Topic D: Lessons 20: Vectors and stone bridges
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Direction of vectors from components: 1st & 2nd quadrants
- [Voiceover] So what we're going to do in this video is look at a series of vectors, and we're gonna draw them in standard form, where their initial points or their tails are gonna sit at the origin. And we're gonna see if we can figure out the angles that they form, the positive angles that they form with the positive x-axis. And like always, pause this video and see if you could figure out what these thetas are going to be on your own, and we're gonna figure 'em out in degrees. So in this first one, I have the vector u. If we were to write it in unit-vector notation, sometimes this is called engineering notation, we would say it's three times the unit vector in the horizontal direction, the i unit vector, plus four times the unit vector in the vertical direction or you could view this as the x-component is three and the y-component is four, and you see that here. If you start at the origin, we're gonna move three in the horizontal direction, and we're gonna move four in the vertical direction. Now to figure out this theta, there's a couple of ways to think about it. We could just construct a right triangle for this one in particular. We could say, okay, the x-coordinate is three. So if we were to create a right triangle here, this side would have a length of three, and then we have a height of four or the y-coordinate is four. So this side has a length of four, and we know just from even our basic SOHCAHTOA definitions of trig functions that what trig function involves the opposite of an angle? So the opposite of an angle, and the adjacent of an angle? Well, tangent does. SOHCAHTOA. So we know, we know that the tangent of theta is going to be equal to the length of the opposite side, which is four, over the length of the adjacent side, over three. And so if we wanted to solve for theta, we could just say that theta is equal to the inverse tangent. Sometimes people say arc tangent of 4/3, of four over ... four over three, and let's evaluate this. And I'm gonna get my calculator out to do it. So I wanna take four divided by three, which is that, and I had already had pressed this button right over here that takes my tangent and makes it into an inverse tangent. So I'm gonna take the inverse tangent of this, and I get, I get roughly 53.1 degrees. So this is approximately 53.1 degrees, which looks about right. This looks like a little bit more, even though I didn't draw it completely super precisely. I hand-drew this. It looks a little bit more than a 45 degree angle, so that feels, that feels pretty good. Now another thing you might have said, "Hey. Well, look. "Four is the y-component. "Three is the x-component, and so maybe tangent "of theta is always going to be "the y-component over the x-component." And that, in fact, is the case. And that actually comes straight out of the unit circle definition of the trig functions, where you could say that if you have a unit circle, if you have, I could draw it right over here. So let me draw the coordinates. So if I were to draw a unit circle right over here, and if I were to have some line, here we're thinking about vectors, that the angle formed with the positive x-axis, the tangent of that angle, tangent of theta is going to be the y-coordinate, where we intersect the circle, over the x-coordinate. And so you could imagine if you made a unit circle right here, if you made a unit circle right over here, the ratio between the y-coordinate and the x-coordinate of this point right over here where we intersect the circle is going to be the same thing as the ratio between four and three. So this is going to be, this also is going to be one and 1/3 or 4/3. So either way, you could think of, when you think of vectors, the tangent of the angle that it forms with the positive x-axis is going to be, is going to be the y-component over the x-component. So it's nice when everything kinda fits together like that. So let's leverage that to figure out this, to figure out this angle right over there. Well, we could say that tangent of theta is going to be equal to the y-component. The y-component over the x-component. Over, do this in a new color or let's see, over negative five. So it's gonna be negative 6/5. Or we could say theta. So let's be a little bit careful. So we could say, we could say that theta is equal to, is equal to the inverse tangent. Let me do that same color. Inverse tangent of, I'll just write it like this, negative 6/5, but I'm going to put a little question mark here to see if we feel good about the answer that you get when we do this. So let's do that. So if we do, if we do six divided by five, which'll equal to that, and then we want to make this negative. So that's negative 6/5, and we're gonna take the inverse tangents. I already pressed this button, so this is going to be an inverse tangent, not tangent. I get negative, roughly negative 50.2 degrees. So this is approximately negative 50.2 degrees. Well, does that look right? Well, no. Theta looks like it's over 90 degrees. Negative 50.2 degrees, negative 50.2 degrees is actually giving us this angle right over here. So that's giving us, that's specifying another vector, if you think about a line, another line that would have the same tangent. The same tangent. So it's really important to visualize this and think about this, and the reason why it is is because the arc tangent or the inverse tangent functions in calculators, they will give you an angle that is between negative 90 degrees and positive 90 degrees, so something that's in the fourth or first quadrant. While here, we have something that's in the second quadrant. So we have to make sure that we're thinking about it right so that we could make the appropriate adjustment. So that would give the case if we were looking at a line like that or a vector like that. So in order to figure out what, in order to figure out the actual angle, what we want to do is add 180 degrees to it. So to get the vector that goes in the other direction. So you want to add, so theta is going to be equal to or I could say approximately equal to negative 50.2 degrees plus 180 degrees. So let's do that. So let me plus 180 is equal to approximately 129.8 degrees. So theta is approximately equal to 129.8 degrees. Did I get that right? I have a very short memory. Okay, yeah; that's right. 129.8 degrees, and that looks much better. That looks clearly is an angle we're gonna go. That take us to 90 degrees, and then we're going above 90 degrees. Now another way you could have thought about this is what you know from SOHCATOA and just right triangles. We could construct a right triangle where, we construct a right triangle using some colors. Well, we know, so I could maybe just draw it like this. So this is the height of it, and what is that height? Well, the height is going to be six, and then the base, the base right over here, what is that length going to be? Well, we know that we're going from the origin, we're going five back, but if we think about just the absolute value, the length of that line, that is going to be five, and so we could figure out just using right triangles this angle. Maybe we call that x, and so we could say that the tangent of x, tangent of x is equal to six over five, opposite over adjacent. Six over five. Or that x, or you could say x is equal to the inverse tangent of 6/5, and what is that going to be? So, if we take six divided by five, 1.2. And then we're going to take the inverse tangent of that, and we get approximately 50.2 degrees. So x is approximately 50.2 degrees, which looks right. But remember, we're not trying to figure out x, we're trying to figure out, we are trying to figure out what theta, what theta is, and you can see that x and theta are supplementary. So theta plus x is going to be 180 or theta's gonna be 180 minus that. So 180 minus that. So let's just put a negative sign in front of that and add 180, and that might look familiar. So that gets to us to 129, roughly 129.8 degrees is what exactly what we got before.