In Exercises 43–44, find the angle, in degrees, between v and w.v...

Question

In Exercises 43–44, find the angle, in degrees, between v and w.v = 2 cos 4𝜋 i + 2 sin 4𝜋 j, w = 3 cos 3𝜋 i + 3 sin 3𝜋 j                3                 3                       2                 2

Accepted Answer

Hey, everyone in this problem, we're given vectors A and B and we're asked to determine the angle between them. We're given that A is equal to six cosine of seven pi divided by three I plus six sine of seven pi divided by three J and B is equal to 12 cosine five pi divided by two I plus 12 sine of five pi divided by two J. We have four answer choices here. Option A pi divided by six, option B pi divided by four. Option C pi divided by three and option D pi divided by two. Now, when we think about an angle between vectors, we want to think about our dot I recall that the dot product A dot B is equal to the magnitude of A multiplied by the magnitude of B multiplied by cosine of theta where the is the angle between the vectors. And so that's what we're looking for. We're looking for the, how we can go ahead and isolate data in this expression. Let's divide by the magnitude of A and the magnitude of B on both sides. We get that cosine of data is equal to the dot product A dot B divided by the magnitude of A multiplied by the magnitude of B. And then we can take the inverse cosine to get beta, which is gonna be cosine inverse of the dot product divided by the product of the magnitude A and B. All right. So we know how we want to calculate data. But in order to do it, we need the dot product and we need the magnitude of each of these vectors now because we're given the vectors, those are things that we can calculate. So let's go ahead and do that. Now we're gonna start with the dot product and we're gonna calculate that in bullet. Let's give ourselves some more space. And let's recall with the dot product A dot B. It's gonna be equal to A one B one plus A two B two. OK? Where the subscript ones indicate the first component. OK? In this case, the I component and the subscript two indicates the second component or in this case, the J component. So substituting in our values for this problem, OK? A one, the I component of A is six cosine of seven pi divided by three. We're gonna multiply that by B OK. So sorry, the first component of B, the I component of B which is 12 cosine 055 divided by two. And then we're gonna add the J components. So we get six fine of 75 divided by three, multiplied by of 55 divided by two. Now we can simplify this a little bit. So these constant coefficients in front we can multiply together. So we have six multiplied by 12 that gives us cosine of seven pi divided by three multiplied by cosine of 55 divided by two plus 72 9 of seven pi divided by three multiplied by sine of 55 divided by two. Now we need to figure out what value each of these trig expressions have. OK. So we wanna look at these one by one and sort out their values. OK? And we're gonna do that on the right hand side here, cosine of seven pi divided by three. This is gonna be equivalent to cosine of two pi plus P K 75 divided by three can be broken up into two pi plus five third. OK? So what we're doing really is going all the way around our unit circle and going a little bit more. Now, when we have cosine of two pi plus an angle, this is actually equivalent to just cosign of that angle, cosine of pi divided by three recall is gonna be equal to one half. So we know a cosine of seven pi divided by three. That's great. Now, we can move on to cosine of 55, divided by two and we're gonna do the exact same thing. We can write this as cosign of two pi what? Bye. That's just equal to cosine of the pie hat. The end of five halfs is equal to zero. All right. So this entire first term is just gonna go to zero. We're gonna be left with just the second term. So what about these trig values for the second term? Let's work those out too in the same way. Nine of 75 divided by three. Oh, that's gonna be equal to of 25 plus 5 which is just equal to sign of five third. OK? Because these functions are too high periodic and sign of pi divided by three, recall whether you wanna use your unit circle or a special triangle is gonna be equal to the square root of three divided by two. All right. And the last one sign of five pi divided by two again, breaking it up sign of Tupac Pi divided by two because these, this is Sinus two piper, we can write this just a sign pi divided by two which we know is equal to one. OK? So now you've evaluated every single one of those and we can simplify our expression. We've already said the entire first term goes to zero. So now we're left with 72 multiplied by sine of seven pi divided by three, which is the square root of three divided by two, multiplied by sine of five pi divided by two, which is one and simplifying 72 divided by two. We get that this is gonna be equal to 36 multiplied by the square root of. All right. So we've figured out our dot pro remember this is A dot B. The other thing we needed to figure out was the magnitude of A and the magnitude of B. And let's go back to our expression. We're looking for the, we found the dot product. Now we need to find these magnitudes. So we're going to calculate the magnitude of A and green, the magnitude of B in red. Let's do that down below, starting with the magnitude of a. Now recall that the magnitude of a vector is gonna be the square root of the sum of the squares of the component. So the magnitude of A is the square root of A one squared plus A two squared A one. Again, this is our I component of A. We use that in our dot product. We know what that is. It's six cosine of seven pi divided by three and that's all squared. And then we add the J component six sign of 75 divided by three. Again, all squared simplify, we know these values. Now, we know cosine and sine is 75 divided by three. We already calculated them. So let's not calculate them again. Let's use what we've already done. We get that this is gonna be seven multiply by one half squared. Oh sorry, that's six multiplied by one half squared plus six multiplied by the squared. Of three divided by two squared. I get the square root of three squared. Yes, three root three squared. Simplifying some more. This is the square root of nine what 27? Which is equal to the square root of 36 which tells us that the magnitude of A is equal to six. OK. So we have our magnitude of A. Now we have our dot product on to the magnitude of B. The last thing we have to calculate before we can get back to our calculation for the, the magnitude of B is gonna be equal to the square root B one squared plus B two squared. OK. Just like a substituting in our values, this is gonna be equal to 12 cosine of five pi divided by two square was 12 sign of 55 divided by two squared. We found that cosine of five pi divided by two was zero. So that first term goes to zero, we get zero squared, which is just gonna be zero nine of five pi divided by two is equal to one. So we have 12 multiplied by one. She gives us 12, we get zero squared plus 12 squared underneath that square root which gives us a magnitude of B of 12. So now we have our dot product, we have the magnitude of A, we have the magnitude of B, we can go back and find the angle between these vectors. And we said that it was gonna be equal to cosine inverse of A dot B divided by the product of the two magnitudes. This is gonna be equal to cosine inverse of 36 multiplied by the square root of three divided by six, multiplied by 12. We can simplify a little bit. This is the inverse cosine of we have divided by 12 multiplied by six numerator and denominator are divisible by 36. So if we divide by 36 we get cosine inverse of the square root of three divided by two. OK. And we know that cosine inverse of the square root of three divided by two is going to be equal to pi divided by six. OK. And so the angle between these vectors is gonna be pi divided by six. We go up to our answer choices. We can see that this corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 43–44, find the angle, in degrees, between v and w.v = 2 cos 4𝜋 i + 2 sin 4𝜋 j, w = 3 cos 3𝜋 i + 3 sin 3𝜋 j 3 3 2 2

Key Concepts

Vector Representation

Dot Product

Magnitude of a Vector

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