Find the direction of the following vector: u ⃗ = ⟨ 5 √ 3 3 , 5 ⟩ u⃗=\langle\frac{5\surd3}{3},5\rangle u ⃗ = ⟨ 3 5√3 , 5 ⟩ .

this problem, we are asked to find the direction of the following vector u, and we are given u in component 4. Now to solve these types of problems, I generally first like to start by drawing this graph. And this small graph allows me to have a general understanding of the direction of the vector. Now this would be our x axis and our y axis, and if I wanna think about where this vector is going to end up, well, I see that I'm given vector u in component form. And looking at vector u, I can see that this is gonna be some positive number 5 times the square root of 3 over 3. And since this is the x component, that means we're going to be somewhere in the positive x direction. I can also see that we have the y component, which is positive 5, which would be somewhere up there. So our vector is ultimately going to look something like this, where this angle would tell us the direction of our vector. Notice we're in this first quadrant here, so all we need to do is calculate this angle theta. Now Now I can find this angle by using this equation that we learned about. The tangent of theta is equal to the y component divided by the x component. Now I can see that we are given both of these components. We have the y component, which is 5, and that's going to be divided by the x component, which is 5 times the square root of 3 over 3. Now what I can do here is I notice that we have a 5 here and there in the numerator denominator that will cancel. So So all we're going to end up with is that the tangent of theta is equal to 1 over the square root of 3 divided by 3. Now what I can do here is flip this fraction in the denominator and bring it to the numerator, and that's going to give me 3 over the square root of 3. And what I need to do is actually rationalize this now that I have a square root in the denominator. So if I multiply the top and bottom of this fraction by the square root of 3, and cancel the square roots there, giving me 3 times the square root of 3 divided by 3. And these threes are going to cancel, so all I'm really gonna end up with is the square root of 3. So we're going to have that the tangent of theta is equal to the square root of 3, which means that theta is going to be equal to the inverse tangent of the square root of 3. And the inverse tangent of the square root of 3 this turns out to be 60 degrees so that means that our angle theta is 60 degrees and that is the direction of our vector so that's how you can solve this problem hope you found this video helpful