In Exercises 47–52, write the vector v in terms of i and j whose ...

Question

In Exercises 47–52, write the vector v in terms of i and j whose magnitude ||v|| and direction angle θ are given.||v|| = 6, θ = 30°

Accepted Answer

Everyone. In this problem, we have a vector V with magnitude 26 an angle degrees. And we're asked to express that vector in terms of I and J, we have four answer choices. Option A V is equal to 13, I plus 13 multiplied by the square root of three J. Option BV is equal to 13, multiplied by the square root of three I plus 13 J. Option CV is equal to the square root of three divided by 13, I plus 1/13 J. And option DV is equal to 1/13 I plus the square root of three divided by 13 J. So we're given the magnitude of this vector. We won't wanna break it into its I and J components. OK. So we can draw this vector. And if we want to break it into its INJ components, we can draw it as a trunk. So we know that the vector V is the hypotenuse of our triangle and it has a magnitude of 26. So that length of our hypotenuse is 26 are also told this vector has an angle of 60 degrees. And so the angle between the horizontal and a vector is going to be 60 degrees right. Now, this is a right angle triangle and we have these two sides. Now, I recall when we want to write a vector in terms of I and J, we can write it as V as equal to something I plus something, right? And we're gonna call these things X and Y. OK. So we're gonna write V as X I plus YJ. Now looking at our diagram, OK. This X is gonna correspond with the horizontal side of our triangle and Y is gonna be this vertical side of our triangle, maybe horizontal and vertical X and Y or I and J components of our vector and those all mean the same thing. So now we're looking at a right angle triangle, we have one angle, we have the hypotenuse, we can use our trigonometric ratios to find these values of X and Y. So let's recall, let's start with X and because X is on this adjacent side to our angle, we wanna use the coast. OK. So again, we're calling that the cosine of an angle theta is gonna be the adjacent side divided by the hypo. Now, in our problem filling out the information, we get that cosine of 60 degrees is equal to X divided by 26 multiplying both sides by 26 X is going to be equal to 26 cosine of degrees. OK? And we're gonna come back and simplify this cosine of 60 degrees in just a minute. Once we write that sign um component as well and we can do them both at the same time. So moving to the Y component, OK. And again, we're gonna relate through the sign because now we're talking about the opposite side to our angle. So we have signed, the data can recall that this is equal to the opposite side invited by the high partners. Substituting in our information, dying of 60 degrees is equal to Y divided by 26. And multiplying both sides by 26 we get that Y is equal to 26 multiplied by sine of 60 degrees. OK. So we want to simplify these, we have coast of 60 degrees and sin of 60 degrees. So let's think about our special triangle that has angle 60 degrees in it. OK? And we're gonna draw that down below. We're gonna say that this bottom left corner is 60 degrees. Now, we call or special triangle with angle 60 degrees. OK. This is a right angle triangle has side lengths of one and the square root of three and hypo two. And you can always double check if you use Python theorem with these two side links, you will get that hypotenuse of two. OK. And it's really, really important to remember that the smallest angle is across from the smallest set. OK. So here we've drawn one adjacent to 60 degrees because it's across from the opposite angle which would be 30 degrees. OK? Smallest angle, smallest set, all right. So this is our special triangle. Now, we just have to look at it and figure out each of these values. So cosine of 60 degrees, we just said that cosine is the adjacent side divided by the hypotenuse in our special triangle. That would be one divided by two. So cosine of 60 degrees is just one half. And so X is going to be equal to multiplied by one half, which is equal to 13. Similarly for Y, and it's gonna be sine of 60 degrees which we've written as the opposite side divided by the hypotenuse. That's gonna be the squared of three divided by two. And so why is going to be equal to 26 multiplied by the squared of three divided by two? And we can simplify this and write it as 13 multiplied by the square root of three. So now we have our X value, we have our Y value and we can write our vector B, our vector B is going to be equal to 13, I pack 12, 13 route three. And that is our final answer here. And I've written them as I hat and J hat in the solution. It's just bolded and those indicate that same thing that those are vectors. All right. So comparing this to our answer choices, we 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 47–52, write the vector v in terms of i and j whose magnitude ||v|| and direction angle θ are given.||v|| = 6, θ = 30°

Key Concepts

Magnitude of a Vector

Direction Angle

Vector Components

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