 In this lecture, we've talked all about the algebraic representation of the vector, and I keep on advertising, it makes life easier. Well, how is that? To prove it, I want to revisit a problem that we have seen previously in this lecture series. Let's look at a problem of static equilibrium. So we have a little boy named Danny, he's five years old, and he weighs 42 pounds. He's sitting on a swing. While he's sitting on the swing, his big sister Stacy is holding him up, and she's exerting some force horizontally. And then the tension of the rope is going to form a 30 degree angle with the vertical right there, and she's holding them there, holding them there. And so you have these three vectors, S, W, and T. And the fact that the swing isn't moving means they're in a state of static equilibrium. The sum of the three vectors is equal to zero. So we take W plus S plus T, this is equal to zero. Now the fact that they're in static equilibrium, we utilize the fact that we could move around these vectors and form a triangle. And then we try to do some trigonometry on that triangle, but there are some issues with that. One, that triangle is not always a right triangle. Is that right triangle? I don't know. Solving oblique triangles is a lot harder than solving for triangle, right triangles, of course. What if it's not even a triangle, right? What if you have some type of like, pentagonal shape like so, that makes the trigonometry a lot easier. What if it doesn't even come into static equilibrium? We saw some examples of that, right? What do you do something like that? The geometry can get very complex very easily. We basically chose Mickey Mouse examples to make these examples doable. But in the algebraic form, these constraints are really nothing for us. We can convert each and every one of these vectors to its algebraic form pretty quickly. So take, for example, W. W is pointing directly downward, and so W is just equal to negative 42 times J. J is the vertical unit vector, and since it's going down, we're going to get a negative magnitude right there. Stacey, we don't know her magnitude, assuming we forgot the previous video what that number was. But we do know that whatever her magnitude is, it's going to be S times I. All of her force is pointing in the positive horizontal direction. T is a little bit more complicated because this angle right here is 30. This angle right here would be 60 degrees. This is the reference angle to, of course, 120 degrees right there. And so we see that T is given as, well, it's tension times cosine of 120 degrees. But in the second quadrant, again, this references to 60 degrees and cosine would be negative. You get I right there. And then you're going to add that to the tension of T times sine of 60 degrees times J, like so. So again, we don't know the magnitude there, but we do know the direction. So whatever that magnitude is, the horizontal component will be 60, cosine 60 of that. And then the vertical component will be sine 60 of that magnitude here. And so when we put all of these things together, right, we can plug in W, we can plug in S, we can plug in T. We see the following, right? We have a negative 42, 42 J. We have a absolute of S times I. And then we have this negative magnitude of T, cosine of 60 degrees is course one half. So I'm going to put negative magnitude of T over two. And then sine of 60 degrees, that's the magnitude of T times that square root of three over two, like so. Don't forget your unit vectors, I and J, like so. This should equal zero. And then combining like terms, let us, we can add together the horizontal components. We can add together the vertical components. This is the same thing as, well, we have the magnitude of S. And then we have minus the magnitude of T over two, that is our horizontal component. And then the vertical component, we're going to end up with the magnitude of T times the square root of three over two minus 42, like so. Times J and this is equal to the zero vector here. Now the only way that these two vector, that this vector can equal zero is if the horizontal components equal to zero and the vertical components equal to zero. So this gives us equations, right? The first equation, the horizontal equation here, the one associated to I, this tells us that the magnitude of S minus the magnitude of T over two is equal to zero. And then the vertical component tells us that the magnitude of T times the square root of three over two minus 42 is equal to zero, like so. We have the system of equations to unknowns. In particular, if you look at the one associated to J, the vertical component, there's only one unknown and that's the magnitude of T. We can solve for the tension very quickly. We can add 42 to both sides. We get the tension times the square root of three over two is equal to 42 times both sides by two. We get the tension times the square root of three is equal to 84 divide both sides by the square root of three. And we see that the tension is equal to 84 over the square root of three, which of course is approximately 48.5 pounds of force. And this agrees with what we got previously for which you can verify. Now that we know the tension, we could substitute that back in to our previous equation right here and solve for S, right? So S here minus we're going to take 84 over the square root of three and times that by one half, this is equal to zero. Well, two goes into 84, of course, 42 times. So we end up with S is equal to 42 over the square root of three. You can rationalize the denominator if you feel so compelled to do so. I won't worry about that. It's just half of the value there. We're going to 24.2 pounds of force. And so this was equal to the force we found previously for Stacey. So Stacey is holding up Danny applying about 24 pounds of force and the rope is holding up Danny applying about 48 pounds of force. And these are exact answers we got previously when we did this with the trigonometry. We didn't have to worry about any of the confusing geometry. It was just a straightforward algebra problem. We have a system of linear equations that we solved pretty easily and we got the answers. And so I would consider and invite each and every one of you watching this video to reconsider those story problems we did previously. If you were to do them in algebraic form, how does that simplify it? And I think you would find that it is simpler. There is the process conversion between trigonometry to algebra and vice versa perhaps. But that conversion is cheap compared to the amount of savings you'll make with solving these problems algebraically.