welcome to calculus three video lecture number five equations of lines and planes now in the past you've found equations of lines that were in a plane and in this situation what was required in order to write the equation of a line was a point and a slope and I want you to start thinking of those two quantities more generically so a point is a location and the slope gave you the direction of the line so similarly in space a line is determined by a point and a direction but if we're working in three-dimensional space the two dimensional ratio known as slope will not be sufficient to provide direction right so instead the direction is going to be described using a vector and we call that vector the direction vector and we usually use V to represent the direction vector and we'll say it has components a B and C okay now suppose I'm given some point P naught with coordinates X naught Y naught new Z naught that's a point on my line and I have a direction vector V what that means is my line l that I'm trying to find the equation that is going to be parallel to this vector V so how would I come up with writing the equation of this line well let's think about what sort of information we have and what's true about the line okay so we're in 3-space okay so here's my x axis z axis y axis okay and I'm given some point P naught I'll put it here okay and then I have a direction vector V and I know that my line is parallel to this direction vector so let's say vector V it's gonna be a position vector maybe it's something like this here's my vector V okay now the equation of the line would contain all points so all XYZ right would be on my line out so it contains all the points P X Y Z such that P not P so the vector from P not to any other point P is parallel to Z so let's draw that out and see what that would look like so think about if you were to have some line out right parallel to this vector V it would look something like this right this is my line out and any other point P here on this line this vector from P not to P needs to be parallel to V well what do we know about parallel vectors so that means P not P is going to be a scalar multiple of Z so I can write the vector P not P as some scalar we'll use T times V so T is a scalar so now I can call this vector here let me darken it a little bit so this vector here from P not to P that vector there is gonna be T 3 right it's a scalar multiple of the vector V they're not necessarily the same length but they're parallel ok well notice here we have an interesting relationship between position vector that goes to P naught and position vector that goes to any other point on the line so I'm going to call the position vector to P naught arm naught and then I can write a vector position vector to any other point on the line I'll take it to P here and I'm going to call that vector R now notice that R naught plus T times V is equal to R can you see that I'll draw in that addition so I have R naught I add T V and the result is the vector R and this holds for any vector R that connects a point on the line to the origin so for any position vector that terminates on the line and this in particular gives us the vector form of the equation of a line and this form it's rarely used but conceptually it's how we build the idea of how to write the equation of a line in 3-space okay so there aren't more useful ways that we're going to rely on primarily the remainder of the course but this is the first one since we understand where it's coming from so suppose we have direction vector V with components ABC then for any point P XYZ on the line we obtain the following so my vector R I'm gonna write it the other way around now so I'm gonna have remember we have R is equal to R not plus TV but this basically becomes remember R was the position vector to any point on the line so I could write it as X Y Z in component form equals R not I could write as X not Y not Z naught plus T times vector V had components ABC okay and then the parametric equations of the line L through the point P naught with coordinates X naught y naught Z naught and parallel two direction vector V ABC is given as follows so you obtain the parametric equations by setting all of the corresponding components equal to each other so we would have X is equal to X naught plus ta y is equal to y naught plus T times B and then Z is equal to Z naught plus T times C and this is probably the most popular form of writing the equation of a line that we're going to work with a B and C are often called the direction numbers beau-line oh ok so let's practice find the parametric equations of the line that goes through the point P with coordinates negative 2 4 5 with direction vector V 3 negative 1 6 and then identify two other points that are on the line ok so here we go so parametric equations x is gonna equal X naught which is negative 2 plus 3 times T don't write T 3 that looks weird y is equal to Y not that's 4 minus 1 times T and then Z is equal to Z naught that's 5 plus 16 so here's our parametric equations you can generate all of the points on the line by changing the values of your parameter okay that box is just not working for me is it ok here we go so T can be any real number and so as you let T vary it'll generate all the points on the line so the problem asks us for two other points that are on the line you could pick whatever you want to plug in for T to obtain them don't plug in T equals zero because that's just gonna give you the initial point that you use so pick something other than zero so say we're gonna use T equals 1 then I'd obtain the point so I'm substituting in 140 now to get the coordinates of another point on the line so negative 2 plus 3 that's 1 4 minus 1 that's 3 5 plus 6 that's 11 and then we need one more so say we do T equals 2 then we'll get negative 2 plus 6 so for next component would be 2 and the last one would be 5 plus 12 so 17 so those are two other points ok now notice that the parametric equations of a line are not unique if instead we have chosen our initial point to be maybe 4 to 17 right which we know lies on the line we just came up with that point right now my parametric equations would look different the direction vectors the same but now I would have a equals four plus three T y equals two minus T and Z equals 17 plus 60 so notice it can be a little tricky because you can have more than one set of equations to represent a line now alignment space has another useful representation which can be found by eliminating the parameter T from the parametric equations so we know that we have x equals x naught plus BT and then y equals y naught plus BT M Z equals Z naught plus CT well if I solve for the parameter T I would get X minus X naught divided by a is equal to T similarly Y minus y naught divided by B is equal to T and then lastly Z minus Z naught divided by C is equal to T and so I know all of these quantities here since they each are equal to T must be equal to each other and that's where we come up with the symmetric equations of a line and they're as follows two you would set them all equal to each other so X minus X naught over a is equal to Y minus y naught over B which is equal to Z minus Z naught over C provided a B and C are different from zero okay now what if one of the direction numbers is zero right that's totally possible so what would you do in that case well you would only use the symmetric equations for the direction numbers that are nonzero so for example I'm just gonna assume B is zero just to give you a hypothetical you would write X minus X naught over a is equal to Z minus Z naught over C and then you would say yes Y is equal to Y naught because if the direction number for that component is zero then that means it's just gonna be equal to every single point on the line is going to have the same coordinate as the initial point that's given and this would be in the case if B is equal to zero okay good so here we go find the symmetric equations of the line that goes through the point one two three with direction vector V 5 negative two zero so notice here since I have a zero for the third component that means separately I'm gonna write the equation Z is equal to three that coordinate will not change for any of the points on the line so the symmetric equations would just be X minus one coming from here divided by five is equal to Y minus two divided by negative two then you put a little semicolon Z is equal to three all right nice now how could we determine whether two lines are parallel what would you do well it's pretty easy to look at the equation of the line whether it's written parametrically or using symmetric equations and determine what the direction vector is so if you have the equations for two different lines all you would need to do is check whether or not the direction vectors are scalar multiples of each other if they are then the direction vectors are parallel which means the lines are parallel now similarly how would you determine whether two lines are perpendicular well if two lines are perpendicular that means their Direction vectors would need to be perpendicular and how can you check if two vectors are perpendicular well the best way is by taking their dot product so the dot product of the direction vectors is zero okay now something interesting in 3-space that doesn't occur when you're just working in a plane is that two non parallel lines may or may not intersect and the definition is if you have two lines they don't intersect and they are not parallel they are called skew lines how is this possible well basically they live in different planes okay so we're going to look at an example regarding how to determine whether or not this is the case so determine whether the lines x equals 1 plus t y equals 2 minus t is the equals 3t so we'll call that line 1 and the other line x equals 2 minus s y equals 1 plus 2 s z equals 4 plus s we'll call that line to determine whether they intersect or are skew and if they intersect find the point of intersection ok so the first thing I want to ask is are they parallel well what's the direction vector for the first line can you tell you just look at the coefficients on T so the first component is gonna be a 1 the next component is gonna be a negative 1 and then the last component is gonna be a 3 so that's the direction vector for the first line what's the direction vector for the second line but the components would be negative 1 to 1 well clearly I can see these are not scalar multiples so I know the lines are not parallel because they're not scalar multiples okay now let's check if they intersect how would you go about doing that well what does it mean for two lines to intersect it means that they share a point in common it means that they would have to have an X a Y and a Z coordinate in common so if they have an x coordinate in common that means one plus tea needs to equal to minus s while at the same time two minus T would have to equal 1 plus 2 s and lastly 3t would have to equal 4 plus s I have two unknowns can I simultaneously solve this system so that XY and Z all intersect let's number these equations 1 2 3 I'm just gonna focus on equations 1 & 2 for now okay and solve that system I have more equations than unknowns so what I'm gonna do is solve just two equations at a time solve for those two variables and then substitute that solution into the remaining third equation if it holds true then I found a point of intersection if it doesn't then there's no solution to the system so I could just add equations 1 and 2 together and notice he's gonna cancel out right so if I add these two together this and this then that T is gone right so you're just gonna have 3 is equal to 3 plus s which would mean s equals zero and if s is equal to zero you could just substitute in to whatever equation here plug in a zero there for us subtract over a 1 that means T is equal to 1 ok I'm gonna check this solution now of s equaling 0 and T equals 1 into the equation that I didn't use check it into equation 3 so it's 3 times 1 is that equal to 4 plus 0 no so that means there's no solution to the system which means the lines do not intersect so they are skinned okay very nice now let's look at an example of a question that's going to come up in several different context so say we want to find the distance from a point to the line that goes through two other points I'm going to draw a picture I know we're working in three-dimensional space but I'm just gonna draw it like we're in a plane to kind of help orient things okay so say we have a line and it goes through the points B and C so I'll put B here I'll put C here and I want to find the distance that a is from that line so whenever we talk about the distance from a point to a line its implied that we want the shortest distance the limit in our sex at a 90 degree angle that hits it at an altitude okay so what are your options here we've got quite a few first thing that we absolutely have to do is we're gonna have to make a vector here from B to a okay and then I have another vector here connecting the two points B and C okay so one way that you could find that distance is by using Pythagorean theorem option one so how would you do it using the Pythagorean theorem well you're gonna have a squared plus B squared equals C squared well a squared or one of the sides squared let me draw it out for you this distance here that's going to be the scalar projection of vector B a along vector BC right do you agree so I'll write that in here this is the scalar projection of vector B a along BC so if I'm using the Pythagorean theorem I'm gonna have that scalar projection squared plus the other leg square is equal to the hypotenuse that would be the magnitude of be a squared and then how would you in practice go about solving it what you'd have to make your little vectors be a which is three four four and then BC which is two negative four six and then you can go ahead and use the definition of your scalar projection find the magnitude of B a squared and then you just solve for the distance square there's option one option two D is the absolute value of the orthogonal projection of BA along BC so you could solve it that way and remember our orthogonal projection is BA minus the vector projection of BA along BC okay and since I want it to be distance I'm gonna take absolute value last way would be using trig but we're gonna end up not needing the angle so I'm gonna set it up right I'm gonna involve the angle theta but notice it wasn't given so I'll show you how you don't actually need it so what we're gonna do is express the ratio of D with one of the other sides here that we can compute so I could write option three that sine of theta is equal to the distance divided by hypotenuse which would be magnitude of B a so that means the distance is equal to sine of theta times the magnitude of BA but remember we have that nifty property of the cross product which tells me that sine of theta is equal to the magnitude of BA crossed with BC divided by the magnitude of each of those vectors so the magnitude of B a times the magnitude of BC so what's highlighted is equivalent this is equivalent to sine-theta that's from the lesson on the cross product and then I still have times magnitude of BA and then notice I can cancel out magnitude of BA and then you're just left with the distance is equal to magnitude of the cross product ba MBC divided by magnitude of BC so I'll leave it up to you actually you should verify that you can solve it all three ways and find the distance and to verify that you did it correctly you should get five rad six over two but I'll leave it as an exercise for you to do okay that concludes our discussion of lines for now now we're going to discuss planes and a plane is also determined by a point which is a location and a direction however if you just have a single vector lying in the plane or a vector that you say is parallel to the plane that's not enough to describe the direction of the plane there's infinitely many possibilities if you just give a vector that's parallel to the direction of the plane instead if you want to uniquely describe a plane its direction you need a vector that's orthogonal to the plane and we call this vector n the normal vector okay now notice that n is going to be perpendicular to every single vector that's in the plane and this leads us to the vector form of the equation of the plane so I'll draw you a little picture here so when you're writing the equation of a plane okay all you need is some point P and then you need a vector that is normal to the plane here's n my normal vector okay now I'm gonna let let's call this peanut actually so let's let P not have coordinates X naught Y naught Z naught that's gonna be a fixed point of the plane okay and I'm gonna say P with coordinates XYZ is an arbitrary point in the plane any other points it's in the plane then the vector P not P is a vector in the plane so imagine he is any other point in the plane right say it's over here and draw it anywhere you want so P not P if I make that vector that's gonna be a vector in the plane what does that mean so it is perpendicular to the vector n the normal vector well what do I know is true about vectors that are perpendicular I know that their dot product is going to equal zero which means n dotted with and we saw on the previous notation for writing equations of minds we're gonna introduce that again here for P not P you could write it as R minus R naught equals zero okay so the vector form is rarely used but it's helpful in obtaining an alternate representation of the plane so say your normal vector has components ABC okay and you want to make your vector P not P right P not P well it would have components X minus X naught y minus y naught and Z minus Z naught right since these coordinates are X Y Z P knots coordinates or X naught Y naught Z naught and then the equation would be ABC dotted with X minus X naught y minus y naught Z minus Z naught is equal to zero is that how we actually write the equation of a plane no what we do is we take the dot product so the scalar equation of the plane is given by the following we have a times five - X dot plus B times y minus y not plus C times Z minus Z naught is equal to zero that's the scalar equation and the linear standard form is given by a X plus B y plus C Z equals D where you clean up and move all the constants to the other side now this is the form that's required for class okay if I'm not your instructor check with them what they want okay so let's look at some examples here find the equation of the plane that goes through point P with normal vector given there so you can just jump straight to the scalar equation and then clean up so you get linear standard form so what do I mean by that so I'm gonna write this as negative 5 times X minus negative 1 so X plus 1 plus 2 times I'm gonna have y plus 6 minus 2 times z minus 4 is equal to 0 and then we're gonna clean up here so we have negative 5x minus 5 plus 2y plus 12 minus 2z plus 8 is 0 and so I end up with negative 5x plus 2y minus 2z is equal to negative 15 okay good that was a pretty straightforward example now we're going to grasp the plane 3x plus 4y plus 6z equals 12 and identify its normal vector well the normal vector I can spot right off the bat by just taking those coefficients off of each of the variables there so the normal vector I know has components 3 4 6 okay and then to graph it what we're going to do is graph each of the traces so what do I mean by that what we're going to do is set a different variable equal to 0 and one at a time and graph the trace in each of the planes so here's X Y Z oxys say I'm gonna graph the XY trace well would be from when I set Z equal to zero so that would give me the line 3x plus 4y equals 12 so I'm just going to graph that in the XY plane I can see that it would have an x intercept of 4 and a y-intercept of 3 okay five six so I'm going to graph that line the XY trace 3x plus 4y equals 12 ok and then let's do another trace so I'll do the XZ trace so that comes from setting y equal to 0 so then I'd have 3x plus 6 Z equals 12 again each of the intercepts there I'd have 4 0 0 and then 0 2 0 so I'd have this line in the XZ plane and then lastly I could draw the Y Z trace and we'll do that in pink so I will have 4y plus 6 Z is equal to 12 so I'm just gonna connect these two points here okay I'll list up what each of these intercepts are so we have 4 0 0 0 3 0 and 0 0 2 okay and if I just want to shade in maybe the portion of the plane that lies within the first octant it would be this little triangular region here but remember the plane extends forever so that that triangle just represents the portion of the plane that's in the first octant okay good now let's move on to some further applications now we know that two points are needed to determine a line uniquely right and similarly three points are required to uniquely determine the plane so we're gonna find the equation of the plane determined by the points P Q and R and let's think about how we would do that so I know for sure I need a point that's in the plane okay no problem I got three of them okay I have more than what I need the other thing that I need is a normal vector normal to the plane if I have three points that lie in the plane how could I get a vector that's normal to the plane so let's see here let's draw a little picture see if it's illuminating okay so here's point P here's point Q here's point R I want a vector normal to the plane well what if I make a vector thank you what if I make another vector P R is there something special I could do with those vectors so that I get a normal vector to the plane why yes maybe you're thinking take their cross product and that's precisely what we're gonna do okay so the vector PQ remember you take Q minus P final minus initial so that would have components 1 2 8 and you know what it doesn't matter like if you make PQ or QP just be consistent in how you create the vectors that you use the same initial point so we have PQ then we have PR PR has components 2 1 negative 2 and then now I'm gonna take the normal vector I'm gonna find it by taking their cross product so PQ cross with PR are you getting good at doing this mentally so you're gonna imagine crossing this out crossing out the IJ hat K hat so you're gonna have negative 4 minus 8 so that's negative 12 then the next component would be 16 minus negative 2 so that's 18 and then lastly you would have 1 minus 4 so negative 3 now I could go ahead and use this vector that's fine but notice I could divide out a 3 and it would look even better if I divide it out a negative 3 that way the first component would be positive and that's just to make our equation more manageable more user-friendly and more visually appealing so what we're gonna do instead is let's use or negative six one four n okay and remember any scalar multiple of that cross product is still going to be normal to the plane it's just gonna be a little bit nicer a vector to work with and then you can use any point as your P not okay so let's just use the first point P here that they gave us but any point will work that's on the plane so we'll have 4 times X minus 2 minus 6 times y minus 0 plus 1 times Z plus 3 is 0 and then we want to go ahead and clean it up so I'm gonna have 4x minus 8 minus 6 y plus Z plus 3 is 0 and then move all your constants to the other side so we have 4x minus 6 y plus Z is equal to 5 okay and you can verify you'll get the same result if you went about it a different way but followed everything correctly ok good next example find the equation of the plane that goes through the point Q with coordinates negative 1 negative 3 2 and that contains the following line ok so let's think about what's going on so we have some plane here's my generic plane that I use for all examples okay we have some plane here and they're giving me a point that's on it and then also I've got some line in the plane ok well just think back what worked before when we were in a similar situation when I have three points well I took two vectors and I cross them so I could get a vector normal to the plane well I want to do something similar I know that the line is in the plane right that's given so its direction vector the direction vector of the lines me do you know what it is can you tell by looking do you remember you take the coefficients on T from the parametric equations the direction vector has components negative two for one that direction vector and the point P the point P has coordinates negative one zero two you might be like where in the world is that coming from that's coming from plugging in T equals zero into the parametric equations so notice if I plug in T equals zero X is going to be negative 1 Y is going to be 0 Z is going to be 2 so I know that point P is on the line and it's in the plane ok so both that Direction vector and that point are in the plane what to do now I'm going to connect P and Q why am i doing that because I'm gonna try to be in a scenario like I was in the previous problem where I have two vectors that are in the plane so I'm gonna connect P and Q so I'm gonna make vector PQ now what are the components for vector PQ well here's P here's Q let's make a vector the components are gonna be 0 negative 3 0 that is also in the plane so do you see now I have two vectors that are in the plane I have vector V my direction vector on the line now I don't know exactly how long it is or where it's going well I do know how long it is but the direction vector V it's somewhere here okay there's a vector in the plane PQ was also in the plane how do I get a normal vector to the plane that's right you take their cross product so n is gonna come from taking V crossed with PQ okay and let's go ahead let's do that maybe off to the side here so you have negative 2 4 1 PQ has components 0 negative 3 0 so this is gonna be imagine crossing this up that would be 0 minus negative 3 so that three next we're gonna have cross this up this out you do 0-0 so the next components zero and then lastly we're gonna have positive six minus zero so this is six okay so we could use three zero six what would be nicer though is if we use one zero two okay and then what point do we want to use I mean they gave us point Q in the plane you could also use point P we know it's in the plane but I'm just gonna use Q and Q negative 1 negative 3 2 and let's go ahead and write the equation so I'm gonna have one this one times X plus 1 plus 0 times y plus 3 plus 2 times Z minus 2 is equal to 0 ok so this whole term is gone so I'm just left with now X plus 2 Z is equal to positive 3 ok and then it asks us to sketch the plane so you know what I'll do that down below here so there's my z axis x axis y axis and if you just want like a nice quick and easy way to sketch it you can just write out well I know X plus 2 Z is 3 so let's just get the intercepts so I know if Z is 0 Y is always 0 right so we're gonna have 3 0 0 as a point on the plane as well as 0 0 3 halves okay so 1 2 3 1 2 3 and then Y is always 0 so we're gonna go 3 0 0 0 0 3 halves so I have this line in the XZ plane I'm not happy with that bear with let's see if this is better ok I have this line in the XZ plane and then why is equal to zero so it's gonna be extending out this way okay and I'm drying it you know as sort of rectangle but remember it extends forever all right good now returning to the lesson keep in mind planes are parallel if they're normal vectors are parallel and two non parallel planes intersect in a straight line you could imagine if you have two planes they're not parallel and they intersect their intersection is going to be an entire line and the angle between the planes is defined as the angle between their normal vectors so let's look at how to apply this find the equation of the line of intersection of these two planes that are given here okay so how would I find the equation of the line of intersection well think first what do you need to write the equation of a line you need a point that's on the line and then you also need the direction vector okay and think about it this way so we've got two planes and they're gonna intersect and their intersection is aligned so here's plane number one and then let's draw plane number two okay and they're gonna intersect and their intersection is a line okay now each of these planes we know what their normal vectors are right so there's a normal vector for one plane and then there's a normal vector for the other plane so let's see can we figure out a point that is in both planes well by observation I can see that the point zero zero one is in both planes okay that's just by observation if you can't easily observe then you just solve you plug in something for X you plug in something for a Y and then you saw up and find a point that's on both lines okay well no just plug in something for one variable and then solve for the other two so you find a point that's on both lines okay in this case I left I didn't have to do that so that means about point P it is on the line of intersection what else is on the line of intersection Wow before I get there I want to think about this actual line of intersection this line lies in both planes so it's perpendicular to both of their normals what are the normal vectors for each of the lines well normal vector for line number one I'm just going to look at the coefficients here on the variables so the normal vector would be 1 negative 2 1 and then normal vector number 2 would be 2 1 1 so if my direction vector for my line is perpendicular to both normals then that means I could find it by taking the cross product those two normals so the direction vector can be found by taking n1 crossed with n2 which would give me negative 3 1 5 all right so that means I have a point I have the direction vector now I can write the parametric equations of the line so the line would have x equals 0 minus 3t so just negative 3t y would equal 0 plus 1 T and Z would equal 1 plus 5t so that's our final answer all right now earlier we examined ways to find the distance from a point to a line and I gave you three different methods to do so now say I want to actually find the distance from a point to a plane if you memorize the formula in the book I will give you no credit on a quiz or an exam ok it's actually easier to think through what would be needed and use projections appropriately so we want to figure out the distance from a point to a plane so let's think about what we would do okay so here's our plane not happy with it okay here's our plane and I have some point P here I want to know what the distance is from the point to the plane okay I do have a normal vector because I have the equation of my plane so what would you do well all you would need to do is find any other point on the plane any points on the plane you're gonna connect that point on the plane to the point P and what you want to do is project that vector along the normal vector if I project that vector PQ or QP along my normal vector then I will get that distance D but I want the absolute value of that so the distance D is going to be the absolute value of the scalar projection of the vector PQ along N and when you go ahead and compute it then you would take the dot product of P Q and N and then divide by the magnitude of n okay good so let's look at an example and this would be exactly how I would want you to show it on an exam so nice little diagram and notation using the projections not some blindly copied formula out of the textbook so find the distance from the point T to the plain 4x minus 6y plus Z equal side which we found previously on the page prior okay so first things first we need to connect T to the plane and in the diagram above I just said we needed one other point that was in the plane well we have an abundance of points we have three points in the plane so have your pick let's use point P why not so I'm going to connect it using point T P so PT would have components 1 negative 2 10 now what do I want to do I want to project that vector along the normal vector to the plane what's the normal vector to the plane it has components 4 negative 6 1 ok so the distance is going to be the absolute value of that scalar projection of PT along N and so I'm going to take the dot product of those two vectors so I have 1 negative 2 10 dotted with 4 negative 6 1 divided by magnitude of n so that's going to be square root of 16 plus 36 plus 1 close up the absolute value so numerator let's see I'm going to have 4 plus 12 plus 10 so that's 26 over Raab 53 so if you do this you would get full credit on an exam all of this is necessary and sufficient work all right that concludes the lesson on lines and planes it can be a little tricky to get the hang up initially so make sure you do your best to visualize and draw as much as necessary to get a grip on what's going on stay tuned for more calc 3 coming your way