in this video we're going to talk about vectors but what is a vector really and let's distinguish it from something called a scalar quantity so what's the difference between a vector quantity and a scalar quantity well a scalar quantity only has a magnitude a vector quantity has both a magnitude but it also has direction so for instance speed is considered to be a scalar quantity however velocity is a vector quantity so let's say if i'm driving a car going 40 meters per second what i told you is my speed i only gave you a magnitude just a number but if i say i'm going 40 meters per second north now i've given you my velocity i gave you both speed and direction so the 40 meters per second represents the magnitude it tells me i mean it tells you how fast i'm going and north tells you the direction of where i'm going and so that's basically a vector it gives you it combines a magnitude with a direction another example would be force let's say if i have a box i can apply a force of 300 newtons directed east or i can apply a force of 200 newtons directed north so let's say the first example 300 newtons that's the magnitude of the force and the direction is that's the second component which makes it a vector so make sure you understand that a vector has both magnitude and direction another scalar quantity would be temperature you can't apply direction to temperature if i ask someone hey what is the temperature today well let's see it is 80 degrees east what that would make no sense you can't say you can't apply a temperature with a direction it doesn't make sense so temperature would be considered a scalar quantity you can't say 80 degrees fahrenheit north or something like that it just doesn't it doesn't apply and so that's a scalar quantity another one would be mass so let's say you want to find out how much matter is in the object so this box has a mass of let's say 50 kilograms you can't apply a direction to that you can't say it's 50 kilograms north or 80 kilograms south it just doesn't work and so mass is a scalar quantity so if you want to determine if something is scalar or a vector ask yourself can i apply direction to this whatever it is that i'm considering so like volume is volume a scalar quantity or is it a vector quantity what would you say so let's say this container can hold 50 milliliters of fluid can i apply direction to that can i say it has 50 milliliters of fluid north that wouldn't make sense so therefore volume cannot be a vector it's a scalar quantity and so hopefully that highlights the difference between what is scalar and what can be a vector another way to think of a vector is to think of a directed line segment so let's say if we have two points point a and point b a is the initial point b is the terminal point so we can draw a vector from a to b that looks like this and so you can write it this way with the arrow pointed at b so that's vector a b the left of the arrow indicates the magnitude now where the arrow is pointed tells you the direction so let's try another one so let's say this is point c and this is point d so based on this arrow which one is the initial point and which one is the terminal point so the initial point is where the arrow starts and where it it ends so to speak that is the terminal point so we can call this vector vector cd with the arrow pointed at the letter d or towards d now there's two common ways in which we could describe a vector and that is we can give its a magnitude and its angle so let's say this is the x-axis this is the y-axis and let's say this vector has a length of 5 and it's directed at a 40 degree angle let's call it vector v so the magnitude of vector v is 5. the magnitude represents the length of the vector how long it is so this vector would have a magnitude of 10 because it's longer and this vector will probably have a magnitude of 2 or 3 because it's shorter and so the length of the directed line segment tells you the magnitude so that's one way in which you can describe a vector using its magnitude and an angle another way in which you could describe a vector is using components let's say we have the vector a and we have two comma three so this tells us that the x component of the vector you can say a sub x that's two and the y component is three so let's represent it graphically so here's the y-axis here's the x-axis so starting from the origin we're going to travel two units to the right and three units up so here is the x component it's two units along the x axis and the y component is three units parallel to the y axis and that will give us this vector vector a so you can describe a vector using the x and y components of that vector now there's something that you need to be able to do and that is you need to be able to distinguish a point from a vector a point has parentheses for example this is the point 3 comma 4. a vector let's say vector b has these inequality symbols like this one let's say this is it has the components 4 and 5. so this would be a vector you wouldn't use parentheses to represent a vector so if you want to represent a vector in component form you need to use those arrows instead of a set of parenthesis keep that in mind so if we were to plot it a point would be just a point so point p is right here it's at three comma four and that's it it's not a directed line segment it's simply a dot in space now a vector on the other hand is different let's graph vector b on this graph so we're going to start from the origin and then we're going to travel 4 units to the right because that's the x component and then we're going to go up 5 units so let me use a different color so this is 4 this is 5 and vector b starts from the initial point which we just based it from the origin and it ends at the terminal point which in this case is the point four comma five but the initial point is zero comma zero and so that's a vector b it has an x component of four and a y component of five now let's work on a practice problem vector v has the initial point one comma negative two and terminal point five comma one part a find the component form of vector v so what we want to be able to do is we want to find the component form of vector v we need to determine its x component and its y component how can we do so given the initial point and the terminal point what we need to do in order to find it is take the difference between the x values of the terminal point and the initial point so v x is going to be five minus one to find v y it's simply the difference between the y coordinates of the terminal point and the initial point so think of the final minus initial we need to subtract the terminal point minus the initial point so 1 minus negative two five minus one is four and one minus negative two that's basically one plus two which is three so the component form of vector v is 4 negative 3. let's illustrate this with a graph so the initial point is that one negative two so here it is and the terminal point is five one i was going to plot six one here's five one so this is the initial point and this is the terminal point now to go from the initial point to the terminal point how many units do we need to travel along the x direction or parallel to the x-axis the initial point has an x-value of one and the terminal point has an x-value of five so we need to travel from one to five parallel to the x-axis so we have to travel four units to the right and that gives us the x-component as you can see it's four so that's a v-x now the initial point has a y value of let's see if i can fit it here negative two and a terminal point has a y value of one so to go from negative two to 1 we need to travel up 3 units and so this is the y component this is the x component now how can we determine the magnitude of vector v remember the vector is right here i'm going to highlight it in red how can we determine the length of that vector since we have a right triangle we could use the pythagorean theorem to determine the length of the hypotenuse of the right triangle now if you recall the hypotenuse is c and the legs are a and b and so we have the formula c squared is equal to a squared plus b squared so c is the square root of a squared plus b squared therefore if we wish to find the magnitude v which is the length of the vector which is the same as c we could use this formula it's the square root of v x squared plus v y squared in this example v x corresponds to a v y corresponds to b so to finish part b the magnitude of vector v is going to be the square root of four squared plus three squared and four squared is sixteen three squared is nine sixteen plus 9 is 25 and the square root of 25 is 5. so basically we have the 3 4 5 right triangle and so that's how you could find the magnitude and the component form of the vector v if you're given the initial point and the terminal point let's try this problem vector v has initial point a negative four comma one and terminal point b eight comma six and vector u has initial point c negative seven comma negative fifteen and terminal point d three comma nine determine if the two vectors are equivalent so how can we determine if two vectors are equivalent well they need to have the same magnitude and the same direction so let's begin by finding the component forms of vector v and u so the component form for vector v let's start with v x is going to be the difference between the x values of the terminal point and the initial point so it's going to be 8 minus negative 4. and v y the y component of v is going to be the y component or the y coordinate of the terminal point minus the y coordinate of the initial point and so that's going to be 6 minus 1. and so we have 8 plus 4 which is 12 and 6 minus 1 which is 5. so this is the component form of vector v now let's do the same for vector u the x component of the terminal point is three and the x component of the initial point is negative seven so we're going to subtract three and a negative 7 together now the y component of the terminal point is 9 and the y component of the initial point is negative 15. so it's nine minus negative fifteen so we're gonna have three plus seven and nine plus fifteen three plus seven is ten nine plus fifteen is twenty four so this is the component form of vector u here is a question for you what is another way in which we can describe vectors v and u based on the information that we were given another way in which we could describe it is using the initial and terminal points vector u has the initial point c and the terminal point d therefore vector u can be called vector c d and the vector v has the initial point a and the terminal point b so vector v is also vector a b which is equal to 12 comma 5. so now that we have this information let me get rid of a few things now that we have the component form of vector v and a vector u can we say that these two vectors are equivalent so looking at them the numbers are different which means that the vectors will be different but if we want to determine if two vectors are equivalent we need to determine the magnitude and the direction let's start with the magnitude first let's find the magnitudes of vector v and vector u the magnitude of vector v is going to be the square root of 12 squared plus 5 squared 12 squared is 144 5 squared is 29 and 144 plus i think i said 29 5 squared is 25. the 144 plus 25 is 169 and the square root of 169 is 13. so that's the magnitude of vector v let's do the same for vector u it's going to be the square root of 10 squared plus 24 squared now 10 squared is 124 squared that's 576 i believe so 100 plus 576 is 676 which i'm going to use a calculator at this point the square root of 676 is 26 so the magnitudes of these two vectors are different because the components are different but what about the direction do they have the same direction a quick way to determine if the two vectors have the same direction is to determine the slope of each vector so if you recall the slope is the change in y values divided by the change in the x values now for vector v the change in the y values looking at the initial point and the terminal point it's six minus one and that basically gave us v y which was five now the change in the x values for vector v it's eight minus negative four so you could say x two is eight x one is negative four and so that gave us twelve which is the x component of v so basically the slope of vector v it's simply v y over v x so i'm going to write m sub v the slope of vector v so it's v y over v x which is 5 over 12. the slope for vector u is going to be u y over u x so the y component of vector u is 24 and the x component of vector u is 10. now we can reduce 24 and 10. they're both even numbers so if we divide 24 by 2 that's twelve if we divide ten by two it's five and so these two vectors they don't have the same slope this is five over twelve this is twelve over five so they're completely different in order for two vectors to be equivalent they need to have number one the same magnitude which these two vectors do not have and at the same time they need to have the same slope which these two vectors also do not have so right from the beginning you can tell if they're equivalent just by looking at the component forms of the two vectors but if you don't have that if you have the magnitude and the direction then you can look at that so if two vectors have the same magnitude and the same slope they will be equivalent or if they have the same components they will also be equivalent to each other let's talk about adding vectors so let's say this is vector a and this is a vector b and let's say c is the sum of vectors a and b how can we add the two vectors together so what you need to do to graphically add the vectors is connect them from head to tail so if we draw vector a and then we draw vector b right after a so basically we want to take the initial point of b and connect it to the terminal point of a and to draw vector c start from the initial point of a and let me use a different color draw the vector towards the terminal point of b and this is vector c that is the sum of vectors a and b now you could start with a or you could start with b so let's say if we started with b if we drew b first and then afterward we drew a we would still get the same vector c which is going in the same direction and has the same left so the order in which you do it doesn't matter now let's say if we want to draw a vector d and we're going to define it as being b minus a how can we draw this vector well let's start with b we have positive b which is the original vector and when you see b minus a it's best to view it as b plus a negative a so we're going to add negative a to vector b so if this is a what's negative a when you multiply a vector by negative 1 the direction of the vector will change it's going to change by 180 degrees so all you need to do is reverse the vector so vector a is going to be going this way so now to draw a vector d we're going to start from the initial point of vector b to the terminal point of vector a and so that's going to be vector d so that's how you can subtract vectors b and a from each other now let's say that vector e is 2a plus b go ahead and draw this vector what's going to happen if we multiply vector a by a scalar quantity which is just a number in this case 2. so what is 2a all it is it's basically doubling the length of vector a so if this is a 2a is going to be twice the length and then we need to add vector b to it now to draw vector e we are going to draw it from the initial point of the first vector to the terminal point of the second vector and so this is vector e it's the sum of two a and b so now you know how to graphically add and subtract two vectors together and you could do it with three vectors if you want to so let's say that vector d is the sum of vectors a b and c and we're going to say this is a this is b and let's say this is c so let's start with a and then connect it to b and then connect b to c so starting with the initial point of the first vector we're going to draw a vector to the terminal point of the last one and so this is vector d that's how you can add three vectors together now let's work on some vector operations so let's say if we're given vector a and it's going to be 4 negative and let's say that vector b in its component form is negative three comma five so perform the following operations let's calculate the value of 2a plus 3b and also let's say 5a minus 4b go ahead and try those two problems feel free to pause the video so let's start with the first one 2a plus 3b so the first thing i'm going to do is replace a with what it's equal to so a is equal to 4 comma negative 2 and then i'm going to replace a b with the stuff that it's equal to which is negative three comma five and then i'm going to multiply a by two so i'm going to multiply the x component by two two times four is eight and i'm gonna multiply the y component by two two times negative two is negative four now i'm gonna do the same thing with b so three times negative three that's negative nine and three times five is fifteen now when adding vectors together you need to add the corresponding x components together and the corresponding y components together so eight plus negative nine is negative one and negative four plus fifteen is eleven so the answer for the first part that is part a is this it's negative of one comma 11. now let's do the same for part b so we have 5 a minus 4b so let's begin by replacing a with four comma negative two and let's replace b with negative three comma five so let's begin by distributing the five five times four is twenty and five times negative two is negative 10. now i'm going to distribute the negative sign so i'm going to put a plus here negative 4 times negative 3 is 12 and negative 4 times 5 is negative twenty if you don't use the negative sign it will look like this four times negative three which is negative twelve and four times five is twenty so this is what you should have if you don't change the negative sign to a positive sign but i like to put a positive sign here so i'm gonna distribute the negative to the stuff that's on the inside that's just the way i like to do it to avoid making mistakes so now let's add the corresponding x components together 20 plus 12 is 32 and negative 10 plus negative 20 is negative 30. and so this is the answer this is equal to 5 a minus 4 b that's it now let's talk about position vectors what is a position vector how would you describe it a position vector is any vector that has its initial point placed at the origin so the initial point has the coordinates 0 0. let's call this vector v and let's say the terminal point of vector v is 3 comma 4. so vector v is a position vector but notice that vector v its components is the same as the coordinates of the terminal point because if you subtract the x coordinates three minus zero will still give you three and if you subtract the y coordinates four minus zero will still give you four so when dealing with position vectors the coordinates of the terminal point will be the same as the components of the position vector due to the fact that the initial point is placed at the origin and that's basically what you need to know about position vectors so to review remember a position vector is any vector that has its initial point placed at the origin now what about unit vectors what is a unit vector think of the keyword units what is the basic unit of any number most numbers are based on a unit of one and think of the unit circle when you hear the word unit circle what do you think of the unit circle is basically a circle with a radius of 1. likewise a unit vector is a vector with a magnitude of one so some textbooks will describe a unit vector as just being u sometimes you'll see this symbol associated with a unit vector but what you need to know is that the magnitude of any unit vector is always one that's what you want to take from this now how do we go about finding unit vectors from other vectors so for instance let's say if we have some generic vector vector v and let's say it has a magnitude of 4. what do we need to do to find the unit vector of vector v now recall that every unit vector has a magnitude of one so to find the unit vector of vector v we just need to divide vector v by four because four divided by four is one and so that will give us a shorter vector that's in the same direction as v but with a magnitude of one and this is going to be called the unit vector of v so therefore to find any unit vector you need to take the vector let's say if we want to find the unit vector v take that vector and divide it by its magnitude and that's how you could find the unit vector of any generic vector now it's important to be familiar with this equation because if we multiply both sides by the magnitude of vector v we are going to get this vector v can be described as the product of the magnitude of vector v times its unit vector so what does this mean recall that in the beginning of the video we said that vectors can be described using two things a vector has both magnitude and direction so this represents the magnitude of the vector therefore the unit vector describes the direction of vector v it tells us where the vector is going and we could extend the length of that vector to any other number we can increase the left decrease the left we just need to multiply by some magnitude so all vectors can be described as the product of the magnitude of the vector times its unit vector so it's a combination of magnitude and direction now let's work on this problem find the unit vector in the same direction of v so how can we do that what's the first thing that we should do what i recommend doing is write in the formula so the unit vector u is going to be the vector v divided by the magnitude of v and v in its component form is 4 comma negative 3. and now the magnitude of v is going to be the square root of four squared plus negative three squared and we know four squared is sixteen negative three squared is positive nine 16 plus 9 is 25 and the square root of 25 is 5. so this becomes 4 negative 3 all divided by 5. and now what you could do is divide each number by five so four divided by five is just four-fifths and then we have negative three divided by five and so this is the unit vector of vector v it's four over five comma negative three over five and if you want to check it you need to show that the magnitude of this vector is one and let's do that let me make some space first so if i take the square root of 4 over 5 squared plus negative 3 over 5 squared i need to get 1 in order to prove that this is indeed a unit vector four squared is sixteen five squared is twenty-five negative three squared is nine and sixteen plus nine is twenty-five so we have twenty-five over twenty-five which is 1 and the square root of 1 is 1. so we can confirm that this is indeed is a unit vector so now that you know how to find a unit vector i think it's a good time to talk about standard unit vectors so what exactly is a standard unit vector there's three of them that you need to be familiar with perhaps you've seen them before i j and k each of these are unit vectors with a magnitude of one so i has an x value of 1 but a y value and a z value of 0 when dealing with 3 dimensional systems j is a unit vector with a y value of one and k is a unit vector with a length of one along the z axis so make sure you associate i with the x coordinate system j with y and k with z and remember that each of these vectors have a length of one so for a two-dimensional system where this is the x-axis and this is the y-axis i is going to have a length of 1 along the x-axis so that's i and j is a unit vector with a length of 1 along the y-axis and z is the same thing but along the z axis so how do we apply this well let's say that we have vector v and in component form it's four comma negative seven how can we represent this vector using the standard unit vectors so we can also say that vector v it's four times the unit vector i and then plus negative seven times the unit vector j or we could simply say it's four i minus seven j so this is the x component of vector v and this is the y component so the x component v x is basically 4i and the y component of vector v is negative 7 j and so you could represent any vector in its component form or using standard unit vectors so let's try another one let's say we have vector w and it's 3 comma negative 5 comma 4. go ahead and represent this vector using the standard unit vectors so vector w can be written as 3 i minus 5 j plus 4 k so wx is 3i wy is negative 5j and wz the z component of vector w is 4k but this is the answer though now let's say that vector a is 3i minus 5j and vector b is 2i plus 6j and let's define vector w as being 4a minus 3b go ahead and find vector w using the standard unit vectors so go ahead and perform the vector operations so what we're going to do is we're going to replace a with what it's equal to 3i minus 5j and then we're going to replace b with 2i plus 6j and so we have 4 times 3i which is 12i and 4 times negative 5j which is negative 20j and negative 3 times 2i that's negative 6i negative 3 times 6j that's negative 18j and now all we need to do is add like terms 12 minus 6 is 6 and then negative 20 minus 18 is negative 38 and as you can see when using the standard unit vectors it's very easy to perform vector operations it reduces to simple algebra now you do have to be careful not to make mistakes so i recommend doing everything one step at a time because it's easy to make a mistake or miss the negative sign but that's how you can perform vector operations with the standard unit vectors now let's talk about the unit circle and how it relates to unit vectors but let's focus on the first quadrant of the circle now the unit circle is a circle with a radius of one so the distance between the center and any point on a circle will always be one which is the same as the length of a unit vector so basically the radius of the unit circle we could say is the unit vector now what is the x component of the unit vector and what is the y component of it now this unit vector will have an angle with the x axis and if you recall from trigonometry the x component of the unit circle is cosine and the y component is sine so we can say that u x is equal to cosine theta and u y is equal to sine theta so as we represent the unit vector u as being ux comma u y in its component form we can also say that it's basically cosine theta and sine theta now earlier we said that we can represent any vector as being the product of its magnitude times the unit vector and so the unit vector is basically just cosine and sine so the vector v is going to be the magnitude of v times cosine and since that's the x component we're going to attach one of the standard unit vectors associated with the x component that's i and then plus we multiply v by sine and that's the y component so we're going to attach a j to it so we could find any vector if we know the magnitude and the angle so we can express it as the sum of the standard unit vectors i j and if we want to k as well now let's work on number five write vector v as a linear combination of the unit vectors i and j given that vector v has a magnitude of 16 and an angle of 30 degrees with the positive x-axis so let's say this is the y-axis here is the x-axis and here's the vector v with a magnitude of 16. and it makes a 30 degree angle with the positive x-axis let me make this longer so what we're looking for are the x and y components we need to determine v x and v y but let's use the formula that we wrote earlier so vector v is going to be the magnitude times cosine that's going to give us the x component and to get the y component it's going to be the magnitude times sine so make sure you understand this v x it's going to be v times cosine theta and v y is v times sine theta so we have v that's 16 we're going to multiply it by cosine 30. and to get the y component we're going to multiply sine 30 by 16. and this should be a j what is cosine 30 cosine 30 is the square root of 3 over 2. let's not forget to put the unit vector i and sine 30 is one half so now 16 divided by 2 is 8 so i'm going to have 8 square root 3 times i and half of 16 is 8 as well so that's going to be plus h8 this is the answer so that's how you can express vector v in its component form using the unit vectors i and j if you're given the magnitude and the angle now let's work on another problem so let's say that vector v is 3i minus 8j find the magnitude of the vector shown below and determine the angle that it makes with the positive x-axis so this problem is the reverse of number five we're given a vector and we need to determine the magnitude and the angle well we know how to find the magnitude it's simply going to be the square root of 3 squared plus negative 8 squared 3 squared is 9 negative eight squared is sixty-four nine plus sixty-four that's going to be seventy-three so the magnitude is seventy-three now how do we find the angle well if we were to plot this vector as a position vector starting from the origin we would have to travel three units to the right and then we would travel down 8 units now this graph is not drawn to scale but vector v would be going in quadrant 4. so how do we find the angle that the vector makes with the positive x-axis and also let's find this angle going counterclockwise from the x-axis but before we find that angle let's find what is known as the reference angle that is the angle inside the triangle the reference angle will always be a positive angle between 0 and 90. and what we can use is the tangent function if you recall from sohcahtoa intrigue tangent theta is equal to the opposite side divided by the hypotenuse in this case opposite to theta is v y and adjacent to it is v x so tan theta is v y over v x now to find a reference angle it's going to be the arc tangent of the y component divided by the x component now when looking for the reference angle even though v y is negative i recommend using the positive value or the absolute value of v y it's going to make your life a lot easier so we're going to take the arc tangent of 8 and divide it by three and make sure your calculator is in radian i mean at radian but degree mode mines was in radian mode just now so you should get this angle now this is the reference angle it's 69.4 degrees now sometimes your teacher simply wants this angle between the vector and the x-axis and that would be the answer however if you want the angle measured from the x-axis but in a counter-clockwise direction as opposed to a clockwise direction if you want to go counterclockwise we need to do some extra work here but all we need to do if the angle is in quadrant four is subtracted by 360. or more specifically subtract the angle from 360. so it's going to be 360 minus 69.4 and so the angle measured from the positive x-axis going in a clockwise i mean a counterclockwise direction it's 290.6 this is the angle that we want that's the standard angle and so we have a magnitude of 73 and an angle of 290.6 so if your vector is in quadrant one the angle that you want is the same as the reference angle in quadrant two the angle that you're looking for it's going to be 180 minus the reference angle in quadrant three it's going to be the reference angle minus 180 and in quadrant four the angle measured from the positive x-axis going in a counter-clockwise direction will be 360 minus the reference angle now this is going to be the last problem of the video go ahead and work on it find the resultant force of two vectors take a minute pause the video and see if you can get the answer now for those of you who wish to subscribe to this channel that is of course if you like this video don't forget to click the notification bell if you wish to receive any updates on new videos that i'm going to be posting in the future now let's begin let's graph the vectors or at least a rough sketch of it so the first force vector f1 it's going to be going in this general direction it has an x component of five and a y component of two now f2 will be going it's towards quadrant three we need to travel two units to the left and down eight units so we could say that for the most part it's going in this general direction now the resultant force is the sum of these two force vectors in which quadrant do you think the resultant force which we'll call fr is located in so if we take f1 and graphically add it to f2 where is fr located so fr is going to be going in this direction so if we move that vector to the origin of this graph we can say that approximately it's going in quadrant four that means that the x component should be positive and the y component should be negative so what we're going to do is we're going to find the magnitude of the resulting force vector and also its angle so let's go ahead and begin the resultant force vector is simply the sum of f1 and f2 and f1 is 5i plus 2j f2 is negative 2i minus aj so what we need to do is add the x components together so 5i plus negative 2i that's 3i and then add the y components together 2j plus negative 8j is negative 6j so that's the results in force vector in terms of the standard unit vectors i and j now once we have that our next step is to find the magnitude of the resultant force vector and so it's going to be the square root of 3 squared plus negative 6 squared 3 squared is 9 6 squared is 36 9 plus 36 is 45. now if you want to you can simplify the square root of 45. 45 is 5 times 9 and the square root of 9 is 3 so you get 3 square root 5 which i'm going to write here so that's the magnitude of the resultant force vector in this problem now let's go ahead and find the angle that it makes with the positive x-axis going in the counterclockwise direction but let's graph it first so to graph this vector we need to travel three units to the right and then down six units so as we can see this force vector is indeed in quadrant four so let's find a reference angle first so the reference angle is going to be the arc tangent of the absolute value of v y over v x so v y it's absolute value it's positive six we're going to get rid of the negative sign v x is three so we need to take the arc tangent of two and so it will give us this angle which is 63.4 degrees so that's the reference angle inside here let me erase this so this is 63.4 degrees now we need to find this angle the angle that is counterclockwise measured from the positive x-axis now because our resultant force vector is in quadrant four the angle is going to be 360 minus the reference angle and the reference angle is always an acute angle between 0 and 90. so our reference angle in this example is 63.4 so if we take 360 and subtract it by 63.4 we get this angle which is 296.6 degrees and that is the answer so now you know how to add two vectors to get the resulting force vector and you could describe it using unit vectors or using the magnitude and the angle so that's it for this video thanks for watching if you like it feel free to comment like and definitely subscribe to this channel