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Current time:0:00Total duration:10:16

Vector components from magnitude & direction

Video transcript

so we have two examples here where we're given the magnitude of a vector and its direction and the direction is by giving us an angle that it forms with the positive x-axis what we need to do is go from having this magnitude in this angle this direction to figuring out what the x and y components of this vector actually are so like always pause this video and see if you can if you can work through this on your own alright now let's let's work through this together and it's really just going to involve a little bit of trigonometry so we want to break this vector down into its horizontal and it's a vertical components or its X&Y components so I could draw it's I could draw its horizontal component just like this so it's going to look something like let me draw it I can do a better job than that so it's going to look something like that that's its horizontal component or its X component and then its vertical component is going to look something like this it's going to look like that that's its vertical component and notice if you add the horizontal component and the vertical component you are going to get your original vector now I've just constructed a right triangle and in this type of a right triangle I could just use actually some of my most basic trig definitions or the simplest form which is the sohcahtoa definition of the trig functions I might want to break into the unit circle definition later on but if I want to figure out the magnitude of this base right over here we see that that is adjacent to this 50 degree angle it's not the hypotenuse it's the other side that forms the angle and so what trig function deals with adjacent and hypotenuse well we could just put a little reminder here so Toa well cosine deals with adjacent over hypotenuse so we could say that if I call if I call this X the length of our X component we could say that the cosine the cosine of 50 degrees of our angle is going to be equal to the length of the adjacent side X over the length of our hypotenuse which is the magnitude of the vector over 4 and so if I want to solve for x just multiply both sides by four so I could get four times cosine of 50 degrees four times cosine of 50 degrees is equal is equal to X now what about the Y component so the Y component what is that going to be well what that that side the length of that side is opposite to the angle the 50 degree angle so what trig functional deals with opposite and hypotenuse well that is signed so we know that the sine of 50 degrees is going to be equal to Y over the length of the hypotenuse Y over 4 and so we can multiply both sides by 4 and we get we get 4 times sine of 50 degrees sine of 50 degrees is equal to Y and so if I don't have a calculator I could just write that this I could this vector I could write as if I write it in the component form its X component is 4 cosine of 50 degrees and its Y component is 4 sine of 50 degrees and you might notice something interesting here and I have cosine full of the angle the angle we form with the positive x-axis I have that for the x coordinate I have signed on the y coordinate and then I just multiply it by the magnitude of the vector can I always do that well it turns out you can and this comes out of the unit circle definition of the trig functions the unit circle definition of trig functions cosine if you have a unit circle so the unit circle looks like this a unit circle has a radius of 1 cosine is the x-coordinate of where you intersect the unit circle and sine is the y-coordinate or if you had a vector of magnitude 1 it would be cosine of that angle would be the X component for if we had a unit vector there in that direction and then sine would be the Y component well we don't have a unit vector our vector has a magnitude of 4 it's 4 times bigger than a unit vector so each of the components are going to be 4 times bigger so that's why we multiply the X the cosine of 50 by 4 to get the X component and we take the sine of 50 and we multiply it by 4 to get the y component that's going to come in handy when we think about this one over here but we could get our calculator out and approximate what these are going to be so let me get it out where's my calculator all right there we go so 50 I'm in degree mode got to make sure so 50 degrees I'm going to take the cosine and I'm going to multiply it times 4 so times 4 is equal to 2 approximately two point five seven so this is approximately equal to two point five seven is our X component and then our Y component our Y component if I take 50 degrees and if I take the sine of it and then multiply it by 4 I get approximately three point zero six three point zero six and we see that over here this X this X component looks like it's a little more than two and a half and this Y component looks like it's slightly more than three so it all worked out even though this is kind of a sloppy hand-drawn graph all right now let's tackle this one and this is interesting because this the terminal point when we draw it in standard form is in the second quadrant so what would be the x and y components here you can immediately tell because we're in the second quadrant our X component is going to be negative and our Y component is going to be is going to be positive well we could just resort exactly to what we did just there we could say well this vector this vector is going to be the magnitude its X component is going to be its magnitude times the cosine of the angle that it forms with the positive x axis cosine of 135 degrees and its Y component is going to be the magnitude times the sine of the angle that it forms with the positive x axis and we'd be done and we can evaluate each of these things so we could get let me get my calculator out again so if we take 135 degrees and if I take the cosine of that cosine of that and then multiply it by 10 times 10 I get approximately negative seven point zero seven so that's approximately negative seven point zero seven and then if I take the sign and we're going to see something very similar so 135 degrees I take the sign of it I get positive point 707 and so let me multiply that times the magnitude times 10 is equal to seven point zero seven so seven point oh seven and we see that over here this is roughly looks like it's a little more than seven looks like it's a little bit more than seven in that direction and we'd be done and you might say well once again how did this work well if I had a unit circle if I had a unit circle right over here and if I had a unit vector so its terminal point would sit on the unit circle that went in the exact same direction that still formed 135 degrees this point this point right over here it would have the coordinates cosine of 135 degrees sine of 135 degrees let me make it a little bit more visible at this point right over here cosine 135 sine of 135 would be its coordinates or for this unit vector right over here to going in that direction its X component would be cosine of 135 and it's Y component would be sine of 135 well the vector that we care about has 10 times the magnitude of a unit vector in that direction so it's X component is going to have ten times the magnitude and so is its Y component and you could even resort to your sohcahtoa definitions to figure this out we could construct a right triangle if we like we could say we could say here is we could say here is my X component that is my X component it's going to be something like that and then there and then my Y component is going to look something like I'm drawing a little bit imprecise but I could draw that let me draw it a little bit better my Y components going to look something like that right triangle and so if I want this magnitude the magnitude of this bottom side I would say well look I know I know if this is if this is 135 degrees this angle right over here supplementary to that so it's a 45 degree angle and so what trig function deals with adjacent and hypotenuse will cosign so I could say cosine of 45 degrees is equal to the length of this so let me just call the length of that I don't know let me just call it X so the length of it x over the hypotenuse over 10 multiplied both sides by 10 you get 10 times cosine of 45 degrees is equal to X and cosine of 45 degrees is square root of 2 over 2 so this is going to be 5 square roots of 2 is equal to X and you might say wait wait I thought this is supposed to be negative well the way we just solved it right now with our right triangle we just figured out the magnitude of the side and then we would just have to reason okay but we're not going we're not going 5 square roots of 2 to the right we're going 5 square roots of 2 to the left so the magnitude in the X direction or not the magnitude our X component I should say is going to be negative 5 square roots of 2 and you can do the exact same reasoning to say well 10 cosine of 4 it's 10 sine of 45 degrees is going to give us the magnitude of the of the Y component and that's actually going to be our Y component so same exact thing this is going to be 10 sine of 45 degrees which is 5 square roots of 2 and if we took our calculator out they would evaluate to approximately these things right over there