Welcome back, everyone. So in the last video, we got introduced to something called the dot product, which is an operation you can do to multiply 2 vectors, where you take their like 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 going to 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. 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 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 2 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. 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 5. 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 going to do from here is simplify these 2 square roots by combining them into 1. So I'm going to have the square root of 5 times 13 and that's all going to be multiplied by the cosine of 29.7. Now what I'm going to do from here is multiply 5 and 13, and that comes out to 65. So I have the square root 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 7. This is rounding our answer here, but this is what the solution should come out to. Now because we got the same solution we got over here, it turns out the 2 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 2 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 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. That's going to be equal to a, which we can see is 4, multiplied by b, which we can see is 8, multiplied by the cosine of the angle between them and the angle is what we don't know. Now, 4 times 8, that's equal to 32. So we're going to have 32 times the cosine of our angle is all equal to 16. What I can do from here is divide 32 on both sides of this equation. That's going to get the 32s to cancel there leaving me with the cosine of theta is equal to 16 over 32. Now, 16 over 32, this is a fraction that actually reduces to 1 half. Because 16 can go into 32 two 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 2 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 we hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles39m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
8. Vectors
Dot Product
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