 When one is working with vector quantities, it's helpful to introduce a geometry to the model. In particular, a type of coordinate system like there's an x-axis, there's a y-axis. And if you have an x and y-axis, then there's necessarily an origin. Since the location of a vector in the plane doesn't change the vector quantity, we often talk about putting a vector in standard position. A vector in standard position means that we've placed the tail of the vector on the origin of our coordinate system. When a vector is in standard position, then it makes sense to talk about its components. For these vectors, there are two components, the horizontal component, which we'll call vx, because x is the horizontal variable. And then there's the vertical component, which we'll call vy, because y is the vertical variable right there. And these are two vectors which represent pieces of the original vector v. If you were doing vectors in three dimensions, you'd have three components for the variables x, y, and z. But in our lecture series, we'll be exclusive to two-dimensional vectors. So the horizontal vector is gonna be a vector that runs along the x-axis that you see right here. The vertical component vy will be a vector that runs along the y-axis. And in particular, it'll have the property that vx plus vy adds back together to be v itself. And so by the parallelogram rule, notice here that we have vx, we have vy. If we were to copy vx and copy vy, we form a parallelogram because vx and vy are actually perpendicular to each other. This parallelogram turns out to be a rectangle, right? But the parallelogram rule still applies, v is the diagonal of this parallelogram, and therefore it's the sum. And so this gives us the vertical and horizontal components of a vector. Notice though that when you look at this vector, it's sort of like if you have the sun shining over here, then vx is just the shadow of v cast onto the x-axis. In the other direction, if you think of vy, right, we're gonna place the sun over here, right? And it casts a shadow now, v onto the y-axis. That's exactly what this vy is. It's the shadow cast onto the x-axis or the y-axis by the vector v. But with this parallelogram rule here, we can move vy over here. And notice now if we put vy on this side instead, we now have this right triangle, like so, for which then using that just right triangle trigonometry, we can relate together the quantities, the vector quantity v with its two components, vx and vy. All right? So in particular, the magnitudes of these three vectors, which are scalar quantities, will satisfy the Pythagorean relationship. You're gonna have that the magnitude of v is equal to the magnitude of vx plus the magnitude of vy, assuming we put squares on each of these things. That's the Pythagorean relationship. The hypotenuse squared is equal to the sum of the legs, sum of the leg squared right there. So that gives us this Pythagorean relationship that connects the two. If we're interested in this quantity theta right here, this direction, well, you can use a tangent relation there. Tangent of theta is equal to vy over vx. So the tangent ratio connects the direction of v here with the magnitudes of these friends over here. So I should put these magnitudes here because those are giving us the scalar quantities. But also using other trigonometric relationships with respect to theta right here, if we think of the cosine relationship, you're gonna take the adjacent over hypotenuse. Like so, clearing the denominators, this tells us that the magnitude, the, excuse me, the horizontal component is gonna equal the magnitude of the original vector times cosine of its direction in standard position. And similarly, if we use the sine ratio in this situation, you're gonna get the magnitude of vy, the vertical component divided by the magnitude of v. If you clear the denominators, you get this formula right here. And so these formulas tell you how you can compute the vertical and horizontal components of a vector if you know its magnitude and its direction. Now, when it comes to the horizontal component, we don't really need to know its direction because it's the horizontal component. It's always pointing in the right word direction. We need to know this value though, how strong it is. And we do, of course, need to pay attention to the sine. If it's a positive value, it points to the right. If it's a negative value, it points to the left. With vertical components similar, we don't have to worry about the direction of a vertical component because it's always gonna point up and down, depending on whether it's a positive or negative value. But the magnitude of that vertical component does matter. And so these formulas come into play for that situation. So imagine that a cannon ball is shot from a cannon with an initial velocity of 53 miles per hour at an angle of 60 degrees with the horizontal. So we have this, we have the ground. Think of it that like though, so. And we've launched a cannon ball of some kind. So it's gonna follow a path for which we're just gonna look at the initial vector at the start of this thing, this initial velocity vector. So this vector V here, it has a direction of 60 degrees. It's good shot above the ground. And it starts off at 53 miles per hour, our cannon ball. That's how it's launched. Can we find the magnitudes of the horizontal and vertical components of this thing? So be aware that this velocity vector we're gonna call it V. Its magnitude is given as the 53 miles per hour and its direction theta is given as this 60 degrees right here. So if we wanna find the horizontal component, that is we're looking for the magnitude of Vx because we know the horizontal component goes in the right direction. It's gonna look something like this, right? We're going some direction to the right, some direction up. This is Vx, this is Vy. We need to know the magnitudes of these things. So by the formula we saw previously, the magnitude of Vx, this is gonna be 53 times cosine of 60 degrees. The magnitude of Vy is gonna equal 53 times a sine of 60 degrees, which 60 degrees is one of our special angles. So we know this without our calculator here. Cosine of 60 degrees, that of course is going to be one half. So we get 53 over two. Taking half of that, you're gonna end up with 26.5 and this will have the same units as the magnitude of the original vector. So it's traveling in the horizontal direction at 26.5 miles per hour. Sine of 60 degrees, that's of course equal to the square root of three over two. You might wanna use your calculator to help you with the approximation here. You'll end up with 45.9 miles per hour that it's traveling vertically. So notice how this thing is angled more upward than downward. So you see that the horizontal is actually a smaller magnitude than the vertical because it's pointing more upward than it is to the right. All right, like we discussed earlier, we can go the other way around, right? If we know the vector's magnitude and direction, we just saw how we can compute its vertical and horizontal components. But what if we know the vertical and horizontal components of a vector? Can we compute its magnitude? Can we compute its direction? And yes, by the Pythagorean relationship, the magnitude of the original vector, you can compute as the square root of the sum of squares of the vertical and horizontal components, magnitudes. Again, that's just a consequence of the Pythagorean equation. And then by a tangent ratio, if you take the ratio between the vertical component with the horizontal component, that gives you tangent theta, take arc tangent to get the direction and pay attention to the quadrant, the direction you're probably in. So changing the direction this time, we have an arrow that's shot into the air. So that's vertical velocity is given as 25 feet per second. That's horizontal. And its vertical velocity is given as 15 feet per second. So it's not gonna be quite as long there. 15 feet per second, like so. So we're looking for the vector that connects these two together. We wanna find this vector V, like so. And so let's find the velocity of the arrow. How fast is the arrow going? Well, we need to find the magnitude of this thing. We would take the absolute value, excuse me, the square root of these values, 25 squared plus 15 squared. Continuing on here, 25 squared is 625. Oops, sorry about that, 625. 15 squared is gonna be 225. Add that together, you're gonna get 850 inside of the square root. I am gonna simplify this here because 825 is the same thing as 25 times 34. 25 is a perfect square. So you could get the exact value of five square root of 34, but approximating this thing, we're gonna end up with 29 miles per hour. That's how fast this thing is going. Well, what if we wanna ask about the direction? What's the direction of this thing? Now, it makes sense that we'll represent the angle with respect to the horizontal. So what angle above the ground is this thing traveling? Using the tangent ratio, we'll recognize that tangent theta is equal to 15 over 25. 15 and 25 both have a common factor of five in there. So I can actually reduce this to three-fifths. So theta is gonna equal tangent inverse of three-fifths, like so, and that when we consult our calculator, we'll give us 31 degrees. Make sure you are in degrees on your calculator for this one. So because of the vertical and horizontal components we were given for this arrow's velocity vector, we were able to determine its speed, the arrow is traveling at 29 miles per hour, and it's shooting at an angle of 31 degrees above the horizontal.