In Exercises 17–22, find the angle between v and w. Round to the ...

Question

In Exercises 17–22, find the angle between v and w. Round to the nearest tenth of a degree.

v = 2i - j, w = 3i + 4j

Accepted Answer

Hey, everyone. Welcome back in this problem. We're asked to calculate the angle between the two vectors A and B. We're told to express the answer in degrees and round it to one decimal place. Fear that A is equal to 10, I minus three J and B is equal to eight, I plus 12 J. You have four answer choices all in degrees. Option A 81.2, option B 73 option C 104.9 and option D 68.6, we're looking for the angle between these vectors and let's recall that we can relate the angle between vectors to the dot product. And so we can write that A dot B is equal to the magnitude of A multiplied by the magnitude of B multiplied by cosine of beta. And this data is the angle between those factors that we're looking for. We're given A and B. So we can calculate the doc product, we can calculate the magnitude of each of those vectors. And so we have everything we need to calculate the. So let's start by calculating the dot product and we're gonna do this in bolu and So we have a dotted with B can recall that this is equal to A one B, one plus A two B two in the case where we have two components like we do in this problem. And A one is just gonna be the first component of vector A B one, the first component of vector B. And same thing for those subscript twos substituting in our values. And we're gonna take the first component to be those I components. So we're gonna take the I component of A which is 10 multiplied by the I component of B which is eight, can we add the J component of a negative three multiplied by the J component of B 12? And this gives us 80 minus which is equal to 44. OK. So we have our dot Park A dot B. Now let's move to the magnitude of A. Now we're called that the magnitude of a vector can be written as the square root the sum of squares of the components. So in this case, we have two components. So we have the magnitude of A is equal to the square root of A one squared plus A two squared substituting in our values. The first component of A is that I hat component or the I component which is 10. So we have 10 squared plus the J component which is negative three and again squared OK. If we simplify this, we can write this as the square root of 109. The last thing we need to calculate before we can calculate data is the magnitude of B. So just like we did with the magnitude of A, we have the magnitude of B is the square root of B one squared plus B two squared. This is gonna be equal to the square root of that I component eight squared. What's the J component 12 squared? Simplifying our square root gonna be the square root of 208. And we can actually simplify this square root of 208. A 208 is divisible by 16, which we know is a perfect square. So we can break 208 up, we can write it as 16 multiplied by 13. The square root of 16 is equal to four. And so we can write this as four multiplied by the square root of 13. OK. So we've just simplified a little bit there. And you'll see in the next step that that allows us to simplify our expression for coast data. So now we have our dot product, we have the magnitudes of each vector. We're gonna get back to our dot Product equation with the cosine in it in order to calculate this angle. OK. So starting with the dot product, we have 44 that's equal to the magnitude of a, the square root of 109 multiplied by the magnitude of B four multiplied by the squared 13 multiplied by cosine of data. And again, 44 comes from our blue calculation square root of 109 from our green calculation and four multiplied by the square of 13. From our red calculation, we wanna isolate for the. So the first thing we're gonna do is divide by both of these square roots. So we have cosine of theta on one side, everything else on the other, we get cosine of data is equal to divided by the square root of multiplied by four square root of 13. Now you can see we can divide the numerator and denominator by four. So in the numerator, we have 11 and in the denominator that four goes away. And so it allows us to simplify one step further there and isolating for theta, we're gonna take the inverse cosine of everything on the right hand side. So 11 divided by the square root of 109 multiplied by the square root of 13. And this gives us a state of value of approximately 73.1 degrees. OK. So the angle between vectors A and B is gonna be about 73.1 degrees. If we compare this to our answer choices and we round to one decimal place, we can see that this corresponds with answer choice B. Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 17–22, find the angle between v and w. Round to the nearest tenth of a degree. v = 2i - j, w = 3i + 4j

Key Concepts

Dot Product

Magnitude of a Vector

Cosine of the Angle

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