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so we have two vectors here vector a it has a magnitude of three so the length of this blue arrow is three and its direction it forms a 30 degree angle with the positive I guess you could say the positive x-axis didn't I haven't drawn that here and vector B has a magnitude of two the length of this arrows too and it forms 135 degree angle with the positive x axis and what I want to think about in this video and maybe the next one depends on on how we do with time is what is the magnitude and direction of the sum of these two vectors what is the magnitude and direction of the vector a plus B and I encourage you to pause this video and try to work this through on your own before I work through it well to think about this I'm going to decompose each of these vectors into their horizontal and their vertical components so for example vector a could be viewed as the sum of this horizontal pointing vector plus this vertical plus this vertical pointing vector another way of thinking about this horizontal vector this right over here this is going to be a scaled up version of the unit vector I and so this is going to be this is going to be something times I and this vector right over here is going to be something times the unit vector in the vertical direction the univer Terkel direction might look something might look something like might look something like that or let's see if this is three then the unit vector he actually would look something like that so that's what the J unit vector looks like and this is what the I unit vector this is what the I unit vector would look like would look like so this is a scaled up version of it so it's something times I and this is something times J and the same exact argument right over here right over here this is going to be something times the I unit vector it's horizontal component and it's vertical component is going to be something times the J unit vector so what we really need to do is figure out the magnitudes of these horizontal and vertical components and then we know how much to scale up I and how much to scale up J so let's think through this a little bit so this one might jump out at you immediately is this is a 30-60-90 triangle so this is a 30-60-90 triangle then this side right over here is going to be half the length of the hypotenuse so this is going to be this is going to be half of three so it's three over two and this one right over here let me do in that same color so this is going to be three that's not the same that's a different color this is going to be three over two and then this length right over here this is going to be square root of three times the shorter side so this is three times the square root of three over two I just took this and multiply it by square root of three and once again that just came out of 30-60-90 triangles now another way that you could tackle this is to use your what would you know about your trig functions you can say okay I know this angle right here is a 30 degree angle this is the opposite side so you could say well the opposite over three is going to be equal to the sine of 30 degrees let me write this down so Chi Toa sine of 30 degrees is going to be equal to all right up is he going to be equal to the opposite side over the length of the hypotenuse over three or we could say that the opposite side or we could say that the opposite side is equal to three times the sine of 30 degrees three times the sine of 30 degrees and if you put this in your calculator sine of 30 degrees is one-half and so you're going to get three halves here similarly for this side this side is adjacent to the 30-degree angle so you could say cosine cosine is adjacent over hypotenuse you could say that cosine of 30 degrees is equal to adjacent over the hypotenuse or multiplying both sides by three the adjacent is going to be equal to is going to be equal to three times the cosine of 30 degrees times the cosine of 30 degrees and the cosine of 30 degrees if you type in your calculator you actually going to get some type of a decimal but it's square root of three over two so three square roots of three over two and you can you can verify this if you if you like so we see here sine of 30 is indeed one-half and cosine of 30 well you get this kind of crazy decimal but notice that is the same thing as square root of three divided by two the exact same value right over here but we were able to figure that out just using what we know about 30-60-90 triangles now this triangle you might say well what's the angle what angles do we know about it well this angle right over here this angle right over here is this is supplementary to this 135 degree angle so this is a 45 degree angle this is 90 so this is 45 so this is a 45 45 90 degree triangle now you could you could say that using the trig identities and what we did right over here that this is this length is going to be the hypotenuse times the cosine of 45 degrees times the cosine of 45 degrees and you could say that this length right over here is going to be the hypotenuse times the sine of 45 degrees times the sine of 45 degrees using the exact same logic here and as you get more practice with finding the components you'll realize okay you take the hypotenuse times the cosine of the this angle you're going to get or you're going to get the adjacent side if you do the hypotenuse times the sine of this angle you're going to get the opposite side but we could use this sine of 45 degrees once again if you put in your calculator you get a crazy decimal but we can figure that out we know 45-45-90 triangles sine of 45 degrees is square root of 2 over 2 cosine of 45 degrees is also square root of 2 over 2 so I know my my writing is getting a little bit messy and actually I should have done all of this in green square root of 2 over 2 so what's the length of that green vector well it's two times square root of 2 over 2 that's going to be square root of 2 what's the length of this orange vector it's two times square root 2 over 2 so it's going to be square root of 2 so now we know the magnitudes of the component vectors the horizontal and vertical components of each of these so now we can we can write these out as some of these horizontal and vertical vectors so vector a vector a we can write as square root of three are three square roots of three over two times I that's this vector right over here we've scaled up the I unit vector by 3 square roots of 3 over 2 plus three-halves times j that's this this right over here is this vector right over your scaled up version of the J unit vector and if you add this orange vector to this green vector you get vector a similarly similarly release simply a vector B is equal to the length of the horizontal component and we got to be very careful it's length is square root of 2 but it's going in the leftward direction so we're going to put a negative on it times I if it was just square root of I is doing something like this square root of 2 times I would look like that so negative square root of 2 would point it to would point it to the left so this is negative square root of 2 times I and then we're going to have plus plus square root of 2 square root of 2 times J and so now that we've broken up them up in their components we're ready to figure out what at least broken up into its components what a plus B is so a plus B a plus B plus B is equal to its is what's going to be the sum of all of these things so let me just write that down so it's going to be that so copy and paste that Plus this Plus this let me just copy and paste it and you're going to get that but of course we can simplify this we can add the I unit vectors to each other if I have 3 times the square root of 3 over 2 eyes and then I have another negative square root of 2 eyes I can add that together so this is going to be equal to 3 times the square root of 3 over 2 minus square root of 2 times I and then I can add to this and I'm going to get plus three-halves plus square root of two times J so it looks a little bit complicated but we could type it into our calculator and get approximations of each of these two values and then we could and we essentially have a at least a broken down into its components representation of a plus B in the next video we're now going to take this and figure out the actual magnitude and direction of a plus B