’ and multiply them together. So 2 times

this ‘a’ value of 1 and add that with the b values of 4 times -1. 2 minus 4 is negative

two. And that is it! You know. As it stands alone, that just ends up being a scalar multiple

of negative two and we are done. Now if I take vector v and I do a dot product with

itself and just say, ok let’s take vector v and find the dot product with vector v that

is going to be the two ‘a’ values multiplied together…so 2 times 2 plus the two ‘b’ values

multiplied together and v is being dotted with itself so I am using the same ‘a’ and

the same ‘b’ values. So 4 times 4. Here I have got 2 squared plus 4 squared. See what

this is starting to look like? That is a^2+b^2 which is exactly what I had over here when

I wrote out the magnitude formula. And that ends up becoming 4 plus 16 which is equal

to 20. So that is just a couple of quick examples of dot products. It is just a little bit of

multiplication and division. Let’s see how that helps us find the angle between two vectors.

ALRIGHT! So here we have the formulas for finding, there is two versions of it, we are

going to have these formulas for helping us find the angle between two vectors. These

formulas, you probably have the derivation in your textbook, they come from or are derived

from the Law of Cosine. And it says that the dot product between two vectors is equal to

the magnitude of the first vector times the magnitude of the second vector times the cosine

of the angle that is in-between them. Now if we solve that for theta, we are going to

divide both sides by these magnitudes. Then to get the cosine function away from theta

we will have to do the inverse cosine. So this is actually the format that you are going

to be using most of the time which is just… you are going to be given the vectors and

then asked for the angle measure. Thus you are probably going to be using the one that

is set up to equal theta more often than the one set up to equal the dot product. So here

we have a couple of vectors and let’s run through this formula. Theta is equal to the

inverse cosine of the dot product between the two vectors…so we have 2(1)+4(-1)…

that is going to be over the magnitude of v which is going to be the square root of

‘a’ squared plus ‘b’ squared times the square root of the magnitude of vector w… excuse

me… times the square root of 1 squared plus -1 squared or the magnitude of w. Now be aware

that I wrapped my -1 with a set of parenthesis. It might not be a bad habit, I usually do

all of my substitutions wrapped in a set of parenthesis just to make sure I don’t make

any sign errors. Because let’s remember there is a difference between -1^2 and (-1)^2. This

comes out to be negative one, and this because the negative is wrapped in the parenthesis

is -1 times -1 which is positive one. This just squares the 1. So let’s not let our fancy

calculators give a wrong answer because it understands the difference in that notation.

Now when you get all of this typed into your calculator, there really is not much more

to this depending on how many times or how many steps you take getting this into your

calculator, but it is going to come out to be inverse cosine of -.316. This comes out

to a theta value equal to 108.4 degrees. Now let’s take a look at this drawing really quickly

and see why we can just that I am right from the calculator right off the bat. Maybe we

can have a little insight as to why this formula is derived from the Law of Cosine. So let’s

get a different colored chalk here. Draw an x y axis and put a few tick marks out here.

Ok, so vector v says that it is 2,4… We are going to the right two and up four and

I am placing this on the origin only for convenience. You do not need to have vectors always have

their initial point on the origin. Vector w is identifies i-j, so a is 1 and b is -1.

That vector is right here. A rather short one. We are finding the angle between those

two vectors. This angle right here is what we just figured out to be 108.4 degrees. Well,

if we are looking for that angle between the vectors we have enough information to find

the magnitude of v. Indeed the magnitude of v, that is 2^2 is 4, 4^2 is 16 so this length

is the square root of 20. And this other vector with just the a and b values of 1 and -1 has

a length of square root two. So if you just got done studying the Law of Cosine, you know

that… now I don’t have a full triangle set up… but you knew that with the Law of Cosine

that if you had side-angle-side… if you had two sides of a triangle and the included

angle, you would solve that triangle first, or the first step you would be required to

do was the Law of Cosine to figure it out. There you can see that side-angle-side, and

we are looking for the angle between those two vectors. That set up is what we use or

when we use the Law of Cosine when solving oblique triangles. One more page. OH! One

more comment. The 108.4. Remember that inverse sine with your calculator will only give you

answers between -90 and 90. It is the same with tangent. But inverse cosine with your

calculator will give you values between 0 and 180 degrees. And we are always going to

be looking for the angle that the… the short way around the circle. When vectors are opened

up like this, we want the inside angle. We do not necessarily need to wrap around the

long way around those two vectors. If we need to know that, we can just take this entire

360 degree rotation and subtract 108.4 from it. So when you do this formula to find the

angle between two vectors, you are going to be finding an angle that is on the more closed

side of those vectors… an angle that is less than 180 degrees since that is what you

get out of our calculator for the inverse cosine since it has a restriction to act as

an inverse cosine function. But, if I babble on more I am going back to an old concept.

Next screen! Well if we are finding the angle between vectors, what does it look like when

those vectors are perpendicular or parallel. Well if they are perpendicular you can stop

with the dot product. If your dot product comes out to be zero, then those vectors are

perpendicular or orthogonal as we like to say in vector language. So what does it look

like when they are parallel. Well when you have vectors that are parallel, if the numbers

are easy enough like these are you might notice that they are scalar multiples of each other.

4 times 1.5 is 6 and 14 times 1.5 is 21. So just the fact that I can see that these are

scalar multiples of each other automatically tells me that they are parallel, and all the

signs are the same. So I also know that the angle between those vectors is zero. Thus

they are going to be parallel or going in the same direction. If my signs had changed

then I would have parallel vectors but going in opposite directions. Well that is fine

and dandy, but what if you can’t just glance at and see that they are scalar multiples

of each other. You apply the formula. So we have theta is equal to once again the inverse

cosine of the dot product between the two vectors. We are going to have 4 times 6 plus

14 times 21 over the magnitude of v…that is going to be the square root of (4)^2+(14)^2…

times the magnitude of vector v… the square root of (6)^2+(21)^2. Now again you can work

this out in tiny little pieces with you calculator and show the many steps at a time. I am just

going to give you the final answer. When you get this typed into your calculator, you are

going to get the value of…. oh! We don’t even need a calculator for this, right? I

already noticed that these are scalar multiples of each other and they are going the same

direction. So I hope that when we get done this answer comes out to be one. So theta

equals the inverse cosine of 1, and won’t notice that to begin with maybe… but if

you do notice this relationship you probably are not trying to find the angle between them.

And the inverse cosine of 1, well the inverse cosine of 1. What do you put into an inverse

trig function? You put into an inverse trig function, the sides of a triangle and you

get out an angle measurement. So around the unit circle, where is your cosine value equal

to one? The cosine value is equal to one, your x/r value, is equal to one with a rotation

of zero. So, you know, thus again as I noticed ahead of time these vectors are parallel and

they are going in the same direction. Whereas if I made this negative four and negative

fourteen, that would be -4 here and -14 there, it would be -4 squared and I am running out

of space here… plus a negative -14 squared. My answer, these negatives are going to get

squared away anyway and my top number instead of being positive will become negative. Where

are on the unit circle now is inverse cosine of negative one? What rotation gives me cosine

value of negative one? That is going to be at -180 degrees. So that is just a last little

step there for your special cases of when your vectors are either perpendicular with

a dot product of zero, or parallel. There are two cases of being parallel. You might

want to be able to remember or note which it is. Are they going in the same direction

or are they going in opposite directions. I am Mr. Tarrou. BAM!!! Go do your homework:D

awesome video! helped me review for my Calc III exam coming up – Rutgers Student

@foshrizzle Thanks. I hope you do great on your exam!

couldn't theta also equal │arctan(b1/a1)-arctan(b2/a2)│?

And THANK YOU for watching, learning and liking:)

What book do you teach with Mr. Tarrou?

I just wanted to say thanks for your great videos. Math can be a tough subject, and the way you go over subjects really is different than most math teachers. You explain the formulae and HOW they are derived. As a student, there is a huge difference between memorizing a formula, and understanding a formula. It makes the difference between a "C" and an "A". You are that difference. Thank you for taking the time to make videos like these to teach others.

And THANK YOU for taking the time to share that complementary comment with me:) I teach the same way that I was taught by a teacher who I felt the same way about…and most of my students appreciate the extra time taken to "understand" that difference:) Please continue to share your experience and my channel info with others you meet who might also appreciate the "extra" help, as you continue to support my channel growth:D

I use Blitzer Pre Calculus book in my classroom. It is a very good book:)

What's the point of the i and j vectors?

I think my books author uses them consistently to torture me…but I am being consistent with my book for my students. They are component vectors which means they have a length of 1. i is horizontal and j vertical. 2i+4j can be written, and in my opinion written 2,4 with pointy parenthesis, not round ones. I tried to use them and they were identified as HTML tags for the comment!

after 4 hours of math lecture. I come to you for a nice refresher….thank you for your work, I will always stay subed and I always refer students to check you out. Don't stop what you're doing!

THANK YOU for your support! I'm happy to hear that my videos are being watched and shared so much by you and your friends:) Maybe you could even let them know about my channel at your student tutor center:D I do have a printable postcard on my Facebook Fan Page if you even want to take that a step further…My wife & I are working very hard to groW my channel & asking viewers for help is that next step to keep this going and growing…THANKS:)

I actually do. Usually when I run out of time and the tutor(s) are busy with someone else I always tell them. They always get surprised

That's AWESOME! Thank you for the help:)…I really do appreciate any and all support from my dedicated supporters and viewers like yourself, so keep it up and maybe I'll make it to the big time #'s someday:D

@8:30 I was always told during vector mathematics to use radians. When do you know the formula's are for radians or degrees? Because I always get confused. When looking at a huge table of formula's.

I only introduce vectors and work with them for a couple of sections in my Precalculus book…and some of the questions are done in degrees. I am sure classes like physics and more advanced math classes go much more indepth with vectors. I also work with trig equations that only work with angles measured in radians. So, I would say that you should work with your vector problems like your professor instructs and know the conditions required for use of the forumulas you are working with,

You also need to make sure your calculator is in the correct mode for the given information. If you are doing a question with angles given in degrees and your calculator is radian mode you will get the incorrect answer. My trig students forget the check the mode of the calculator all the time.

Thank you for the fast reply. I guess I through two questions in there. You answered one of them. I think what I'm trying to get at here is, "When do you know to use radians over degrees or vise versa?" I'm currently not in university anymore. I went up to calculus 3. I seemed to cap out on my math when I arrived in that location mostly because you realize then how many gaps you actually have because public schools don't teach you everything you need to know when graduating university.

I didn't answer this question because it is a bit difficult. I tell my students if the information is given in terms of degrees, then work the problem in degrees and make sure that your calculator is in that mode. At the same time lets say you are working on arc length of a circle, you have s=r*theta which only works for radians and s=2pi*r*(theta/360) which is only good if you are working in degrees. Oh and by the way, I teach at a public school.

Its hard to teach a subject with so many path ways. In a public school you have a structured set of items you need to teach the students to move on. The problem with that is over the course of lets say 5 to 10 years. The teaching changes and the regulations and standards change. Which means what you taught may or may not be valid anymore. When I was back in school I was taught a lot of things that are no longer valid anymore. Math is just a complex subject ever changing over the past 20 years.

I still make the same mistakes now. The problem comes from mixing degrees and radian problems in the same set of practice problems. They should be separated as in Math they are separated too. 1 sheet for radian and 1 sheet for degree problems. The field that I was in until the economy crashed made it hard to figure things out as a mechanical engineer. It was hard to know when to use radians or when to use degrees. It still is. The time I have spent on the subject tells me to use radian measure.

That is true in a way. Technology changes, curriculum experts can change what topics should be taught in what classes, vocabulary can change a bit, and so on….at least the Pythagorean Theorem will always be Pythagorean Theorem as an example. I am glad I don't teach science at least! New discoveries are being made all the time, which is great but changes the curriculum even more:O

Sorry to hear about the negative impact you experienced from the bad economy. I have a lot of friends struggling too. A couple of people I know who are engineers remind me of how much math I have forgotten after teaching high school for 18 years, and how much I never learned or understood from not taking classes that applied more of the math I did learn.

oh thank you so much sir

you are very very very great maths teacher

Yes this is the current problem with curriculum's in the United States. They change from year to year or with new teachers to new substitute teachers. So, the students always miss something. Later on in your college years you start to realize what you have missed. Then by panic you start to learn slowly the things you need. This is just my own personal experience of course. But I think everyone will find this to be true. I still have big holes in my knowledge. With each day that passes I learn

You're still watching more?!

Thanks again:)

Thanks for commenting. That's a great positive outlook to have…if we looked for one new thing to learn daily from someone or something, we would all be rich with knowledge.

yes sir because by watching your videos i am getting interested toward maths subject.

and thank you so much sir for reply

I am very excited to hear that my videos have been so successful in peaking your interest in math! I have over 400 videos on my channel to choose from so I hope that means we will be having many more message exchanges in the future:)

thank you and keep them coming

You're welcome and keep watching and sharing! Thanks for supporting by liking and subscribing too:)

Damn, I wish i had you in linear algebra but still, this is better than nothing. I might even finish my course now thanks to you!

Happy to hear that my videos are helping you. Good luck with your class and thanks for subscribing!

Dont EVER stop uploading these, you have no idea how lifesaving they are!

With the help of my viewers subscribing and liking my channel, it will continue to groW to it's fullest potential soon:) Thanks for watching and glad they are helping you so much:)

So LIKE, SUBSCRIBE and SPREAD THE WORD!…BAM!!!

Thank you soooooooooooooooooooooooooooooooooooooooooo much!!!!!!!!!!!

You're soooooooooooooooooooooooooooooooooooo welcome…and thanks for supporting by subscribing too!

nice

This guy writes better with chalk than I do with pen

what is scalar?

Another great video as always. I'm taking calc III right now and needed some review of trig. You explain this much more clearly than my professor who has a doctorate. You should be a trainer for other math instructors. Great job!

Professor, I know you said hwat ever I get out of my calculator will be my answer for questions like angle between vectors. But however I got 71.6 in my calculator, I know after I do 180-71.6 I get my answer, but if your calculator gives an angle between -90-90 always, even for cosine and sine. How do you ensure what quadrant the angle is in to apply the reference angle rule? For this example, if I didn't plot it out, I wouldn't know that the angle is in 2nd quadrant and would have kept the answer 71.6, how do I prevent that from happening？

Doing this in my clac3 course! I love it. thank you for my journey from Trig, Clac1, and Calc2.

May the Mathematics God bless you everyday!!!! Thanks.

Your writing is awesome. I wish you were my teacher. You explain everything so precisely and in detail, so that I understand everything. Great video, keep your work, we are going to support you and your great Chanel, how you support us in math.

you're a great teacher because of you im getting an A in math

I'm sorry I could not pay attention because I was too absorbed in your amazing flick handwriting at the top

Hey Mr. Tarrou, do you have any videos on polar coordinates?

ur mom is going to take a look at the dot product

Bro youre like a unit circle cheerleader when you wave your arm around.

Thanks teach this concepts of math, every time I have trouble I am looking for you video. It is very easy to fallow and understand.

good lecture and your handwriting is just awesome!!

can I ask you ?

who did write in the blackboard?

because the hand writing is so so so so amazing,

the way you explain to us i so amazing

For a moment I thought it was a demonstration of the force of gravity 😉

This video really made me understand vectors products so much more, thanks!

Thanks for sharing videos like this. These are making me consider physics a cakewalk. You are a very very philanthropic being.

I also wish, if I could have been enough fortunate to have handwriting like yours.

Thank you professor !

Hello Prof Rob,

I just wanted to say that I sincerely appreciate your videos and I truly respect your ability to consistently put out videos and help hundreds (probably thousands) of students/learners monthly. I am taking a precalc II course at my local community college currently this summer and we are learning this material quickly and in a condensed-plethora-of-notes manner(you probably know how it is). Watching your videos has helped me gain better understanding of some of the toughest concepts that I have encountered. It, more or so, feels that you have been my professor these past few weeks and so I wanted to take the time to say thank you for being an awesome professor to me this quarter haha!

Keep inspiring,

JR Pagdanganan

if the cosine angle is 0 or 180 then they are parallel ?

im working on a study guide for a test im taking tomorrow and this literally went down with each target thanks!!!

Thanks Mr. Tarrou for making my Linear Algebra course a lot easier 🙂

Wow Mr tarrou, i'm finally in my physics class, Mechanics of Solids and Fluids and i'm still going back looking at your videos! They always have and always will be helpful, thanks again and again!

This video and your other 40 min video on vectors are invaluable. After classes I always do 1-2 pages of concept notes from YouTube and by using these videos I made some very classy notes in pen. I'll be looking at them for the whole summer.

You were able to condense the material into it's smallest and most simple form. Even bringing it back to Trig with the cosine law.. Thanks a lot!!

BAM!!!!!!!!!!! Bob Ace Maths!!!!

You are the most kind,humble and awesome person i had ever seen

Thanks For The Support Sir…….

From India with Love.

PS:Your T-Shirt Rocks

You are brilliant. Thanks !

is that a picture of a formula one car on top of the board?

Shortcut for finding parallel…

Take slope if both slopes are equal then dot product is parallel

Like

A<1,2> B<3,6>

Slope of A=2/1=2

Slope of B=6/3=2

Bam!!

woah never seen this classroom before!