in this article we're going to focus on vectors what is the difference between a scalar quantity and a vector quantity what would you say what's the difference between these two a scalar quantity has the magnitude only but a vector quantity has magnitude and direction so for example mass is a scalar quantity but force is a vector quantity for example let's say you have a mass of 30 kilograms you can't say you have 30 kilograms directed east or 70 kilograms directed north that doesn't make sense so the 30 kilograms is the magnitude but there's no direction that's relevant to it so that's why mass is a scalar quantity you can't have direction with mass however force is the vector quantity i can push an object with a force of 200 newtons directed east or I can lift up an object with a force of 300 units directed up so therefore force has a magnitude and direction so if I direct the force of 80 newtons at an angle of 30 degrees above the x-axis the magnitude is the 80 newtons the angle the 30 degrees that's the direction so other examples of vector quantities are velocity velocity speed with direction and acceleration which is how fast velocity is changing speed is scalar and temperature is also a scalar quantity let's say if we have an initial point point a which is at one comma two and a terminal point point b which is going to be at four six and let's say vector v is the directed line segment from a to b how can you express vector v in terms of its components i and j and also how can you find the magnitude of vector v well first let's draw a graph the majority of the graph will be in the first quadrant so a is at one comma two so this is point a and b is at four six which is about that point the vector v extends from the initial point a to the terminal point b the left of that line is the magnitude of vector v now if we take the difference in the x values that is if we subtract four and one we're gonna get 3 which means to go from a to b we have to travel 3 units to the right and we need to go up 4 units those are the components of vector v so vector v can be expressed as three i plus four j i is associated with the x component of vector v so you could say v x is three the y component of vector v is 4 which is associated with the letter j now to find the magnitude of vector v you can use this equation v is the square root of v x squared plus v y squared this comes from the pythagorean theorem c is equal to the square root of a squared plus b squared so vector v is going to be the square root of 3 squared plus 4 squared and that's going to give us the length of the hypotenuse by the way when you see this inside an absolute value it represents the magnitude of v by the way 3 squared is 9 4 squared is 16. 9 plus 16 is 25 and the square root of 25 is 5. so the magnitude of vector v is five so this line is basically five units long let's try another problem so let's say that point a is located at negative four negative five and point b is located at one comma seven go ahead and express a vector v in terms of its components i and j and find the magnitude so let's call this x1 and x2 and this is y1 and y2 if you don't want to graph it you can use this equation so vector v which is the directed line segment from a to b and let's say a once again is the initial point and b is the terminal point it's going to be the difference in the x values and that's going to give us the i component or the x component and then we have the difference in the y values that's going to give us the y component which is associated with j so x two minus x one that's going to be one minus negative four and y two minus y one so that's seven minus negative five so one minus negative four is the same as one plus four that's five seven minus negative five that's 12 so we have five I plus 12 g so that's the vector v expressed in its components now we can also represent vector v this way as well you can say it's 5 comma 12. so that's another way you can express it now let's find the magnitude of vector v so it's going to be the square root of 5 squared plus 12 squared 5 squared is 25 12 squared is 144 and 25 plus 144 is 169 so this is going to be 13. so that's the magnitude of vector v now you can also get the same answer by graphing it as well so a is at negative four negative five and b is at one seven so here's b so here's negative four and this is positive one so we have to travel five units to the right and then from a y value of negative five to seven we had to travel 12 units up so that's why the vector is 5 I plus 12 j and then the magnitude or the distance between the initial point a and b is 13 units apart so the magnitude of vector v is 13. now here's a question for you let's say if we have the initial point a which is at negative one negative three and you're given the vector v which is the directed line segment from a to b and let's say it's three i minus two j what is point b if a is the initial point and b is the terminal point how c an you use vector v to find the answer well you can get the answer graphically first let's plot vector a so it's at negative 1 negative 3 and then we can use vector v to find point b so notice that we have positive 3i that means from a we need to travel three units to the right so we're at an x value of negative one if we add three to it it's going to take us to this point here so we travel 3 units to the right now notice that there's a negative 2 in front of j which means we need to go down 2 units so therefore this is point b point b is located at an x value of two and a y value of negative five so basically if you have the initial point a and you want to find the term point b all you got to do is simply add 3i or 3 to negative 1 and also add negative two to the y value negative three and then you'll get the point two negative five so that's how you could find the terminal point from the initial point let's try another example like that so let's say the initial point a is 2 3 and vector v let's say it's 5 I minus 4 j find the terminal point in b so a quick and simple way is you could use a form if you want vector v is x two minus x one and it's also y two minus y1 times j so keep this mind this is x1 y1 that's the initial point we need to find x2 and y2 so vector v is five I minus four j and that's equal to x two minus x one and x one is two and y two minus y one we're looking for y two y one is three so what we need to do is set five equal to x two minus two because they're associated with the same variable i or the same letter I and we have to set negative four equal to y two minus three so five is equal to x two minus two so basically we have to add 2 to 5 to get x2 the terminal point so x2 is 7. next let's set negative 4 equal to y2 minus 3. if we add 3 negative 4 plus 3 that will give us a y2 value of negative one so therefore point b is seven comma negative one so if you add two plus five you get seven and three plus negative 4 you get negative 1. now let's say if we're looking for the initial point a and let's say the vector v is negative 3 i plus 4j and the terminal point is 5 3. find the initial point a so keep in mind you can use the formula vector v is equal to x 2 minus x 1 i plus y 2 minus y 1 j and it's always good to define what everything is so this is x 1 and y 1 that's what we're looking for in this problem and the terminal point b is x2 y2 so the vector v is negative 3i plus 4j and that's equal to x2 which is 5 minus x1 times I plus y2 which is 3 minus y1 times shane so this time we're going to set negative 3 equal to 5 minus x1 that's associated with the i values and for the j values we're going to set 4 equal to 3 minus y1 and let's find x1 so what i'm going to do is subtract both sides by 5. so negative 8 is equal to negative x1 and then if we multiply both sides by negative 1 x1 is equal to positive 8. now let's set four equal to three minus y one and let's subtract both sides by three four minus three is one and if we multiply both sides by negative one negative one is equal to y one so therefore the initial point a is eight comma negative one now let's say if you're given an initial point in three directions let's say you have the x value the y value and the z value and you're given the terminal point as well find the vector v and express it in component form that is an I j and k and also find the magnitude of vector v so vector v is going to be the difference in the x values that's going to give us the x component and then the difference in the y values will give us a number in front of j and the difference in the z values will give us the number in front of k so this is x1 y1 z1 and this is x2 y2 and z2 so x2 minus x1 that's going to be 2 minus 3 y2 minus y1 that's 8 minus 4. and then we have z2 minus z1 that's three minus negative two so two minus three is negative one so we got negative one times i and eight minus four is four three minus negative two is five so the vector v is negative one I plus four j plus five k which we can represent as negative one comma four comma five now to find the magnitude it's simply the x component squared plus the square of the y component plus the square of the z component all within a square root so it's going to be this square of negative 1 plus the square root of 4 plus the square of 5. negative one squared is just one four squared is 16 five squared is twenty five one plus 16 is 17 and 17 plus 25 that's 42 and we can't simplify the square root of 42 so this is the answer that's the magnitude now let's say the vector v is positive four I plus six j this time find the magnitude of vector v and also find the angle the direction of vector v relative to the positive x axis so we know how to find the magnitude it's going to be the square root of 4 squared plus 6 squared 4 squared is 16 6 squared is 36 16 plus 36 is 52 now 52 is not a perfect square so we need to simplify it however 4 goes into 52 13 times so we can write it as four times thirteen