Hey everyone and welcome back. So in recent videos, we've been talking about operations you can do on vectors. And one of the operations we discussed was the product of a scaler and a vector. When multiplying a vector by a scaler, what this does is actually stretches or shrinks your vector in some kind of way depending on what that scaler is. But the result you get is another vector. Now what we're gonna be discussing in this video is another operation called the dot product and the dot product is actually a way to take a vector and multiply it by another vector. Now the results you get when you do A dot product are actually quite interesting. So without further ado let's jump into some examples that you'll likely see in this course. So let's say we have these two vectors, vector V and vector U and we wish to find their dot product. Well, when doing a dot product, what you want to do is multiply the like components together and then you want to add their results. So let's say we have these two vectors right here. What I can do is I can multiply the X components together and the Y components together and then add them all up. So let's go ahead and try that with the two vectors we have here. So if I wish to find the dot product of VNU what I can first do is multiply their X components and their X components I can see are one and three. So we're gonna have one times three and we're going to add this to the product of their Y components, which is going to be two times two. Now, one times three is equal to 32 times two is equal to four and three plus four is seven. So the dot product and the solution to this problem is seven. So this is how you can find A dot product. It's a pretty straightforward process. But I want you to notice something interesting notice when we took the product of these two vectors, we ended up getting a scaler. And that's the interesting result I was discover before. Now what this number tells you is this number generally represents how close the vectors are to pointing in the same direction. So a higher number would mean that the vectors are more close to a line. Now to make sure we really understand this concept in operation. Let's actually try a few more examples. Now, in this problem, we are asked to complete the following vector operations below. And we're going to start with example, a example A asks us to find the dot product of vectors U and W now I can see that vector U and W are given to us on this graph. So we can just go ahead and find their dot product. Now vector U I can see is three I and three, I is the same thing as three I plus zero J. All this zero J tells us is that we have no direction in the y direction because this vector is completely along the X axis. And what I'm going to do is find this dot product with vector W which is two I plus J. Now, when doing this operation, all we need to do is multiply the light components and add the result. So we're going to have three times two which is six and that's going to be added to zero times one which is zero. So we're gonna have six plus zero, which is six. And this right here is the result for our dot product. Now, what exactly does this number mean? Well, it tells us how aligned vectors W and U are. And since I can see that the vectors actually are close to pointing in the same direction, we can see that they do have alignment. So we get a positive result and that is nonzero. But now let's go ahead and try example, B example B asks us to find vector V dotted with vector W. Now I can see that we have these two vectors right here. So let's go ahead and find their dot product vector V I can see is negative two, I plus three J. That vector is right here. And I can dot this with vector W which is two I plus J that can multiply the light components. So we'll have negative two times two which is negative four. And I'm going to add this to three times one which is three and negative four plus three is negative one. So this right here is the result for our dot product. What exactly does it mean when we get a negative result in the dot product? Well, what this tells me is we have negative alignment. That means the two vectors are not aligned at all. They're actually pulling against each other. And since it appears that these two vectors are actually pulling opposite ways, it makes sense that we got a negative result for our dot product. But now let's try example C and example C is a bit more complicated because you can see that we have vector U but this is going to be dotted with vector W plus V. Now my first step here is going to be to figure out what W plus V is, I can see vector W is two I plus J and I can see that vector V is negative two I plus three J. So all I need to do is add these vectors together to get this result now negative two plus two, that's going to be zero. So we have zero in the I direction. And then I'll add this to three J plus J which is four J. So this is the result that we get for vector W plus vector V. And what I need to do is find this dot product which is vector U, which I can see here is three I plus zero J and this is going to be dotted with vector W which is zero, I plus four J. Now, what I can do is multiply the light components and add the results. So we're going to have three times zero, which is zero plus zero times one which is zero and zero plus zero is zero. So that's what we get for our dot product. Now, we've talked about what a positive result means and what a negative result means. But what does a zero result mean? Well, a zero means we have no alignment whatsoever. These vectors do not pull on each other nor do they point in the same direction in any way they're completely not aligned. And so what this means is if we have a dot product where the result is zero, the only way two vectors could not be aligned at all is if they were perpendicular. Now, if I were to draw vector four J on our graph, it would look something like this. It would point in the Y direction and notice how these two vectors are perpendicular. And because these vectors are perpendicular, that means their dot product is zero. So that's what it means to get a zero dot product. A negative dot product and a positive dot product. So I hope you found this video helpful. Thanks for watching and please let me know if you have any questions.
2
Problem
Problem
If vectors v⃗=⟨4,3⟩ and u⃗=⟨9,1⟩, calculate v⃗⋅u⃗.
A
39
B
⟨36,3⟩
C
33
D
⟨4,27⟩
3
Problem
Problem
If vectors v⃗=12ı^ and u⃗=100ȷ^, calculate u⃗⋅v⃗.
A
1200k^
B
1200
C
0
D
100
4
Problem
Problem
If vectors a⃗=13ı^, ⃗b⃗=5ı^−12ȷ^, and c⃗=24ȷ^, calculate b⃗⋅(a⃗−c⃗).
A
353
B
132
C
65
D
−223
5
concept
Find the Angle Between Vectors
Video duration:
4m
Play a video:
Welcome back everyone. So in the last video, we got introduced something called the dot product, which is an operation you can do to multiply two vectors where you take their light components, you multiply them and then you add the results to get your answer. This is all review from the previous video. Now, what we're gonna be talking about in this video is an alternate way of doing this dot product that involves the magnitudes of vectors as well as the angle between them. Now, this might sound like it's unnecessary since it's just some other equation to do something we've already learned. But it turns out there actually are examples that you're going to come across in this course where you actually have to use this alternate method of finding the dot product to get a solution. So without further ado let's just jump right into an example to see what this equation looks like. And then we'll talk about some of these special cases that you'll see in this course. Now let's say we have two vectors and we're given the magnitude of our first vector V and the magnitude of our second vector U. And we're also given the angle between them. If you wish to calculate the dot product between these two vectors, the dot product is going to be equal to the magnitude of V multiplied by the magnitude of U multiplied by the cosine of the angle between those vectors. So let's go ahead and see what we get. So if I want to find the dot product of V and U, that's going to be equal to the magnitude of V, which we can see here is the square root of five. I'll multiply that by the magnitude of U which is the square root of 13 and then that's all going to be multiplied by the cosine of the angle between these vectors, which I can see is 29.7 degrees. Now, what I'm gonna do from here is I'm going to simplify these two square roots by combining them into one. So we're going to have the square root of five times 13 and that's all gonna be multiplied by the cosine of 29.7. Now, what I'm gonna do from here is multiply five and 13 and that comes up to 65. So at the square of 65 times the cosine of 29.7 degrees, all you need to do is put this into your calculator and make sure your calculator is in degree mode when you do this. So if you plug in this value exactly as you see it, you should get that. Your result is about equal to seven. This is rounding our answer here. But this is what the solution should come out to now because we got the same solution that we got over here. It turns out the two vectors we have, these are actually the same vectors. In each case, just in one situation, we were given both vectors in component form. And then the other situation, we were given the magnitudes and the angle. So notice how this dot product result comes out the same. But then this begs the question, why exactly would we need this equation if we're already familiar with this operation? Well, it turns out this equation allows us to find the smallest angle between two vectors. So if you ever want to find the smallest angle, all you need to do is rearrange this dot product formula. And that will actually allow you to, to find a missing angle. So let's actually take a look at some examples that you might see in this course to make sure we know how to do this. So in this example, we're told if the dot product between these two vectors and we're given both the magnitudes of these vectors is equal to 16, then find the angle between vectors A and B. So to solve this problem, what I can do is use this equation. So we're going to have vector A dotted with vector B is equal to the magnitude of A times the magnitude of B times the cosine of the angle between them. Now, we can see here the dot product of A and B is 16 and that's going to be equal to A, which we can see is four multiplied by B which we can see is eight multiplied by the cosine of the angle between them. And the angle is what we don't know. Now, four times eight that's equal to 32. So we're gonna have 32 times the cosine of our angle is all equal to 16. And what I can do from here is divide 32 on both sides of this equation. That's gonna get the 30 twos to cancel there leaving me with the cosine of theta is equal to 16/32. Now, 16/32 this is a fraction that actually reduces to one half because 16 can go into 32 2 times. So this reduces to here. Now, what I need to do from here is take the inverse cosine on both sides of this equation to get the angle by itself, that'll get the cosine to cancel. Leaving me with just our angle, theta and our angle theta is going to be equal to the inverse cosine of this fraction, which we said is just one half. So all you need to do is figure out what the inverse cosine of one half is. And you can use the unit circle or even plug this into your calculator. Just make sure you're in degree mode and you should get that this angle is equal to 60 degrees. So there is a 60 degree angle between vectors A and B and that is how you can solve for the missing angle or the smallest angle between two vectors. So this is what the new dot product formula is right here. And this is a situation where you would need to use this formula to find an answer. So hope you found this video helpful. Thanks for watching and please let me know if you have any questions.
6
Problem
Problem
If vectors ∣a⃗∣=3 and ∣b⃗∣=7, and a⃗⋅b⃗=14.85, determine the angle between vectors a⃗ and b⃗.
A
21°
B
1.0°
C
45°
D
14.85°
7
Problem
Problem
If vectors a⃗=4ı^ and b⃗=3ı^−2ȷ^, determine the angle between vectors a⃗ and b⃗.
A
46.10°
B
13.90°
C
43.90°
D
33.69°
8
Problem
Problem
If vectors ∣v⃗∣=12, ∣u⃗∣=100 and the angle between v⃗ & u⃗ is θ=6π, calculate v⃗⋅u⃗ .
