Vector is going to be the topic in this last lesson in the first chapter of my brand new General Physics playlist and this is going to be foundational so a lot of the true physics problems we're going to begin doing are going to be one dimensional to start but the moment they become two-dimensional vectors are going to become super super important but before we get there we've got to do a little bit of a trigonometry review because I really want you to understand vectors not just kind of memorize some rules of how to break them into components and add them and stuff I really want you to understand it you're much more likely to own it that way my name is Chad and welcome to Chad's prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on chatsprep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat okay so we're going to start this off with a little bit of a trigonometry review here and a reminder of sohcahtoa here and just reminder what this stands for so the so here so sine equals opposite over hypotenuse so sine of an angle Theta is equal to the opposite over the hypotenuse K cosine equals the adjacent over the hypotenuse and then TOA the tangent of that angle is equal to the opposite over the adjacent that's socatoa and that's going to become relevant here in just a little bit so we're going to go back and take a look at a right triangle we're going to find out that looking at vectors and think of a sword breaking up the components adding them together all this stuff is based on right triangles and in those right triangles the two legs of the triangle are always going to line up with the x-axis and the y-axis typically how we're going to make this work and then you'll have some sort of hypotenuse associated with it as well now the right triangle I'm going to start with is this one right here it's going to be a classic right triangle and maybe you're already familiar maybe like oh Chad I see it it's a three four five right triangle but if you don't we can use the Pythagorean theorem to figure that out anyways so if you recall with a right triangle we've got the Pythagorean theorem a squared plus b squared equals c squared and so in this case that's going to be 3 squared plus 4 squared equals c squared so 9 plus 16 equals c squared or 25 and I'm writing this out all the way on purpose here just to make it explicit here so we'll take the square root of both sides there and we can see that square root of 25 is indeed five and so this hypotenuse here is equal to 5 and indeed it is a three four five right triangle it's a very convenient triangle to use for this example because it just makes the numbers easy at least for the lengths of the sides now what if on the other hand what if I was curious about what this angle was right here or what this one was right here now it's this one that I want and we'll see why that is when we get to vectors specifically but what might I use to figure that out well it turns out I've been knowing all three sides I could use any of these trigonometric identities to figure that out however what if I said before we figured out that part right there using that Pythagorean theorem what is that angle and based on what I currently know what would I use to figure that out well I've got the opposite side I've got the adjacent side I don't have the hypotenuse so how do I figure out angle well in this case because I've got the opposite side in the adjacent but not the hypotenuse well not the hypotenuse means I can't use sine and I can't use cosine but I can use the tangent to figure it out and so in this case we're going to have tangent of we'll call this Theta is equal to the opposite over the adjacent so 4 over 3. and so then to solve for Theta we take the inverse tangent of four thirds all right we'll definitely use our calculator to figure this out now one thing to note make sure the mode of your calculator is in degrees I'm going to be doing all these calculations in degrees that's principally what you're going to be doing in your course as well there will be a couple places somewhere towards the end of first semester maybe the beginning of second semester where we do some things with radians but for now pretty much everything is going to be in degrees and you want to make sure your mode of your calculator is adjusted accordingly all right so we're going to do the inverse tangent of four thirds in parentheses there and we're going to get 53 .13 degrees okay so that's that angle right there 53.13 degrees and if we wanted this one well we could go use one of our trigonometric identities or we could just remember that the other two angles in a right triangle are complementary they have to add up to 90 since that guy's 90 and they all three out to 180 so we could just subtract 90 minus that to figure it out but I don't care about that angle it turns out it's not the one I want that's relevant to what we're going to do here with vectors I just want in principle show that we could determine it if we needed to all right so now we've got this lovely angle and so now I want to go and make some other assumptions let's kind of flip this on its head so we got this three four five right triangle and I want to say well what if I didn't know the three and the four what if I only knew the hypotenuse now but I wanted to figure out what these two sides what their magnitudes were so how could I do that based on what I know and the trigonometric identities we have so a couple different ways so if I look I've got the hypotenuse and the angle and I want to find the opposite side to the angle and the adjacent side to the angle and you can quickly see that sine and cosine are going to be Revel in here relevant here so with side angle I've got the sine I've got the angle so I've got the hypotenuse and I could use it to figure out that opposite side right there so let's do that real quick so on that opposite side we're going to have the sine of our angle which again in this case was 53.13 degrees is equal to the opposite which we want over the hypotenuse which in this case is 5. and so I don't know how to make that look more like an O than a zero but it's an O for opposite how about there there we go and that opposite side is going to equal now 5 times the sine of 53.13 degrees all right so let's multiply that out so 5 times the sine of 53.13 I'm not worried about sig figs right at the moment FYI we will worry about that for most true physics problem calculations just not right here and it comes out to 3.9999 ie4 just like we already knew it was a second ago all right now if I want this adjacent side right here you might be like Well Chad you could get it from the Pythagorean you're right I could but I don't want to again I want to get it from that what was originally given in this part of the problem just the hypotenuse and the angle and now I want the adjacent side right here and again I've got the angle I've got the hypotenuse and I want to solve for that adjacent side right there and so in this case we're going to have cosine of 53.13 degrees is equal to the adjacent over the hypotenuse so we'll move that 5 over again and so 5 times the cosine 53.13 degrees is equal to our adjacent side once again we'll take the time to do this out but this should come out to three kind of just a prediction so five times the cosine of 53.13 it's going to give me 3.000 in a little bit IE three cool and the big thing we want to demonstrate here is that if I've got the two sides I can figure out this angle and the hypotenuse if I've got the angle in the hypotenuse I can go back and figure out the two sides and the key is if I got the two sides well getting the hypotenuse is just Pythagorean theorem and getting the angle we use the tangent since we had the opposite and the adjacent but if I've got the hypotenuse and the angle then I can use the sine function to get the opposite side and I can use the cosine function to get the value of the adjacent side and this is going to be crucial to when we break vectors into components and when we take those components and assemble them back up into vectors and again I'm not assuming you know what a vector isn't yet we'll definitely talk about that in just a second okay so now let's talk a little bit about what a vector is so and it turns out we talk we contrast two different terms they're vectors and scalars and so a scalar is a term that only has magnitude and no Direction whereas a vector has both magnitude and Direction so I'll give you the difference here the most common example is speed versus velocity and let's just say I was going down the highway doing 50 miles an hour so unless that you know let me let me let me rewind that let's not even say going down the highway because the highway is going to have a specific Direction at a certain point in space so let's just say I was moving in my car at 50 miles per hour okay that's a speed and it's a scalar you have no idea what direction I was heading but if I told you that I was traveling in my car at 50 miles per hour due north that both the magnitude 50 miles per hour and the direction due north combined is what we call a vector and it turns out velocity is a vector whereas speed which is just the magnitude is a scalar and we'll review that again we see that when we encounter those back in chapter two again so so that's the difference between a vector and a scalar a scalar is just pure magnitude a vector is going to have both magnitude and Direction and so it turns out things like energy it's just a scalar there's no you know Direction associated with an energy but things like uh we'll find out displacement but velocity acceleration force things of that sort are vector quantities they have both magnitude and Direction that's going to be important throughout the course and it turns out when you add vectors that direction matters so because if you add things that are you know traveling let's say in opposite directions you want to know what the resultant velocity is or something like that or net velocity or net force when they're pushing in opposite directions or something like this you've got to take into account that they are in opposite directions and then oftentimes that means they're going to be canceling each other out rather than being additive and what if they're not even in you know the same plane in their motion we got to deal with that and that's where this Vector stuff is going to become very significant all right so in principle you got to know what the difference between a vector and a scalar is hopefully that's clear now so we also got to talk about how we break vectors up into components and how we add vectors and it's going to be crucial that we know how to break them up into components first so a vector is written usually with an arrow on the Cartesian coordinate plane like this so and if a quantity is a vector it's customary to put over that quantity an arrow whether it's a half headed Arrow or a full headed Arrow same diff you'll see it both ways so but if you put a little arrow on top of that quantity it signifies that it is a vector it has both magnitude and a direction associated with it and when we draw these on a Cartesian plane typically the magnitude is going to be in some way shape or form proportional to how long the arrow we draw is and so if I drew another Vector on this with a shorter Arrow we'd assume it has a shorter magnitude and then it's going to have some sort of Direction associated with it and that direction is usually defined relative to the x-axis so typically it's this angle right here relative to the positive x-axis specifically that that you're going to need and that you're going to use to identify that Vector now why can't I give you this angle right here relative to the positive y-axis well I can however and you can do all sorts of different trigonometric stuff it turns out to figure out the components and all that stuff however I highly recommend that even if you're given the angle relative to the y-axis that you look at that be like oh yep I'm not going to use that one I'm actually going to know that it's complementary with the angle relative to the x-axis and subtract from 90 and find that one instead and kind of get used to this process of using the angle specifically relative to the x-axis it's customary kind of universally throughout all the physics could you do it with the angle relative to the Y you could you're just going to find that you're going to have trouble understanding certain textbooks like all of them and how they set this up so just kind of keep that in mind that's why we're going the way we're going all right so it turns out a vector can be looked at as a combination of X and Y components and the way this works we'll draw in those X and Y components so we often draw right on the x-axis an X component that takes me as far in the X direction as the very tip of that vector and then another one in the y direction so as well and so this Vector here represents or this line here I should say this Arrow here represents the X component of the vector and the red arrow here represents the Y component of the vector and if you notice they along with the vector itself form a right triangle and that's why we started off with our little trigonometry lesson so because the vector itself what we call the resultant Vector is going to be represented by as the hypotenuse of a right triangle and then the X component will be along the x-axis and be one of the legs the Y component will be along the y-axis and be one of the legs and it forms a right triangle and so the whole basis of dividing a vector up into its components or assembling the components of a vector is based on the same trigonometry we just reviewed all right so it turns out if you take a look at what we often Define we say like to say that the X component which we'll signify as ax is equal to the absolute value in this case it turns out that represents the magnitude of the vector some people write a without the brackets here and stuff but you'll see it both ways and then times the cosine of the angle with respect to the x-axis we've got like well where does that come from well great question where does that come from we'll go back to sohcahtoa here and the cosine of your angle is equal to adjacent over hypotenuse and so if you look if we keep using this terminology here what we wanted was that adjacent side right there and if we solve for that adjacent side we can see it's equal to the hypotenuse times the cosine of the angle well again the hypotenuse is always the magnitude of what we call the resultant Vector so and that's why it's showing up there and that's why it's showing up right here and then times the cosine angle gets you the value of that adjacent side which we're calling the X component similarly on the Y component we like to say that the Y component of that Vector equals the absolute value I.E the magnitude of the vector times the sine of the angle with respect to the x-axis well again where does that come from well that's going to come from the so and socatoa and again sine of an angle is equal to opposite over hypotenuse and again so we don't think that's a zero so if I solve for that opposite side it's going to equal the hypotenuse times the sign of the angle well again this opposite side is our y component so and your hypotenuse is the magnitude of the resultant vector and then times the sign of the angle and that's where we get that y component cool now here's the deal could you look and and you know look up the sohcahtoa look at all the relationships and assemble it all into place every time you look at this you totally could but you're going to want to get in the habit of being able to fast forward right here so it's kind of like reading I have two young sons that have been learning to read in the last couple of years so we have things that are called sight words and sight words are just words you want your that are commonly going to show up in sentences that you you want your sons to be able to recognize now for larger words they're going to have to sound those out phonetically you know until they become words they're familiar with well in this case I'm just trying to get these to be and actually I should have put a box around this whole equation here but I'm trying to get these to be words you are familiar with that way you don't have to phonetically sound them out when you encounter them so and again this is a little bit dangerous here because I want you to get to the point where you're actually doing this without thinking but not because you don't understand it but because it's become Elementary to you so that's the idea I want you to be able to look at a vector with its corresponding magnitude and its angle and be able to get the X component and the Y component very quickly because it turns out breaking it up into X and Y components will be imperative to being able to add vectors and things of A Sort any kind of vector math you might do all right so here we've got three different vectors you can see the first one here is a vector equal to 18 meters so it's got units of length the next one's a vector of magnitude 18 meters per second it's got units of velocity and the last one is 18 meters per second squared it's got units of acceleration now the magnitude minus the units anyways is the same for all three vectors but we can see that the directionality is different for all three now what we want to do is find the X and Y components for all three of these and we want to use what we've taken home right here to kind of work that out now the length is going to be the one that's going to be kind of the easiest to conceptualize but we'll be able to apply it directly to what we see with velocity and acceleration as well as we'll see but we'll start with this one here and if we take a look at those X and Y components again we have an X component running right along here we have a y component running right along here and for that X component that's the adjacent side relative to the angle with respect to the x-axis if we didn't have the angle with respect to the x-axis then we would find it and so in this case we're going to say that the X component is equal to the magnitude of the vector 18 meters times in this case it's the again the adjacent angle so cosine of 60 degrees and that's going to get us our X component plug and chug that into the calculator so 18 times cosine of 60 and we're going to get 9. well that's nice so if you knew your geometric identities from way back in the day you could have predicted this was going to be exactly half of the hypotenuse so in this case because cosine of 60 is one half but well no big deal we worked it out it came out anyways if we do the Y component now and again it's going to be equal to the magnitude of the vector times the sine of the angle this time because we've got the opposite side we're trying to find so and that's again 18 meters times the sine f60 and this one you can't you know do in your head sine of 60 is not a nice lovely fraction like one half the way uh the cosine of 60 was but again we'll let the calculator do the math for us so 18 times the sine of 60. and that's going to get us 15.59 I'm just rounded to 15.6 oh I should put the units on this that's meters and this is meters and if we put them on the diagram we've got 9 meters here and we've got 15.6 meters here cool so these are the vector components here now if you take a look at this what we've really got here is two ways to kind of take a look at this length Vector originally it was given that if you walk from the origin 18 meters 60 degrees above the x-axis you'll end up right here and that's one way of describing it so but the X and Y components are another way of describing exactly the same situation so with the X and Y components it says if you walk nine meters in the positive X Direction followed by 15.6 meters in the positive y direction overall net result you end up in the same place so it's kind of like if uh you know you're going to the grocery store and you're if you're you know if you live in Phoenix Arizona like I do the entire city set up on a grid so uh there are very few exceptions so the whole thing's a grid Streets Run either north south or east west and so if you wanted to go the grocery store from your house well if you could fly like a bird you could go directly there but the rest of us we're gonna drive they're gonna have to drive take streets that either run east west or north south to get there so but either description gets you there right so if I go nine meters in the positive X followed by 15.6 meters in the positive y I end up right there that's giving you kind of like the XY coordinates of a point whereas if I said go 18 meters at 60 degrees above the x-axis that would be analogous to giving you the polar coordinates if you remember your polar coordinates and stuff like this now I'm not really going to talk about polar coordinates at all I just want you to see for those of you that are comfortable with it who might be a little more intimately familiar with some pre-calcan calc it is analogous to Polar coordinates here when you've got the magnitude of the vector and its direction relative to that x-axis okay but they're both ways of representing that same axis whether you have the magnitude of the resultant Vector we call it with its direction or if you have the X and Y components and you've got to be able to convert from one to the other and again if you got the components you can use the Pythagorean Pythagorean theorem to get the resultant vectors magnitude and then use the inverse tangent to get the angle with respect to the x-axis or again if you've got the resultant vector and the angle we can use our lovely socatoas here in this case sine and cosine to get the X and Y components magnitudes respectively okay so that was the first quadrant now we've got a vector that's in the second quadrant here and we have still got the angle relative to the nearest x-axis but the thing I want to point out here is if we draw in the X and Y component here so there's our X component and there's our y component the thing I want to point out here is that the Y component still points up in the positive y direction but the X components pointing to the left off in the negative X Direction and so before we even do the plugin and chug in here we should recognize that the Y component is positive and the x component's negative so and we'll see why that's important in just a second all right so let's calculate out those components and in this case if we say that the X component is equal to 18 meters per second times and in this case again we've still got the reference angle in that second quadrant is still 60 degrees and we could totally use that so the problem is that if you always use the reference angle which is always going to be between 0 and 90 well then all your Sines and cosines are going to come out positive okay so but in this case you'd have to make the X component negative after the fact just recognizing which quadrant it's in and you can do that that's actually the way I learned it back in the day 20 plus years ago so however the other thing you can do is take and get that angle relative to the positive x-axis right there and if you use that angle the Sines and the cosines will work out nicely every time now the tangent won't so but the Sines and the cosines will and so if we use a hundred and twenty degrees here instead as our angle with respect to the x-axis and so instead of putting in cosine of 60 we're going to put in cosine of 120 degrees we'll find out that this x component is going to come out negative by default I don't even have to remember that it's negative using the angle from the positive x-axis will generate whether it's positive or negative for me all right so in this case we've got 18 times the cosine of 120 degrees and it comes out to be negative 9 this time the same thing for the Y component 18 meters per second now times the sine and again I could use the 60 degrees but again it's going to give me the proper sign without me having to think about it if I use the angle with respect to the positive x-axis the 120 degrees instead now here's the deal sine function is going to be positive whether we do the first quadrant or the second quadrant and things of A Sort so it turns out it's not going to make a difference here but if it was in the third quadrant or fourth quadrant it would make a difference it turns out so but in this case if we do that 18 in fact we already know this is going to work out to be 15.6 meters and it does indeed come out positive and I should say 15.6 meters per second in this case cool now these are Velocity vectors this deals with now the direction in which some object is traveling rather than some distance that was traveled it's a direction and a speed at which something is traveling in this case but it works the same way we can see it's heading more in the y direction than it is in the direct Direction based on the magnitude of those X and Y components so and in this case it turns out it's adding 60 degrees above the negative x-axis or 120 degrees from that positive x-axis all right let's do one more so we've done one in the first quadrant we've done a vector in the second quadrant now we're going to deal with a vector in the third quadrant and see it's going to work out similarly here and if we break this up into X and Y components and form a right triangle again there's your X component you see it's traveling in the negative X direction or in this case it's got an acceleration in the negative X Direction but it's negative and then the Y components we can also see is now in the negative y direction so it could come out negative as well so the X and Y components are both negative now a couple of things we should realize if you were going to do this with a reference angle that reference angle is always with respect to the x-axis and so you'd actually want this angle right here not the 30 degrees that was given and if that angle is 30 then this is the complementary angle it's these have to sum up to 90. so it'd be 60 degrees but in this case I'm saying you know you might even just go all the way around back to the positive x-axis and get the angle with respect to that positive x-axis so here it's 90 Plus 90 plus 60 or you could say it's 270 minus 30 and you're going to get 240 degrees all right so if we find our components now we'll save it ax is equal to 18 meters per second squared times the cosine of 240 degrees and once again I know how this is going to work out so in this case notice it's the same triangle same reference triangle in different quadrants and that's why the X components and Y components are magnitudes are not changing but again if you're going to take the cosine of 240 degrees it will come out to be a negative number and in this case again negative 9 but now units of acceleration instead and analogously with the Y component foreign of 240 degrees so in this case instead of being 15.6 meters per second it's meters per second squared and the sine of 240 is indeed a negative number cool so how'd you use just the reference angle and said oh the reference angle is just at 60 degrees you would have had to remember that oh but I need to make it negative in both cases since it's in that third quadrant so but again if you use the reference angle of 240 the negatives will come out right out of your calculator for you okay so now we've broken all three of these vectors up into X and Y components and the next step is now going to be well how do we add vectors so and that's a great question so let's go take a look all right so now we're being asked to calculate the vector sum of these two vectors now in this case I've presented these two vectors to you on the study guide on two different sets of Cartesian coordinates here so but most likely they're going to be most likely presented on something like this where they're put on the same Cartesian axis now when you're doing this you're asked to add these two vectors we add them what we call tip to tail fashion it doesn't matter which one you start with but you've got to start with one of them and I'm going to start with the one in black here first and draw it from the origin and again that's five meters and an angle of 53.13 degrees so and then from tip to tail it's tip I'm now going to start the tail of that one and kind of transpose it right off over here so this angle right here is 53.13 degrees and then this angle right here is 22.62 degrees see if we can write that a little cleaner all right cool the resultant Vector would be from the start the origin all the way to where that final resultant Vector ended up and that's ultimately how we diagram out a result in vector and I need to know it's magnitude and I also need to know its direction it's angle with respect to the x-axis that's ultimately we're being asked to find here and we're asked to find the vector sum of these two vectors and the way you add two vectors is you break them up into components and it's the components that you're going to actually add together and we'll see why all right so for this first one here we're going to have a next component and a y component and this should look a little familiar with that 53.13 degrees so but we're going to have the X component and I'll call it ax equaling 5 meters times the cosine of 53.13 degrees and again this is where again I said you want to get to the point where this is something you've previously understood it's become Elementary and so you can just kind of regurgitate it so but the X component is going to equal the magnitude of the resultant Vector times the cosine of the angle with respect to the axis and then the Y component is going to equal again the magnitude of the vector times the sine of that same angle cool but this again should look familiar because this x component when you work it out is going to be 3 meters and the Y component is going to be 4 meters this was our three four five right triangle we kind of use as our trigonometry example and we did for a reason and it turns out our second one is actually going to be a famous right triangle as well but we'll break up it up into components in the same fashion so ax is going to equal the magnitude 13 meters times the cosine of 22.62 degrees and then the Y component again the magnitude I.E the hypotenuse times the sine of 22.62 degrees we'll let our calculator do the work here for us but I know how this is going to work out I chose the special right triangle this is coming out of but we'll do 13. times the cosine of 22.62 and we're going to get 11.99999 I.E 12. as the X component and the Y components 13 times the sine of 22.62 and it's going to come out to 5.000 and something so essentially five and notice I am totally not worrying about sig figs here I don't want to get lost in the weeds worrying about it again on some serious problems we'll definitely worry about it but for now let's not get lost on some crazy math or on the sig figs let's get the concept down okay so if we see what's really going on here with our X and Y components and we kind of superimpose it right back over here so for the first Vector the 5 we had an X component right here that was equal to 3 meters and a y component right here that was equal to 4 meters okay and then for the other Vector in red we had an X component that was equal to 12 meters and we had a y component that was equal to 5 meters okay and so if we go back one more time here and all I want this time is actually that resultant Vector we drew in blue that's what we're looking for well if you kind of look at what's going on here we know what the X component is it's this three meters right here plus this 12 meters right here a total of 15 meters we're just adding the components because that's the entire X distance of that resultant vector and same thing in the Y we've got this four meters right here and then this five meters right here and we can see that it is a total of nine meters and we've got a new right triangle now where the X component is 15 meters total the Y component is 9 meters total and now I can use the Pythagorean theorem to get the hypotenuse and I can use the tangent identity to get that angle with respect to the x-axis but this is why when we go to add vectors we break it apart into X and Y components and we add the X components together to get the new X component of our new resultant vector and we add the Y components together to get the new y component of our resultant Vector that's why that works now let's go ahead and do this and so in this case again X component here is 15 meters y component is going to be 9 meters and so let's find what that resultant is and so in this case Pythagorean theorem says 15 squared plus 9 squared is equal to 306 and that's our a squared plus b squared equals c squared let's write this out explicitly so 15 squared plus 9 squared equals c squared and again that comes out to 306 so then we'll take the square root to find our hypotenuse and it comes out that's not the square root let's try that again square root of 306. all right 17.49 won't round it just yet so but that's going to have a magnitude 17.49 and again in this case meters all right and then we need an angle and our angle here is again with respect to the positive x-axis and so we got a couple different ways we can go about it now that we also know the hypotenuse but in the classic way is just if you've got the components 15 and 9 just use that inverse tangent identity and be done with it and so in this case we're going to have the tangent theta equals opposite over adjacent so 9 over 15. so Theta is going to equal the inverse tangent of 9 over 15 and again we'll let our calculator do the work for us here inverse tangent 9 divided by 15. it's going to get us 30.96 degrees cool and that's that angle right here Theta 30.96 degrees and so our resultant Vector has a magnitude of 17.49 meters and a direction of 30.96 degrees above the positive x-axis that is our resultant vector and look at the power of this so a couple different ways we can look at this so if you want to know what this you know how do you determine this resultant Vector well again in this case it's a distance or it's a length I should really say a displacement we'll find out later but it's got units of length in meters and I can get from here to where I need to end up by traveling 17.49 meters in a direction that is 30.96 degrees above the horizontal that's one way to pull this off or I could travel three meters in the positive X Direction followed by 12 meters in the positive X Direction followed by four meters in the positive y direction followed by five meters in the positive y direction or I can scramble this up I could travel three meters and the positive X followed by four meters in the positive y followed by 12 meters in the positive X followed by 5 meters in the positive Y and I still would have ended up in the same place or I can travel five meters so 5 meters at an angle of 53.13 degrees above the x-axis or above the horizontal followed by traveling 13 meters at an angle of 22.62 degrees above the positive x axis above the horizontal and I'm still going to end up in the same place so all of these are equivalent and that again I just want to realize that breaking up a vector into its components is just another way of expressing that vector and again it's kind of like Cartesian coordinates versus polar coordinates so but cool this is how you add vectors if you wanted to subtract vectors so the only difference you do is instead of you know if again these are the two vectors here if I wanted to subtract this one minus this one well then when I drew this out to do it's uh the resultant on this one right here instead of drawing it out in this direction I draw it exactly 180 degrees in the opposite direction when I start doing the tip to tail stuff cool what does that mean in terms of your X and Y components well it means if I was doing this Vector minus this Vector instead of doing 3 plus 12 for the X component it would be 3 minus 12. and instead of 4 plus 5 it would be 4 minus 5. well what if I was doing the red Vector minus the black Vector well then it'd be this one minus this one and so it's kind of like just adding the negative of a vector same thing as subtracting vectors you turns out you can also multiply vectors by scalars so if I want to take this uh let's go back to the original if I want to take this Vector right here and you just want to multiply it by three okay well that just means it gets three times longer when you multiply a vector by a scalar you get a vector that's just augmented by whatever scalar you multiply it by so if I multiply by three instead of being 5 meters it would now be 15 meters but in the exact same direction what would that do to the components it would also multiply them by 3 as well and the X component would now be nine and the Y component would now be 12 and it would still be form a triangle if you will a right triangle in the end that's analogous to a three four five or similar to a three four five right triangle instead being a 9 12 15 right triangle cool this is the deal with vectors it takes some getting used to you're going to want to do some practice fortunately we're going to start doing real problems we're going to start with one dimensional problems and vectors is not going to be the hugest deal but you need to get some practice you need to get some practice breaking vectors up to components you need to get some practice then assembling them back together and then you need some practice adding vectors together we've done two here and some problems you might be adding three vectors together and the key is you're just still going to break them all up into X components and Y components add all the X's together add all the white together and still come out with one overall resultant vector now if you've liked this lesson and you found it helpful then do me a favor and like this lesson and help me Reach This broaden audience as possible happy studying