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### Course: MCAT > Unit 9

Lesson 1: Vectors and scalars# Visualizing vectors in 2 dimensions

Learn about two-dimensional vectors, their magnitude and direction. Discover how to add vectors visually and break them down into horizontal and vertical components. This technique transforms complex two-dimensional problems into manageable one-dimensional problems, a key strategy in classical mechanics.

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Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video.

## Want to join the conversation?

- I don't think it's a good idea for him to use a calculator when we won't have that option on the MCAT. This needs to mirror the MCAT.(61 votes)
- This content was not created specifically for the MCAT, only included in this section because it is conceptually relevant.(9 votes)

- I'm a little confused as to why this is in the MCAT section. How are vectors used in medicine?(0 votes)
- Just look at it this way. You need this knowledge for the MCAT and you need a good mcat to study medicine.(14 votes)

- [5]sin(36.8699) = -3.687

[5]cos(36.8699) = 3.377

Where did he get his answers from? I am so confused...(4 votes)- Sal merely made a mistake when writing out the angle in the sine and cosine problems.

You did set up the problem correctly though. Make sure that your graphing calculator is in degrees and not radians. (They give two completely different answers).

You might want to look into memorizing the typical right triangles that will appear on the MCAT. These triangles are the 3-4-5 triangles and the 5-12-13 triangles. It is said that multiples of these two triangles (the 3-4-5 one more than the other) appears on the MCAT the most (ex. a 6-8-10 triangle would have the same angles as the 3-4-5, etc). Also look into memorizing the sine, cosine and tangent values of the typical angles (0, 30, 45, 60 and 90).

I hope this clears things up!(16 votes)

- USELESS if the video does not show how to answer this question with just calculations in the head. Give some shortcuts on how to solve SIN COS without a calculator. That would be useful(3 votes)
- The MCAT uses common angles, like 0, 30, 45, 60, 90. And those are easy to remember, I don't know why he used a ridiculous number. Plus, it you can recognize that it's a 3-4-5 triangle, you don't even need to do calculation! The MCAT sometimes doe that(7 votes)

- At first he said the angle is 36.8699 but later he said the angle is 36.899.(3 votes)
- It should have been 36.8699 (36.86989764584401...) or more accurately cos^-1(4/5).(3 votes)

- If a+b=c in the beginning, how does 3+4=5 make sense in the end of the video?(3 votes)
- Given that we are not allowed to use a calculator on the mcat exam, how can we do the calculation at10:41without a calculator?(2 votes)
- The MCAT uses common angles, such as 0, 30, 45, 60, or 90. And those are easy to remember.(3 votes)

- Is there anyway you can slow down when you start putting the question and answers down you put them down to where I have to pause it just so I can keep up and also why did you not put the 6 in 36.899 because isn't supposed to be 36.8699?(1 vote)
- If you go to the bottom right corner of the rectangle below the video and look for 'Options' (it is to the left of 'Share'), you can speed up or slow down playback. Best, Kim(1 vote)

- 9:43why the "magnitude of our y component" not just "y component"?(1 vote)
- When sal gave the angle of 36.8699 how did it then become 36.899? I think the time is7:01did he just make an error?(1 vote)
- At9:28a clarification pops up addressing this. The angle originally should have been 36.899.(1 vote)

## Video transcript

- [Voiceover] All the problems we've been dealing with so far have had, have essentially been happening in one dimension. You could go forward or back, so you could go forward or back, or right or left, or you could go up or down. What I want to start to talk about in this video is what happens when we extend that to two dimensions, or we could even just extend what we're doing in this video to three or four, really an arbitrary number of dimensions, although if you're dealing with classical mechanics you normally don't have to go more than three dimensions. And if you're gonna deal with more than one dimension, especially in two dimensions, we're also going to be dealing with two-dimensional vectors. And I just want to make sure through this video that we understand at least the basics of two-dimensional vectors. Remember, a vector is something that has both magnitude and direction. So the first thing I want to do is just give you a visual understanding of how vectors in two dimensions would add. So let's say I have a vector right here, that is vector A, so once again its magnitude is specified by the length of this arrow, and its direction is specified by the direction of the arrow. So it's going in that direction. And let's say I have another vector, let's call it vector B. Let's call it vector B, and it looks like this. And what I want to do in this video is think about what happens when I add vector A to vector B. So there's a couple of things to think about when you visually depict vectors. The important thing is, for example, for vector A that you get the length right, and you get the direction right. Where you actually draw it doesn't matter. So this could be vector A, this could also be vector A. Notice it has the same length, and it has the same direction. This is also vector A. I could draw vector A up here. It does not matter. I could draw vector A up there. I could draw vector B. I could draw vector B over here. It's still vector B. It still has the same magnitude and the direction. Notice, we're not saying that its tail has to start at the same place that vector A's tail starts at. I could draw vector B over here. So I can always have the same vector, but I can shift it around, so I can move it up there as long as it has the same magnitude, the same length, and the same direction. And the whole reason I'm doing that is because the way to visually add vectors, if I wanted to add vector A, if I wanted to add vector A, plus vector B, plus vector B, and I'll show you how to do it more analytically in a future video, I can literally draw vector A. I draw vector A, so that's vector A right over there, and then I can draw vector B, but I've put the tail of vector B to the head of vector A. So I shift vector B over so its tail is right at the head of vector A, and then vector B would look something like this. It would look something like this. And then if you go from the tail of A all the way to the head of B, all the way to the head of B, and you call that vector C, and you call that vector C, that is the sum of A and B. That is the sum of A and B. And it should make sense. If you think about it, if these were, let's say these were displacement vectors, so A shows that you're being displaced this much in this direction. B shows that you're being displaced this much in this direction, so the length of B in that direction. And I were to say you have a displacement of A, and then you have a displacement of B, what is your total displacement? So you would have had to be, I guess shifted this far in this direction, and then you would be shifted this far in this direction, so the net amount that you've been shifted is this far in that direction. So that's why this would be the sum of those. Now we can use that same idea to break down any vector in two dimensions into its, we could say, into its components. And I'll give you a better sense of what that means in a second. So if I have vector A, let me pick a new letter. Let's call this vector, vector X. Let's call this vector X. I can say that vector X is going to be the sum of is going to be the sum of this vector right here in green, and this vector right here in red. Notice if I take, if I, X starts at the tail of the green vector and goes all the way to the head of the magenta vector, and the magenta vector starts at the head of the green vector and then, and then finishes, I guess, well, where it finishes where vector X finishes. And the reason why I do this, and you know this, hopefully from this explanation right here says, "Okay, look." Vector green, the green vector plus the magenta vector gives us this X vector. That should make sense, I put the, I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is if I can express X as a sum of these two vectors, it then breaks down X into its vertical component, and its horizontal component. So I could call this, I could call this the horizontal component, or I should say the vertical component, X vertical, and then I could call this over here, I could call this over here the X horizontal. Or another way I could draw it, I could shift this X vertical over. Remember, it doesn't matter where I draw it, as long as it has the same magnitude and direction, and I could draw it like this. X vertical. And so what you see is that you can express this vector X, you can express this vector X, let me do the same colors, you can express this vector X as the sum of its horizontal and its vertical components. As the sum of its horizontal and its vertical components. Now we're gonna see over and over again that this is super powerful because what it can do is it can turn a two-dimensional problem into two separate one-dimensional problems. One acting in a horizontal direction, one acting in a vertical direction. Now let's do it a little bit more mathematical. I've just been telling you about length and all of that, but let's actually break down, let me just show you what this means to break down the components of a vector. So let's say, let's say that I have a vector that looks like this. Let me do my best to, so let's say I have a vector that looks like this. Its length is five, so let me call this vector A, and I will say, so vector A's length is equal to five, and let's say that its direction, we're gonna give its direction by the angle between the direction it's pointing at and the positive x-axis, so maybe I'll draw an axis over here. So let's say that this right over here is the positive y-axis, going in the vertical direction, this right over here is the positive x-axis going in the horizontal direction, and to specify this vector's direction, I will give this angle right over here, and I'm gonna give a very peculiar angle, but I picked this for a specific reason, just so things work out neatly in the end. And I'm gonna give it in degrees, it's 36.86 99 degrees. So I'm picking that particular number for a particular reason, and what I want to do is I want to figure out this vector's horizontal and vertical components. I want to break it down into something that's going straight up or down, and something that's going straight right or left. So how do I do this? Well one, I could just draw them visually, see what they look like, so its vertical component would look like this, so let me, it would start, its vertical component would look like this, and its horizontal component would look like this. Its horizontal component would look like this. The horizontal component, the way I drew it would start where vector A starts, and go as far in the X direction as vector A's tip, but only in the X direction, and then you'd need it to get back to the head of vector A you'd need to have its vertical component. And we can sometimes call this, we could call the vertical component over here A sub Y, just so that it's moving in the Y direction, and we could call this horizontal component A sub X. And what I want to do is I want to figure out the magnitude of A sub Y and A sub X. So how do we do that? Well, the way we drew this, I've essentially set up a right triangle for us. This is a right triangle. We know the length of this triangle. The length of, or the length of this side, or the length of the hypotenuse, that's going to be the magnitude of vector A. And so the magnitude of vector A is equal to five. We already knew that up here. So how do we figure out the sides? Well we could use a little bit of basic trigonometry. If we know the angle, and we know the hypotenuse, how do we figure out the opposite side to the angle? So this right here, this right here is the opposite side to the angle, and if we forgot some of our basic trigonometry, we can re-learn it right now. Soh, cah, soh cah toa. Sine is opposite over hypontenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we have the angle, we want the opposite, and we have the hypotenuse. So we could say, we could say that the sine, the sine of our angle, the sine of 36.899 degrees is going to be equal to the opposite over the hypotenuse. The opposite side of the angle is the magnitude of our Y component, is going to be equal to the magnitude of our Y component, the magnitude of our Y component over the magnitude of the hypotenuse, over this length over here, which we know is going to be equal to five. Or if you multiplied both sides by five, you get five sine of 36.899 degrees, is equal to the magnitude, is equal to the vertical component, the magnitude of the vertical component of our vector A. Now before I take out the calculator and figure out what this is, let me do the same thing for the horizontal component. Over here we know, this side is adjacent to the angle, and we know the hypotenuse, and so cosine deals with adjacent and hypotenuse. So we know that the cosine of 36.899 degrees is equal to cosine is adjacent over hypotenuse, is equal to the magnitude of our X, of our X component, over the hypotenuse. The hypotenuse here has, or the magnitude of the hypotenuse I should say, which has a length of five. Once again we multiply both sides by five, and we get five times the cosine of 36.899 degrees is equal to the magnitude of our X, of our X component. So let's figure out what these are. Magnitude of our vertical component is equal to three, is equal to three, and then let's do the same thing for our horizontal component. We get it to being four. So what we see here is a situation where we have, this is a classic three, four, five Pythagorean Triangle. The magnitude of our horizontal component is four, the magnitude of our vertical component, the magnitude of our vertical component right over here is equal to three. And once again, you might say, "Sal, you know, "why are we going through all of this trouble?" We'll see in the next videos, if we say something has a velocity in this direction of five meters per second, we can actually say that we can break that down into two component velocities. We can say that that's going in the upwards direction at three meters per second, and it's also going to the right in the horizontal direction at four meters per second. And it allows us to break up the problem into two simpler problems, into two one-dimensional problems instead of a bigger two-dimensional one.