In Exercises 37–39, find the dot product v ⋅ w. Then find the ang...

Question

In Exercises 37–39, find the dot product v ⋅ w. Then find the angle between v and w to the nearest tenth of a degree.v = 2i + 4j, w = 6i - 11j

Accepted Answer

Hello, today we're going to be determining the scalar product of the two given vectors and the angle between the two vectors. So we are told that vector A is defined as three I plus 17 J. And we are also told that vector B is defined as 21 I minus two J. We're going to start with the first part of this problem which asks us to find the scalar product between the two vectors. In order to find the scalar product, we'll need to take the dot product of the two given vectors. In order to get the dot product, we'll multiply the I components together multiply the J components together then take the sum of the two products. So the dot product A dot B will be defined as three multiplied by 21 plus 17, multiplied by negative two, three, multiplied by 21 will give us the value of and 17 multiplied by negative two will give us the value of negative 34. Finally 63 minus 34 will give us a final value of 29. So what this means is that the scalar product A dot B is equal to 29. Now let's move on to the second part of the problem. What's asked us to determine the angle between the two vectors? Now, we have an equation that helps us find the angle between two given vectors. That equation is cosine theta is equal to A dot B over the magnitude of a multiplied by the magnitude of B. So there's three things we're going to need in order to determine the angle theta, we're first to need the dot product A dot B which we did calculate earlier. But then we'll need to calculate the magnitude of A and the magnitude of B. In order to calculate the magnitude of A vector will take the square root of the sum of the component squared. Or in other words, the magnitude of the vector is the square root of its I component squared plus its J component squared. Let's start by calculating the magnitude of A. Now recall that vector A was given to us as three I plus 17 J. So the magnitude of A will be defined as the square root of three squared plus 17 squared, three squared will give us the value of nine and 17 squared will give us the value of 289. Finally, nine plus 289 will give us a final value of 298. So the magnitude of A is equal to the square root of 298. Now let's calculate the magnitude of B recall that the B vector is given to us as I minus two J. So the magnitude of B will be defined as the square root of 21 squared plus negative two squared, 21 squared will give us the value of and negative two squared will give us the value of positive four. Finally, 441 plus four will give us the value of 445. So the magnitude of B is equal to the square root of 445. So we've calculated the dot product A dot B, we've calculated the magnitude of A and we calculated the magnitude of B. If we plug in these values to cosign data, what we end up with is cosine of theta is equal to the dot product A dot B which is 29 over the square root of 298 multiplied by the square root of 445. Finally, we want to solve for the, so we'll need to eliminate the cosine function on the left hand side. In order to do that, we'll multiply both sides of the equation by cosine inverse. And that will leave us with data is equal to cosine inverse of the quantity 29 over the square root of 298 multiplied by the square root of 445. If we plug this into a calculator, we'll end up with an approximate value of data of 85. degrees. And this is going to be the angle between the two given vectors. So just to summarize, we've calculated the scalar product of A and B which was 29 the angle between the two vectors is 85.4 degrees. With that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 37–39, find the dot product v ⋅ w. Then find the angle between v and w to the nearest tenth of a degree.v = 2i + 4j, w = 6i - 11j

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