Direction of vectors from components: 1st & 2nd quadrants

so we're going to do in this video is look at a series of vectors and we're going to draw them in standard form where their initial points or their tails are going to sit at the origin and we're going to see if we can figure out the angles that they form the positive angles that they form with the positive x-axis and like always pause this video and see if you could figure out what these Thetas are going to be on your own and we're going to figure them out in degrees so in this first one I have the vector U if we were to write it in unit vector notation sometimes it's called engineering notation we would say it's three times the unit vector in the horizontal direction the eye unit vector plus four times the unit vector in the vertical direction or you can view this as the X component is three and the y component is four and you see that here if you start at the origin we're going to move three in the horizontal direction and we're going to move four in the vertical direction now to figure out this theta there's a couple of ways to think about it we could just construct a right triangle for this one in particular we could say okay the x coordinate is three so if we were to create a right triangle here this side would have a length of three and then we have a height of four the y coordinate is 4 so this side has a length of four and we know just from even our basic sohcahtoa definitions of trig functions that what trig function involves the opposite of an angle so the opposite of the angle and the adjacent of an angle well tangent does sohcahtoa so we know we know that the tangent of theta is going to be equal to the length of the opposite side which is 4 over the length of the adjacent side over 3 and so if we wanted to solve a for theta we could just say that theta is equal to the inverse tangent sometimes people say arctangent of 4/3 of 4 over 4 over 3 and let's evaluate this and I'm going to get my calculator out to do it so I want to take 4 divided by 3 which is that and I already had pressed this button right over here that takes my tangent and makes it into an inverse tangent so I'm going to take the inverse tangent of this and I get I get roughly fifty three point one degrees so this is approximately fifty three point one degrees which looks about right this looks like a little bit more even though I didn't draw it completely super precisely I had drew this it looks a little bit more than a forty-five degree angle so that that feels that feels pretty good now another thing you might have said hey well look for is the Y component three is the X component and so maybe tangent of theta is always going to be the Y component over the X component and that in fact is the case and that actually comes straight out of the unit circle definition of the trig functions where you could say that if you have a unit circle if you have I could draw it right over here so let me draw the coordinates so and if I were to draw a unit circle right over here and if I were to have some line here we're thinking about vectors that the angle formed with the positive x-axis the tangent of that angle tangent of theta is going to be the y coordinate where we intersect the circle over the x coordinate and so you can imagine if you made a unit circle right here if you made a unit circle right over here the ratio between the y coordinate of the x coordinate of this point right over here where we intersect the circle is going to be the same thing as the ratio between four and three so this is going to be this also is going to be one in one-third or or 4/3 so either way you can think of when you think about vectors the tangent of the angle that it forms with the positive x-axis is going to be is going to be the Y component over the X component so it's nice when everything kind of fits together like that so let's leverage that to figure out this to figure out this angle right over there well we could say that tangent of theta is going to be equal to the Y component the Y component over the X component over just in a new color or C over negative five so it's going to be negative negative six fifths or we could say theta so let's be a little bit careful so we could say we say that theta is equal to is equal to the 10 the inverse tangent let me do that same color inverse inverse tangent of I'll just write it like this negative 6/5 but I'm gonna put a little question mark here to see if we feel good about the answer where you get when we do this so let's do that so if we do if we do 6 divided by 5 which is equal to that and then we want to make this negative so that's negative 6/5 and we're gonna take the inverse tangent so I already press this button so this is going to be an inverse tangent not tangent I get negative roughly negative 50 point 2 degrees so this is approximately negative 50 point 2 degrees well does that look right well now theta looks like it's over 90 degrees negative 50 point 2 degrees negative 50 point 2 degrees is actually giving us is actually giving us this angle right over here so that's giving us that's specifying another vector or you think about a line another line that would have the same tangent the same tan so it's really important to visualize this and think about this and the reason why it is is because the arc tangent or the inverse tangent functions in calculators they will give you an angle that is between negative 90 degrees and positive 90 degrees so something that's in the fourth or first quadrant well here we have something that's in the second quadrant so we have to make sure that we're thinking about it right so that we can make the appropriate adjustment so that would give that would give the case if we were looking at a line like that or a vector like that so in order to figure out what in order to figure out the actual angle what we want to do is add 180 degrees to it so to get the vector that goes in the other direction so you want to add so theta is going to be equal to or I could say approximately equal to negative 50 point 2 degrees plus 180 degrees so let's do that so so let me plus 180 is equal to approximately 120 nine point eight degrees so theta is approximately equal to 120 nine point eight degrees that I would get that right have a very short memory oh yeah that's right 120 nine point eight degrees and that looks much closer looks much better that looks clearly as an angle we're going to go that takes us to 90 degrees and they're going going above 90 degrees now another way you could have thought about this is what you know from from sohcahtoa and just right triangles we could construct a right triangle where let me construct a right triangle using some colors well we know so I can maybe just draw it like this so this is the height of it and what is that height well we sight is going to be six and then the base the base right over here what is that length going to be well we know that we're going from the origin we're going five back but if we think about just the absolute value the length of that line that is going to be five and so we could figure out just using right triangles this angle maybe we call that X and so we could say the tangent of X tangent of X is equal to six over five opposite over adjacent six over five or that X or you could say X is equal to the inverse inverse tangent of six fifths and what is that going to be so if we take six divided by five one point two and then we're going to take the inverse tangent of that and we get approximately fifty fifty point two degrees so X is approximately fifty point two degrees which looks right but remember we're not trying to figure out X we're trying to figure out we are trying to figure out what theta what theta is and you can see that X and theta are supplementary so theta plus X is going to be 180 or theta is going to be 180 minus that so 180 minus that so let's just put a negative sign in front of that and add 180 and that might look familiar so that gets us to 129 roughly 120 nine point eight degrees is what exactly what we got before