Adding Vectors by Components - Online Tutor, Practice Problems & Exam Prep

To calculate the resultant vector from two displacements, first sketch the vectors tip to tail. Break each vector into components using A^cos and A^sin. Organize these components in a table, then sum the x and y components. Finally, use the Pythagorean theorem to find the magnitude and the tangent inverse to determine the direction relative to the x-axis. This systematic approach ensures accurate vector addition.

Vector Addition By Components

Hey, guys. So previously, we saw how to add vectors graphically. A situation like this where you have vectors a and b and you want to calculate the result, so you add them tip to tail and then you count up a bunch of boxes. Well, sometimes, unfortunately, you're gonna have to calculate the resultants without being able to count squares because you're gonna have diagrams that don't have grids. So in this video, we're gonna see that all we just need are all of our vector math equations or triangle math equations, and we're gonna pull them all together to solve these kinds of problems. Now these problems, as we're gonna work one out together, are actually really, really repetitive. You end up doing the same things over and over and over again. So I'm gonna give you a list of steps. It's gonna help you get the right answer every single time. Let's check it out.

So this problem says that we're walking 5 meters at 53 degrees, so that's a displacement, and then another displacement in another direction. We're gonna calculate the magnitude and the direction of the total displacement. Remember, total displacement can just be thought of as a resultant vector. So we're gonna be calculating a resultant vector. Alright. So the first step we're gonna do in these problems is we're gonna draw and then connect the vectors tip to tail. So when we were given them graphically, we didn't have to draw them, but now that we don't have them, we're gonna have to draw them out. And because we don't have a grid, we're just gonna kinda sketch them out to the best of our ability. So we're gonna start from the origin, 5 meters at 53, looks something like this. That looks about right. So this is 5. We know this direction here which is relative to the x-axis is 53 and we've got 8 at 30. So connect them tip to tail and this vector here, 30 degrees is a little bit shallower than 53, a little bit flatter, so it's gonna look something like this. So it's 8, doesn't have to be perfect and we know that this is 30 degrees above the x-axis. Cool.

So that's the first step. The second one is just we have just drawn out what the resultant vector is gonna look like. So what is the resultant? Well, when we did this graphically, we just connect them tip to tail and the resultant was the shortest path from start to finish. It's basically as if we had actually just walked in this direction. The principle is the exact same. It's basically the shortest path from the start of the first one to the end of the last and this is gonna be my result in vector. So this is gonna be the magnitude r and then it's specified by an angle which is theta r.

So we're trying to figure out the magnitude and the direction, which means that we're gonna need the components. Remember how we how do we solve for this? We take this result in vector and break it up into the triangles and we have to figure out what Rx and Ry are. So we have to get these components over here. When we did this graphically, this was pretty straightforward. We have the magnitude, we break it up into its little triangles, we get r x and r y, we could just count up the little boxes here. It was pretty straightforward. This was just 4 and this is also just 4. We don't have boxes in this situation, so we can't calculate them by just adding up all the boxes. There's nothing for us to count.

We're gonna need a new method to figure out what these components are. So one way to think about how we got this Rx component is we can kinda break up the smaller vectors a and b into their small little triangles, so we can kinda think about this little triangle up here, this little x component. Basically, we just think about this a vector as breaking up into its triangles and we have this ax component and then we also have bx component as well. And so we know this ax component is 3, we just count the boxes, you know, this bx component is just 1 and so one way you can think about this is that this 4 is really just the addition of this 3 and this one put together.

Now, this is gonna work the exact same way over here. So what we have to do is we have to break up each of the vectors into the triangle. So this is my ax and this is my ay and then this is going to be my bx, and this is gonna be my by. So now, that leads us to the 3rd step, which is we have to calculate what all of these components are. So that brings us to our equations. How do we take a vector which we have the magnitude and the direction and calculate the components? We're just gonna use our vector decomposition equations. So that's which Our a cosine theta and our a sine theta equations. Now, we have 2 vectors. This is a and this is gonna be b, so the best way to keep track of all of the components is, oops. I'm sorry. I'm supposed to make that green.

So the best way to keep track of all these components is by building a table. You're gonna have a lot of stuff floating around in your papers everywhere. It's good to get organized, so I always highly encourage that you guys build the table. So we want the ax components and the ay components. So this we're just gonna use our a cosine theta and a sin theta equations. So my a vector, the magnitude is 5, so this is gonna be 5 times the cosine of 53 degrees and we're gonna get 3. If you do the same thing for the y direction you use sin...

Adding Vectors

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.

Here’s what students ask on this topic:

How do you add vectors using components?

To add vectors using components, follow these steps: First, sketch the vectors tip to tail. Break each vector into its x and y components using the equations $A_x=\; A\; cos(\theta )\; and{A}_y=\; A\; sin(\theta ).\; Organize\; these\; components\; in\; a\; table\; for\; clarity.\; Sum\; the\; x\; components\; and\; y\; components\; separately\; to\; get\; the\; resultant\; vector\text{'}s\; components.\; Finally,\; use\; the\; Pythagorean\; theorem\; to\; find\; the\; magnitude\; of\; the\; resultant\; vector:$ R=\; \surd (R_{x}2+R_{y}2)$,\; and\; use\; the\; tangent\; inverse\; to\; determine\; the\; direction:$ \theta =\; tan-1(R_y/R_x)$.$

What are the steps to calculate the resultant vector from two displacements?

To calculate the resultant vector from two displacements, follow these steps: 1) Sketch the vectors tip to tail. 2) Break each vector into x and y components using $A_x=\; A\; cos(\theta )\; and{A}_y=\; A\; sin(\theta ).\; 3)\; Organize\; the\; components\; in\; a\; table.\; 4)\; Sum\; the\; x\; and\; y\; components\; separately.\; 5)\; Use\; the\; Pythagorean\; theorem\; to\; find\; the\; magnitude:$ R=\; \surd (R_{x}2+R_{y}2)$.\; 6)\; Use\; the\; tangent\; inverse\; to\; find\; the\; direction:$ \theta =\; tan-1(R_y/R_x)$.$

Why is it important to organize vector components in a table?

Organizing vector components in a table is important because it helps keep track of all the x and y components systematically. This organization minimizes errors and ensures that you correctly sum the components. It also makes it easier to visualize the problem and follow the steps methodically, especially when dealing with multiple vectors. A table provides a clear and concise way to see the relationships between the components and the resultant vector.

How do you find the magnitude and direction of the resultant vector?

To find the magnitude and direction of the resultant vector, first sum the x and y components of the individual vectors. Then, use the Pythagorean theorem to find the magnitude: $R=\; \surd (R_{x}2+R_{y}2)$. For the direction, use the tangent inverse function to find the angle relative to the x-axis: $\theta =\; tan-1(R_y/R_x)$. This angle gives you the direction of the resultant vector.

What equations are used to break a vector into its components?

To break a vector into its components, use the following equations: $A_x=\; A\; cos(\theta )$ for the x-component and $A_y=\; A\; sin(\theta )$ for the y-component. Here, A is the magnitude of the vector, and θ is the angle the vector makes with the x-axis. These equations allow you to decompose the vector into its horizontal and vertical parts, which can then be used for further calculations.