Hey, guys. Let's check out this example together. So we've got these two vectors, \( a \) and \( b \). \( A \) has a magnitude of 10 and it's at 40 degrees. \( B \) has a magnitude of 3 at 20 degrees. And we're going to calculate the magnitude and direction of the resultant vector. But this resultant vector is not \( a+b \), it's \( a - 2b \). So it's still going to be vector addition. We just have to use our rules of subtraction and scalar multiples. So \( a - 2b \) is the same thing as saying \( a + \text{the negative of } 2b \). First things first, we're just going to draw and connect our vectors tip to tail. So, we've got this vector \( a \). It's 10 at 40 degrees, so you start from the origin. We've got this vector like this. This is going to be \( a \). It's 10 and we know that this is a 40 degree angle. Now, vector \( b \) has a magnitude of 3 in the direction of 20 degrees. It's a little bit flatter. So we go from tip to tail like this. And so this is going to be our \( b \) vector and let's make it a little bit shorter. Let's say it looks like this. So we've got \( b \) is 3 and we know this angle here is 20 degrees. So, you know, hopefully you guys can see that.

Now we have the vectors. We draw them tip to tail, but the problem is we have to remember that our resultant vector is not going to be from \( a \) to the end of \( b \) because we actually have to do \( a - 2b \). So even though we've drawn out the vectors, there's actually still a little bit more than we need to do. So this resultant vector, first, we actually have to subtract 2 \( B \) here. So in order to subtract \( B \), we're just going to flip it. Basically, it's going to go completely in the opposite direction with the same length. So if \( b \) points in this direction, then negative \( b \) is going to point exactly in the opposite direction. This is going to be negative \( b \) and this is going to be another negative \( b \) because we actually have to subtract 2 \( b \). So that means that my resultant vector is, again, not from here to here. It actually You have to do \( a \) and you have to go minus 2 \( b \) and go in this direction. So my resultant vector is actually going to be this vector over here. This is my \( r \) vector.

So the second step here is we draw the resultant and its components. So now we have to figure out the legs of this triangle over here. But you know, again, so we have this vector here. We can calculate the components but the resultant vector is actually if I go in this path over here. So let's go to the third step. In order to calculate the legs of this, we have to break all the other triangles out into their components and then we have to calculate them. So this is my \( a_x \), my \( a_y \), and then just for the sake of calculating the components of \( b_x \), we're going to draw them out on this table over here or on this diagram. So remember, we have to use a table in order to calculate all the \( x \) and \( y \) components.

Let's go ahead and do that. This is my \( x \) and \( y \) components. And then just for color coding, I'll just use This is my \( a \) and this is my \( b \). And then I'm going to end up with a resultant vector of \( r \). And then I have to write out the equation for \( r \). So let's go ahead and do that. So now if we want the \( x \) component of \( a \), we just have to stick to our vector decomposition equations. This is where we go to vectors, from vectors, and then we want to decompose them into their components. So we just use our \( a \) cosine \( \theta \) and \( a \) sine \( \theta \) equations. So I'm going to have for my \( a_x \) component, I'm going to use \( 10 \cdot \cos(40^\circ) \). What I end up with is 7.7. If you do the same thing for sine, \( 10 \cdot \sin(40^\circ) \), what you get is 6.4.

Now, if we do for \( b \), \( b \cdot \cos(\theta) \), that's this component over here. So I know this is 7.7, this is 6.4. My \( b \cdot \cos(20^\circ) \) term is going to be \( 3 \cdot \cos(20^\circ) \) and what I end up with is 2.8. So we know this is 2.8. Finally, \( 3 \cdot \sin(20^\circ) \) which is equal to 1. So this is just 1. Alright. So these are all of our components. The table is really great at organizing all of this stuff.

Now let's bring us to the next step. If we want to calculate the resultant vector, we're going to have to combine all of these \( x \) and \( y \) components according to the \( r \) equation. And remember, it's not \( a + b \). You're not just going to add all these components straight down because remember that \( a \) is equal to \( a - 2b \). What that means is that the \( x \) component of \( r \) is going to be the \( x \) component of \( a \) minus 2 times the \( b_x \) component. So if \( r \) is \( a - 2b \), then \( r_x \) is \( a_x - 2 \cdot b_x \). Right? Just follows the same format. So my \( a_x \) is going to be 7.7 and then minus 2 times the \( x \) component which is 2.8. If you work this out, your \( x \) component is going to be 2.1. If you do the exact same thing over here, this is going to be \( a_y - 2 \cdot b_y \), so therefore, it's going to be 6.1 minus 2 times 1, and we get a \( y \) component of 4.4. So now we actually know what these \( r_x \) and \( r_y \) components are. I know this is 2.1 and this is 4.4. And if you look at the diagram, it actually kind of makes sense. It's about 2.1 and to 4.4. It's about twice as high. So this should make some sense to you.

Okay. So now that's the last step. We just have to calculate \( r \) and \( \theta_r \). And now we're going from the components of a vector. We have 2.1 and 4.4, the components, and now we actually want to construct the magnitude and the direction. We use our Pythagorean theorem and our tangent inverse. So that means the magnitude of \( R \) is just going to be the Pythagorean theorem. We've got \( 2.1^2 + 4.4^2 \). If you work this out, you're going to get 4.9. So that is our resultant. That's the magnitude of the resultant. Now, for the direction, that's \( \theta_r \). This is going to be the angle \( \theta_r \) over here. It's kind of messy. But I've got the tangent inverse, the arctangent of my \( y \) component which is my 4.4 over the \( x \) component which is 2.1. So if you work this out, you're going to get an angle of 64.5 degrees. So this over here, \( \theta_r \) is roughly 64.5 degrees. Alright, guys. That's it for this one. Let me know if you have any questions.