uh it's time for another math easy solution uh today we're gonna go over further into the vectors and the geometry of space video series and this time we're going to look at equations of lines and planes all right uh so let's continue further and note as always if you don't have time to watch this whole video uh in this case you can play this video at a faster speed yeah there's no summary conclusions in this video and also also top secret life hack as always your brand gets used to faster speed and here's a tutorial I made on it right here and in this tutorial I used this extension right here mesm slash video Speed Dash extension anyways uh going further so you can also download and read the notes yeah if you don't have time to watch this whole video you can also read the notes on the hive blockchain and uh you can also you know watch the video in parts and all the timestamps of all the parts are in the description as well as a link to all these as well these notes and also new thing I've been doing is adding video sections playlists and in this case yeah we'll have it available and I'll upload uh basically this is gonna be a long video so I'm going to break it down into many sections and then upload that as a playlist and you can watch specific Parts uh very interesting stuff and uh and now let's continue for this with Calculus book reference and note that I mainly follow along the following calculus book this is calculus early transcendental 7th edition by James Stewart and note in the earlier videos I used the sixth edition all right let's continue further so sections and calculus book chapter so I've made a list of the sections in this particular chapter with links to The Hive post of the videos that I've already finished uh note that I've started splitting each long video into sections and made a playlist for age and here's a link to it again the notes for this in PDF and Hive article format in the link description below so this vectors you added the playlist here vectors in the geometry of space and then I added yeah so that I made a video on 3D coordinate systems and I have a link to the hive post and then we did vectors and so on and then these playlists these are indicates the section playlist and that's a YouTube playlist right there and uh continue for this another one we're doing currently is equations of lines and planes current video and I also made a playlist on it no videos aren't there yet but I'll upload the videos as I uh share this video and then split the split this up and share these in the future yeah which usually means uh pretty much a a a upload every three days after this one anyways continue for this so cylinders yeah actually that's the next videos after this and now let's continue for this so topics to cover and now we're falling into the video note that the timestamps will be included in the video description for each topic listed below so in this video we'll go over lines Vector then we'll go go over the vector equation of a line then parametric equations of a line then go for example one then solution to one a and solution to one B then Direction numbers symmetric equations of a line example two and then look at solutions to 2A Solutions 2B note on example two then we'll look at an equation of a line segment then example three and then look at a line yeah and look at planes not lines so now we're going to look at planes and this second part since we're going to look at the vector equation of a plane the scalar equation of a plane and then example four and then linear then look at the linear equation of a plane and example five example six parallel planes right there and then we're gonna go for example seven and solutions to them uh Parts A and B of seven and then look at node one two linear equations can represent the line and no two line of intersection from two parametric uh planes and then example a distance from a point to applying the example nine example 10 and an exercise one is fascinating stuff all right so now let's uh Jump Right In and look at lines so a line in the X Y plane is determined when a point on the line and the direction of the line its slope or angle of inclusion are given so all you need yeah only is a point and then the angle then you have a line uh the equation of a line can be written using the point slope form as we're used to just um yeah here actually let's just draw this as a point-slope form is basically if you have a line of s y minus y1 equals to M the slope x minus x one like that and then if you were you had to graph this out like this uh this x this is the Y so then let's say you have a line and the point on it is going to be the coordinates X1 and y1 and the slope right here is going to be our m like that all right so that's in two D's or and again two dimensions are the X Y plane so now likewise a line in three-dimensional space it is determined when we know a point p uh P naught x x naught y naught Z naught uh on L and the direction of L so yeah if we have a line L in that so all we need to know is a point in the direction of it that brings us to the vector equation of a plane so in three dimensions the direction of a line is conveniently described by a vector so we let uh V be a vector parallel to L yeah so then the vector yeah it could represent the direction of it and then we just need a point on so let p x y z be an arbitrary point on L and let R zero Vector R not vector and R vector be the position vectors of P or P naught and P that is they have representations uh this is from the origin this is uh o p naught and O P that Vector so that is uh yeah these are from the order so if Vector a is a vector with representation P naught p as in the as in the below figure or as in the figure below like writing it this way uh then the triangle law for vector addition gives R uh Vector equals R naught plus a but since a and v I'll draw this a lot a and v are parallel vectors there's a scalar t such that a such that such um yeah such that a equals t v so let's make sense of all of this yeah so let's draw this out yeah so let's say we had a point of p x y z to be an arbitrary Point online and again we're doing in 3D right now so let's say we had like this all right yeah so this is the Z axis oh this is a bit neater and this is the y-axis and this is the x axis like this so let me get a bit more 3D looking okay so that's the x-axis so let's say we get a line L so this is our L and let's say there's a point on this and this is our P so this is a point p on it and this is we'll call this p x y z all right so now we have x y z and arbitrary point on L and that R naught and R be the position vectors of P naught and P yes and uh yeah P naught is uh so this is R naught and R uh and we know that P naught is again just a line on L as well so this is arbitrary point on L and there's the other one our starting point let's say we had a point right here and we'll call this over here P naught and this is going to be X not y naught Z now like that and then the direction vectors R and this is the origin so let's draw the origin oh so the representations are o p naught that's going to be our R naught Vector in other words it goes from uh this origin all the way to here this is going to be R naught right just like that so that's R naught and then this one here is going to be our R so this is going to be our R Vector like that and this is from o to p and this is from o to P naught that's uh these these right here now if a is the vector with representation P naught P yes in other words it goes from P naught to P so that's going to be like this where's that yeah but it's a bit hard to see also I'll draw this in red but this is try to see a bit better this right here is going to be our a all right so this one here I'll call this a vector and that is from uh with the vector a little representation P naught to p and then the triangle law for vector addition gives R equals to uh this is R plus uh a yeah so R equals R naught plus a that's again is the triangle also R naught plus a equals this R Vector I'll just write this down here r no a r Vector equals to R naught plus a like this and then we're also told here so let V be a vector parallel to L so in other words a v is a vector parallel to it so let's draw this like just starting anywhere here so it's parallel to this parallel like that and that's called as this is V so that means if it's parallel to it that means well it's going to be parallel it's going to be it just a scalar multiple of this a right here yeah here's fix that up I just made it like that so it's a scalar multiple so in other words the scalar T is such uh yeah times it by a uh I mean yeah it's a scalar t such that a this Vector is equal to a scalar multiplied by this V it could be double it could be half the length uh whatever it is as long as it's well parallel and it's it's the yeah just a scale of multiples hence it's parallel parallel so that means then uh so we have this and a equals two of a t scalar times V so then we could put this all together R naught equals two I mean r uh Vector equals to R naught Vector Plus T times a uh times a scalar t t scalar T Times by a parallel vector V of the two one all right uh continuing further so this above equation so the above equation is called the equation is called a vector equation of out so each value of the parameter T gives the position Vector R of a point on l in other words as T varies as you change it the line is traced out by the tip of the vector R yeah so then as this Vector here this this Vector goes along this LL line based on whatever the parameter key is and in other words as T varies yeah the line is traced out by the tip of the vector R so as the figure below indicates a very positive values of T correspondent points on L that lie on one side of P naught uh where whereas negative values of T correspond to points that lie on the other side of P naught again depending so for example here let's graph this out all right so all right so we have right here so move it over uh like this this is the z-axis and move it up and this is the Y this is the X like that so let's say we had our line L and and then we had you know there's their origin and then there's our Point uh right here let's say this is the um at T is equal to zero yes if you go back to our vector equation so T equals zero or not is this equal to r i mean r is just equal to R not that just so let's point this out this is at our our not like that and is that t is equal to zero and then if you if T was let's say greater than uh yeah greater than zero uh you're gonna have vectors like this this would be the r this would be the r Etc and this is again that's our L this is four uh this is for T is greater than zero likewise you'll have it going back like this if T is a scalar less than zero so this is going to be like this Drop Like This this is for T is less than zero yeah so yeah these values are all the vector equation R just put the other way down and then yes becomes R just if fluctuates based on the value of T and if T is equal to zero just becomes R naught to the starting point all right yeah let's go uh further and look at now parametric equations of a line so if the vector v that gives the direction of the line L is written in the components in component form as V equals uh this uh a yeah the triangle bracket for Vector components a b c then we have t t v equals to uh t a t b t c and again remember the the vector v is this direction right here parallel to the line all right uh we can also write R equals to x y z and then R naught equals to the component is X x naught y naught Z naught so the earlier vector equation becomes so this vector equation right here and write these all in component form uh it's going to be R naught equals to let's write down for minus right the whole thing t v ector like that and then all in component form this is the same as writing X y z is equal to 2 fix up a z the x y z equals two this is X naught versus the starting point y naught and then Z naught like that and then plus uh T let's write this for completeness and plus and then this is the T times V where V is ABC and then T at T times V is going to be t a t b etcetera DC so t a t b t c and those are the components like that and since they are components we can just add these all up so this is going to be and I'll write this out so we can box this whole equation x y z like this what these commas better equals 2 and then add these absolute vector addition I just add X naught plus ta Etc X naught plus t a plus t a and then this uh nothing let me put a comma yeah the next one's will be y naught plus t b comma then the Z component Z naught plus T C like that box this whole thing in and uh actually I'm just going to rearrange this because my cockles book rearranges it writes a T and B T Etc again let's just copying my chemicals book and yeah just flip it around like that it doesn't change anything but just the notation just how it looks like all right so that's written in vectors uh but now uh two vectors they are equal if and only if they're corresponding components are equal so therefore these two vectors are equal each component is equal to it so this x is going to equal to this this Y is equal to this Etc and the Z is equal to this right here equals to this so therefore we have the three scalar equations as opposed to writing in this Vector component form you can write these as scalar so there's no vectors involved yeah so x equals two well this is going to be this part so X naught plus a t and then the next one is y equals to Y naught plus b t and then the next one is Zed uh equals 2 Z naught plus c t like that and let's just make this in the center moves over so yeah these are three scalar equations and the T value is defined for all real numbers so T is an element of the set of all real numbers could be anything positive negative and so on as long as the real number not a complex and we've defined it for in this particular case and now these equations are called parametric equations of the line L through the point P naught X naught uh y naught Z naught n parallel to the vector v with components a b and c again the parametric equations because they all have this parameter T there and each value of the parameter T gives a point x y z or XYZ on the line L is very very interesting all right and uh now let's continue for the take a look an example on this so example one and this States find a vector equation and parametric equations for the line the pass through the point p135 and is parallel to the vector I plus 4J minus 2 K and this is the written in standard basis Vector form and then Part B says find two other points on the line so uh solution support one a we've got to find the vector in parametric equations so let's grab graph out what we are given first hand yeah just to uh just to grasp it all right so what we have is let's just draw this Z it's like this then we have the Y like this and let's try the Z the x is going to be something like this that's our X and now the points that we are asked to graph is five one three so we have five let's go one two three four five is the first point then one it goes down like this one and then it goes up three one two three so it should be somewhere around like here ish let's say this is the point uh this is a point right here five one three that's the point and it's going to have all these position vector are not like that that's the aims over there and it's parallel to the vector I this is going to be I four uh and then negative two yes in other words uh the V is going to be well one so if we start from here it's gonna be parallel to it so let's go one uh X y's and then we're gonna go four on the y direction I'm gonna go one two three four like that so it should be somewhere above above here and then it's going to go minus two it's going to go in minus two direction so let's say this goes like this let's go one two three four then minus two one two so it's going to be going down this Vector like that let's call this as our V vector no here is the erase that stuff there's those vectors like that to be pointing down and this is going to be at our I plus 4 J minus uh 2K and that's parallel to it so then let's just it's going to be right on it so we're drawing it right on it like that and there's that line L like that it's going downwards like that all right so that's what we have is well we got uh r r naught is equal to and write this in Center basis forms to make it all the same well I'll write this a first like this and component form is going to be five one three and then this equals 2 5 I plus uh J plus three k like that all right we have that and the next one is we'll just write our our regular parallel vector v this equals two oh look for a completeness sake is right in both forms write this as one or negative two components like that this just equals two uh I plus 4 J minus two k like that all right so the vector equation becomes R naught I mean I just R uh this is going to be equal to well we got to write write the vector equation uh R naught yeah r a vector plus r naught I mean equals to R naught plus t v like that and this just equals 2 and the accountable just write them in this this format so I'll write everything like that so then R naught is going to be 5 I plus uh J plus three okay and again it's written those components because that's what we were given in here in this Vector form so let's just continue along that and then there's going to be plus T and now the the parallel Vector is going to be this times it by I plus 4 J minus 2 K like that and this is a t not a plus and then combine these all together so multiply this inside and then add them all the terms up so we're gonna have a five plus this is going to be t i is so 5 plus I'll just write this as if I plus t this is going to be T times I and then bracket this is the I standard basis vector and then the next one's the J that's going to be one plus four t and this is going to be I I mean J J like that all right and then this is going to be plus the next one's going to be 3 and minus 2. times T So 3 2 t like this and this is going to be k like this all right so now we have this with the K there and yeah so either of these will work because we're asked to find a vector equation and parametric equation so this one is uh is solved so I'll just put this here as uh just for completeness R Vector is equal to this so that's the vector equation and just box it out simplified either or you can just highlight this one too as well but just simplify to that all right so now that we have this vector equation and now we can just get the parameter parametric equations easily just uh by looking at each of these components because uh remember if we scroll back up if we have it in this form so if we write the parametric equation like this and then write it out in like this component form then we can equate the components x equals to this component and Y and Z Etc so parametric equation is pretty straightforward so X is going to equal to 5 plus t x equals to five plus t y equals two one plus four T and then um here uh next one's going to be Z equals to three minus t now so those are the parametric equations and that's pretty straightforward and then now as I double check uh we could uh yeah double check the graph using the Gog broth 3D graphing calculator so here you can write this in a vector form as R naught five one three like this here's inputs I did and then you could put the the vector right here this is for the parallel Vector you have to you have to equate it or actually uh yeah I just I think it uses a double check I just forgot so you could write Vector here this is 0.1513 to the point five uh then is added T is equal to one there so now so you go five plus one that's five yeah t is one one plus four times one is yeah like this and then it just it correctly solves it as one four negative two exactly as this anyways and there's the line uh right here this is from the point here five one three and then with the parallel vector v and it writes it out like this this parameter of Lambda is just t this is just equals to T in the calculator and it graphs it out like this pretty cool you can even graph out the vectors as well there's a vector in green here one four negative two and there's the red five one three and it goes there and it fits perfectly absolutely amazing and if you click this let's see what happens if I click this and it's opening up yeah so you can play around with it and see what you can find is a pretty epic calculator and let's just see how it is it's 3D graphing calculator it's loading yeah let's see how long it takes to load all right got pretty epic guys yes the mods you can see it all the stuff this is absolutely amazing and you can play around with it you could turn this Vector off on this line off and on anyways uh have fun with that all right let's continue for this now let's look take a look at the solution to one B so choosing the parameter value T gives well again let's see if the we're asked we're asked find two other points on the line so basically we just add the parameter change of value of T and find points on it all right so let's say we have t equals to one and let's go back to our uh our parametric equations to five plus t one plus four T three minus two t so we got 5 plus 3 so x equals to Pi yeah this is a 5 plus t and the other ones can be um yeah that's five plus T so in other words when you plug in T is equal to one that goes like that it's equals to six uh this is the double check again uh yeah this is gonna be yeah five plus t one plus four t then three minus two t so it's going to be Y is equal to one plus four times by one that's going to be five like that and then the last one Z is equal to yeah three minus two t yeah let's see three minus two t three minus two t and t is one this is equals two three times three minus two is gonna be one like that yes uh so thus this point is just six five one so the six five one is a point on the line and similarly choosing T equals negative one gives the point four negative three and five and we could see that here so x equals to five minus one that's equals to four that's correct the next one y is equal to one plus four negative one that's going to be one minus four that equals to negative three that's correct the next one is said equals to three minus uh two uh negative one that's going to be minus plus becomes a plus two so that's equals to five like that it's amazing at 4 negative three and five all right there continue further now let's look at a direction numbers and this involves a note on these examples that we just covered so the vector equation and parametric equations of a line are not unique if we change a point or the parameter yeah yeah the parameter t or choose a different parallel Vector then the equations change for instance If instead of five one three in this example there we choose the point six five one otherwise choose this point here this is also on the curve uh in example one then the parametric equations of the line become instead of this five one three it becomes x equals to six plus T remember that other one's going to be the five I mean not the 5 is going to be this is y is equal to one let's go back here so one um yeah one plus four T and this one's three minus two t but then there's not gonna be one anymore it's gonna be five that's gonna be five now because we're switching up this one with this five so it's going to be five zero y equals five uh and this was going to be minus that's minuses plus four t four T like and then Z is equal to uh this is going to be now one instead of three one minus two t and so or instead of uh switching up the starting position uh or if we stay with the 0.513 yeah so we stay with this one but choose the parallel Vector 2 I Plus 8 J minus four K and again this is just a scalar multiple of this we just multiply everything by two here by two four two times two is four and two times four is eight then and then right here negative two times two is uh is negative 4. if we do if we do all of that we arrive at the equation so x equals to starting point is 5 plus and set a t we're going to multiply by 2 is going to be two t the next one's be Y is going to be uh five one and this is going to be plus four instead of 4 is going to be eight J I mean not eight T we'll just switch up the parameters the parametric equations and lastly Z is going to be 1 minus four T like that let's see whoops that's the one over there that says that's this one we're actually this is three three minus 4T so in other words these all represent the exact same line because this is another parallel vector and there's another point on that line already so yeah we just we get the same line just different equations for it but same thing all right so uh in general if a vector v equals to uh components a b c is used to describe the direction of a line L then the numbers a B and C are called Direction numbers of L since any Vector parallel to V could also be used we see that any three numbers proportional to a b c could also be used as a set of Direction numbers for L yes I could switch them up to whatever they are as long as they're proportional or in other words scalar multiples of each other and you can use this as the direction numbers and then with any point at any starting point on in this equation that's on the line all right now let's continue further so symmetric equations of a line so another way of describing a line is to eliminate the parameter T from the parametric equations yeah so if none of a b or c is zero we can solve each of these equations 4T uh and equate the result or equate the results and obtain so what I mean by this is we can get rid of these parameters T like that so yeah so if we go back to our vector equation or just our parametric equations uh we had before the example we have x equals x naught plus a t and then y equals y not plus a b and this is uh this is actually a t i mean and B T get that mixed up yeah so anyways if we have this let's go back here and yeah so if none of the more that we could uh rearrange it so if we have X is equal to X naught plus a t now we could uh solve this 4T so move this over to this side and then divide it out and then divide this out like that let's put a division sign so this one we subtract this one we divide so we get a t is equal to x minus X naught over a like that and then the next one is y is equal to Y naught plus b t and then likewise do the exact same thing move this over and then divide this out we get a t equals to Y minus y naught over B and the last one Z is equal to Z naught plus C T and then uh rearranges moves over strictly let's put the arrows T is equal to Z minus Z not over C all right and then let's put these all together we get uh T is equal to and then equate all of these together and there's symmetric because they're all the same uh they all equal t x minus X naught divided by a is equal to T well these are all T if y minus y naught over b equals to Z minus Z naught over C and then we can box these in like that yeah so uh these equations are called symmetric equations of the line out so notice that the numbers a b and c that appear in the denominators of equation uh three or I'll just call it the equation above equation three uh that was labeled by the coggles book but I just like writing this the equation above here are Direction numbers of L yeah so note that these are the direction numbers remember the vector v a b c those are Direction numbers of L that is components of a vector parallel to L so if one of a b or c is zero because then you can divide by zero we can still eliminate t for instance if a equals to zero we could write the equation as follow so instead of this we can't do this but if a equals to zero we could just start we could start off with x equals x naught and then the other ones uh leave that in so we could write so if we have uh for example if a equals 0 we have x equals X naught plus a t so this goes to zero because a is zero uh I'll just leave it like that so then this equals two uh this will get this just be a x naught plus zero like that in other words we get x equals two x nine or I'll see I'll do that after this make it for completeness this equals 2x not so what we get is the equations x equals to X naught and then we have the symmetrical equations y minus y naught over B because these ones are non-zero and then Z minus Z naught over C and there's our new symmetric equations and now this means that L lies in the vertical plane x equals x naught and uh here um I'm actually going to draw this I'll just illustrate this so it lies on the x equals x not a vertical plane so for example if you had a just X Y graph like this this is the Y is the Z and I'm just going to draw a plane like this and then this is going to go through it this is the X no I mean this is X and then this plane right here goes across like that and let's say at this point right here uh yeah let's say this this point crossed through it this is at X not so this means that the line is going to be on this plane and it's going to be something like this that's going to be our L like that or it could be like this Etc uh it doesn't matter but it's going to be on this plane like that and that's how you write this yeah x x equals x naught yeah so it doesn't matter what Z or Y value there R you're always going to be on this x not vertical plane all right so let's continue further and look at example two and this one States find parametric equations and symmetric equations of the line that passes through the points a which has a coordinates 2 4 negative 3 and the point B which has coordinates three negative one and one and then Part B say it states at what points does this line intersect the X Y plane all right so let's go over number a first a question answer solution to two a let's first graph out the lines to better uh visualize this this is going to look something like this and this is our Z and this is our y it's going to look something like this uh I mean our Y is gonna be like this and actually extended it past on the negative side and let's say we have our uh X is going to be like this like that that's our X and so now let's graph out Point uh the point a and so the point a goes to 2 4 negative three so it goes one two and then goes to four one two three four and it's gonna go in on the negative side now so I'll draw this uh like that and it's gonna go it's gonna go uh one two three so it's gonna look somewhere well I guess we draw it across there it's going to be like this this is going to be our point a like that and then a point B is going to be 3 negative one and one so then go to three here and then it's going to go to uh negative one so it's going to go back one and it's gonna go up one it's gonna be up like that so it'd be somewhere around here in 3D is gonna be this is our point B like this and then the line is going to be across like that all right and this is going to be our line L so I just fixed it up right a right here and a is again it's going to be at 2 uh two four negative three like that and then the B is going to be the coordinates are three negative one and one like that that's B and then uh and also in part A says at what points does this line intersect the X Y plane so it's gonna be somewhere it looks like somewhere around here ish like that that's I believe something like that but anyways there's all erase this and then until we get it exact so now let's go further all right so going further so again so we're asked to find parametric equations and symmetric equations of this line pass through these two points so we're not explicitly given a vector parallel to the line but observed that the vector v with representation a b so from here to B is parallel to the yeah it's parallel to this line V to this vector or the direction Vector actually no I mean parallel to the line yeah this is the vector yeah so it's parallel to this line right here and we can uh write that out this is going to be V is going to equal to well the difference it's gonna be your position vector uh V is equal to so it's going to be from uh yeah A to B so we're going to go three minus two like that and then negative one minus four and then one minus three I'm a one minus negative three like this this is going to be 1 minus negative three yeah so this one minus this last one and then then it's solving this is going to be three minus two is one negative one minus four is negative five then one minus negative three is going to be one plus three it's gonna be four like that all right uh so thus the direction numbers are just the components of this parallel Vector so V A is equal to one B is equal to negative five and C is equal to four and then taking the point a as a starting point right here is 2 4 negative three as P naught we see that the parametric equations are and remember this one the parametric equations are just going to be x equals to X naught plus a t then y equals to Y naught plus v t and then we have Z is equal to Z naught loss uh this is going to be c t and then where these Z nulls Etc it's going to be this PO and that's going to be choosing our a point a so this means that X is going to equal 2 it's going to be 2 is X naught is right here then plus a a is one t so that's going to be plus d okay one two uh like that and then Y is going to be this way up yeah so Y is equal to Y naught there's going to be four plus I mean it's gonna be negative 5 right here so negative five t and then lastly Z is equal to Z naught is going to be negative three and then it's going to be plus C which is four T oh yeah this is a 4 right there and yeah this is a negative three right here all right and box this and this is our parametric equations of that line all right so that's a parametric equation now the symmetric equations is just when we make t equals to each other so T so we'll just make this uh move this over x minus X naught over a then divide this out etcetera do that for all of them and essentially we just get that same setup here so we get T is equal to again x minus X not over a and this equals to the same thing for the Y do the same thing here y minus y naught over B this equals 2 Z minus Z naught over C and this will equal 2 well we know X is X is General X naught is just two so there's two then four to the negative 3. and then this is divided by a which is one here like that this equals to Y minus four over negative five this equals to Z minus negative three that's going to be plus three negatives become positive yeah with two negatives and then divide Us by uh this is C4 oh yeah so there is our symmetric equations all right so now let's take a look at the solution to B and uh yeah let's go uh over here so again yeah we solve the parametric equation symmetric equation solution part A now the part B says at what point does this line intersect the X Y plane on it so that's a key factor so the X Y plane like this this means it's going to be this whole flat section actually so that means that to pass it is when Z is equal to zero yeah so it's going to be actually oh somewhere here yeah so that's actually the point it goes through the Z equals to zero setup across it so what we could do is solve that so the line intersects the X Y plane when Z is equal to zero so we put Z equals zero in the symmetric equations and obtain so we have these symmetric equations put Z equals zero there so we get x minus two yeah divided by one and just ignores it divided by equals to Y minus 4 my divided by five is equal to well Z is zero zero plus three over four well again it's just equals to three over four like that yeah so then we solve it for each one so we'll have x minus two is equal to three over four and then move this 2 over to this side we'll get well plus two and multiply top and bottom by four so we have same common denominator this equals two actually we'll do that after so we moved it over and then we're going to have x equals two three over four plus two again Times by four over four like this write this a bit neater the two here times the four over four I got now I'll just move that up uh so then two times four is eight plus three is eleven so eleven over four so I have the same common denominator like that and box this and so that's our x value and then the Y value is well we equate the other side y minus 4 over negative 5 is equal to 3 over 4. it's equal to three over four we have three over four let's fix that up and uh yes move this around so then multiply this by negative five yeah we move over this negative 5 multiplied out and then move the 4 in we'll get an equation for y so in other words we'll get Y is equal to we'll put the negative there so negative three over four times by 5 and then minus uh well the minus becomes plus plus four like that and then again same common denominator sometimes by four over four and three times five is fifteen so we're gonna get a negative 15 plus 16 over 4 uh that just becomes uh this is negative 15 like uh that just becomes positive one 16 minus 15 is 1. uh so one over four like all right yeah so thus this means that well we have the equations X is going to be equal to 11 over 4. that's just right here and then we have y is equal to one over four where again Z is zero because it's crossing the X Y plane so then the point we'll call this P has coordinates and this is going to be 11 over 4. 1 over 4 and 0 like that yeah so uh that's a lot intersects the X Y plane at the point p and then it has a coordinates eleven over four one over four and two and uh we could draw that here so this is our Point p this is eleven over four one over four and zero like that yeah so eleven or four is going to be well 12 and 4 is 3 so it's going to be between three and two and then there's this one over four it goes uh pretty much like this around there it's gonna be somewhere around here actually let's just erase that it's gonna be somewhere around here and move this over and yeah all right so that's the point and let's continue further we can graph this out so let's double check the graph using the geogebra 3D graphing calculator so here you can make a line just from two points two four negative three to three negative 1 1 and it gives you the actual equation of the uh the vector equation as well and then I put the points on it and there and it shows the and then I graph the point as well 11 or 4 1 over 4 and 0 and see where it lands so there's the line it looks exactly like what we had there's our a there's our B and there is the P value right there on it that's how the line Looks like again you can click the length player around with it so let's continue further and look at a note on example two so in general the procedure of example two shows that the that direction numbers of the line L yeah through the points p p naught which has a coordinates X naught y naught Z naught n p One X One y one Z one or so the direction numbers are x minus X naught y minus y naught and Z minus Z naught and uh and so the symmetric equations of l r and again you can see that here because uh we were looking at the direction numbers the a b and c so this right here a b and c and we solve them by the difference of these two points so this is three minus 2 that's going to be well we can consider this as as uh this is going to be yeah as X1 minus X naught and then X2 minus I mean and then then y to my I mean y1 minus uh y naught and then Z1 minus Z naught and those are the direction numbers there so in other words we could write this equation as instead of a we could write x 1 minus X naught Etc so we could write it like that because that is what it was so 3 minus 2 that's just X1 minus X naught all right so let's write these out yeah so we could write this all out so we're going to have x minus X naught over now the direction numbers are X1 minus X naught and likewise it's going to be y minus y naught over y1 minus y naught this equals to Z minus Z naught then Z one minus Z naught like that so epic stuff all right now let's continue further and look at the equation of a line segment as opposed to just a full infinite line so often we need a description not often an entire line but just a line segment or but or oh yeah but of just a line segment how for instance could we describe the line segment a b an example two so how could we describe this right here from here to here as opposed to drawing this entire line like that so we just want the line from A to B oh yeah so that line segment all right so going further so if we put uh T equals zero in the parametric equations of example 2 a so those are right here this parametric course if we put a zero here and we're just going to get X we're just going to get the uh yeah X naught and then y naught Z naught and then the next one we just put one and then we're going to get actually uh interesting results so let's take a look right here so if we put it in we get the point well 2 4 negative three that's just the point a and you could see that again this is just x equals 2 and and let's just put that line again plus uh I believe this was 2 plus T so going back here uh this is yeah two plus d then 4 minus five T and this is negative 3 plus 40. all right and then Y is equal to four minus five T and Z is equal to negative three plus four t like that then when you plug in 0 well we get uh so let's write this here for completeness at T equals to zero everything vanishes we're just going to be left with x equals to two Y is equal to four and then Z is equal to negative three which is again uh this point right here is just our a is a negative two four three like that and then if we put T equals one we get the point B because this is just becomes the one and that's just going to be the coordinates that you're subtracting the coordinates of of actually no this is the direction number so if we if we remember these equations yeah just a Teleport up here these equations right here that we got for example this was a t and then a was one where one was three minus two so in in other words in this first example we have yeah this three minus two this is our X1 you can consider that as x one so we have uh we'll have it right over here so X let's write it here x equals two two and then the two right here is just our X naught this equals x naught plus and then the a value which was going to be X1 minus X naught T so then if this is equal to and now we that we have this written I'm going to copy and paste this and then let's put it down here yeah we get here and if we look at it if you plug in t goes to 1 uh we're gonna get uh X X at T equals to 1 is equal to X naught then it's going to be plus X1 minus X naught and then these just cancel the left is X1 so in other words and then do the same thing for Y and Z we get the point B so in other words here at T equals to one for R for our scenario we get T equals to one plug in 1 here so 2 plus 1 4 minus five so write this here so this is going to be yeah it's going to be 3 negative one and one so you could plug this in two plus uh one two plus one is equal to three so that's correct the next one is y is equal to those boxes and this is the correct answer and uh the next one is going to be uh right here four minus five times one is one yeah yeah five times T is uh if five times T is just one it's negative five this equals negative one and that is correct yeah and lastly the Z is going to be equal to Z equals to negative three plus four T so negative three plus four that's just equals to one so four minus three is one yeah so the line segment uh a b is described by the parametric equation because again uh the it ends up at B here at uh at T equals to one it ends up at B at A T equals zero it ends up at a so now in other words all we need to do is solve it from T equals zero to T equals uh one so in other words the equation I'll write this here r of T the line segment this equals two and this is a vector equation form and that is just putting these all together so this is the parametric equation so 2 plus T is the X component and then we have y is equal to four minus five t and then we have this negative 3 plus 14. negative three plus four t and this is 4 T is less than equal to one and greater than equal to zero so this is the line segment from A to B exit this all right so going further let's generalize this so in general we know from the earlier vector equation through the through the tip of the vector are not in the direction of a vector v is R equals two R naught plus T times V that's just the vector equation yeah of a line so if the line also passes through the tip of our one vector then we can take V is equal to R1 minus r naught this uh this vector and so it's vector equation is and before I write that I'm going to actually draw it out just to illustrate this graph uh this can be near Z actually like that sets or Zed and make this a bit less steep and then this is going to be our y and this is going to be our X like that so let's say we had a line L like this let's make it like that this is our line L and it has this from the origin let's say we have our R naught Vector so from the the tip here is on the line and now yeah let's say we had uh right here is going to be our r so let's say this is our r and this is our R Vector this vector equation and before we draw the T V let's say we had R1 is somewhere here so this is another point on the line is R1 and now I'm going to draw this in red so let's say we had a vector this is the vector v this is going to be V like that yeah this Vector goes to like that and uh we'll we'll from this Vector you'll see that R naught is equal to R um yeah this is going to be R naught plus v so in other words to solve this V we just move this over to the other side and we subtract it we get this equation right there and this is the triangle law like this R naught plus v is equal to R1 like that and again this just means that V is equal to R1 minus r not like that and uh here just change it to block uh felt like that just to make it easier to see so anyways so that's that part and this is there or R oops this is going to be our r or R is this one right here and then the TV is going to be just a scalar times this V and that's going to be across here it's going to be like that whoops uh that's just going to be it's going to be actually I just make this a bit neater so put the r naught here this is going to be uh t v not or actually not v v not I mean uh just V and I'll erase that again I'm just going to put this in blue or this is not blue this should be blue here and let's make this in blue and that's going to be like this all right let's draw this again yeah so this is going to be T vector and uh here I just made the blocks because it's easier to see all right so that's what we have so going back to this we're going to solve the vector equation so using what we know using this V equals to r 1 minus r naught just like the just like this setup that we did yeah for the line segment right here yeah in which we found a difference between a and b in the components and so on so uh let's go back here so this is going to be now R factors equal to well R naught so R naught plus triangle lot t v right there TV vector and this equals two let's write it all over here this is going to be R naught plus and then T then the vector v is equal to this part right here so that equals two of uh this is going to be e times r 1 minus r naught vector and now we can just simplify this out to multiply this at T inside and separate the like terms and what we get is I'll write this like this this equals 2 r naught I mean I'm an R Vector not R not a e this R naught equals two this is going to be well we're going to factor out well let's do this all on completeness yeah let's do it on completely I'm going to erase this we're going to write this all later so this equals to that this equals 2 here uh r no you are not plus T R1 Vector minus t r naught and now we could separate the light the factor out the r naughts the like terms that's going to be and then you factor out it's going to be one let's put I'll put it over here that's going to be 1 minus t like that and then Plus T R1 vector like that and I'm just going to box this in like that and now this setup here if you look at it it's the same things we had before so if we put in t equals to zero this vanishes this becomes one so we're left with this R naught likewise if you put in uh T equals to 1 this van this whole thing vanishes uh yeah this this whole thing vanishes oh and this becomes one and you're left with R1 in other words it's exactly what we had earlier but in that scalar format I wrote here so x equals to X now plus uh X1 minus X naught like that t and then when you plug in uh T equals to one uh the X naughts vanish which is separate so it's exactly what we just had but here I'm writing it in Vector form so the line segment from R naught to R1 is given by the parameter zero is less than or equal to T Which is less than equal to one so there's T is from zero to one thus we can summarize as follows and this summary is right here so the line segment from R naught to R1 is given by this should be bold given by the vector equation RT I mean r of t so R is a function of t uh equals to this right here one one minus t are not and then plus T R1 like that where T is less than or equal to one greater than or equal to zero like that all right and again uh as I just alluded to notice that if we plug in t equals zero or t equals one we get well R of zero this just equals two uh this is going to be one times by R naught and I'll just put it all together one minus zero R naught plus zero times R1 like that this just becomes well this is going to be one that's going to be our not vector like that in other words that's our where is going to that a value or the starting point on the vector on the line so and then R1 so T equals one we get one minus one R naught plus one times y r one let's leave that one out and it's going to be r one this is equals to this vanishes and we're left with r one oh my God box this boxes and this one is the vector to that a second point on the line that b the B coordinate I mean the B point with the um yeah so that's the R1 Vector to the B the B point on the line all right uh now let's continue further and look at example three and this one States uh show that the lines L1 and L2 with parametric equations shown below are skew lines that is they do not intersect and are not parallel and therefore do not lie in the same plane so here's line L1 this is x equals one plus T then we had y equals to negative two plus three T and Z is equal to 4 minus t and likewise x equals a 2s y equals 3 plus s and Z equals to negative three plus four S that's better all right so let's take a look at the solution and the solution here uh so the lines now the lines first of all they're not parallel so so that they don't intersect and and are not parallel so they're not parallel because the corresponding Direction vectors one three negative one you can see the directions vectors here's one three negative one and two one fourths two and four are not parallel in other words their components are not proportional or scalar multiples of each other and I'll just illustrate this further I'm just going to write down both equations so if we have uh R1 of T this equals two just in a vector format uh the starting point is going to be one negative two and four one negative two and four like that and then the next one is going to be plus t and then it's going to be one three negative one like that and these are the direction numbers and then likewise for the second one R2 I'm going to call this of s and that is because the yeah the parameter here is s instead of t so the point here is well this 2s that's the direction number the and the uh the starting position number the X component is zero it doesn't show there so zero and then it's going to be the next one is three next one is negative 3. and then the next thing we're going to do is plus s the parameter and then it's going to be 2 one so two one and then four like that and then you can see here so these aren't all proportional for example this is a Times by two now let's do this better so this like that that's 2x this right here is divided by three so it's not the same one this right here is going to be divided by Times by negative 4. so this is negative multiplied by negative 4. so yeah so they're not all proportional otherwise yeah to be proportional you need to be all to do the same thing either multiply by 2 or divide by 2 or divided by three etcetera all right so we show that there are not parallel so the next thing we need to do is show that they do not intersect in other words they don't intersect at a point where they're share the same value so if L1 and L2 at a point of intersection there would be values of T and and S such that well they're going to be the same so we'll have the all the x y z coordinates to be the same so if we write that out that means that we must have a scenario such that X equals to 1 plus T equals two s so the X ones have to be equal 2s and then the next one is uh y so we'll have y is equal to negative two plus three t say negative two plus three T equals two this equals to three right here plus t plus s that's three plus s the next one is four minus t so Z equals two uh the Z is going to be four minus t equals to negative 3 plus 40. and 4S and that's four minus t then negative three plus four S like that all right so if it had a point of intersection they should have uh these set up where they're equal and to show this while uh and and also yeah and we're going to show that they do not intersect so then they shouldn't have these numbers that equal each other and uh so to do that let's just solve it and see what we get so if we solve the first two equations we get so solve these ones for S and T so we get this equation one this is equation two and then this is equation three like that and I'll put it like this right here let's move this over for the voice it doesn't look like it's a part of the equation so we have that uh so then we're going to get one plus T equals to 2s the next one is me negative two let's put these all in line three T equals 2 well s plus three so let's uh cancel out some terms let's cancel out the t's but to do that I'm going to do is we'll multiply this entire uh equation by three and then we're going to do subtract this entire equation here I mean it's not multiply we'll just uh subtract it and we'll get an equation where the t's are canceled so we're going to get a 3 . okay so three times one and then it's going to be minus this would be negative 2 so that's gonna be plus 2. like that put a bracket I got the next one here is going to be 3T minus 3T so those just cancel and yeah over here is going to equal 2 well 2s 3 times 2s is 6s minus s then and then we're gonna have a minus 3 right here so put these all together uh we're gonna get uh this equals two five it's three plus two is five this one cancels equals to 6s minus s it's going to be 5s then plus minus 3. and then move this 3 over to the other side and divide out by The Divide this 5 out or just move it over so we're going to get a uh five plus three is eight so we're going to get again I get an S equals two eight and then divide out divide this out s equals two eight over five like that all right so we have S equals five we could plug this inside equation one so we get one plus t yeah 1 plus T is equal to is equal to 2s right there this equals to 2 times a by eight over five and here's put this inside like that so inside the inside the s and now the next step is we'll move this one over to the other side so we're going to get a t is equal to uh two times a to sixteen sixteen over five minus one multiplied by that five of five like that so that'll be the same common denominator so 16 minus five is just equal to 11. 11 over 5. all right so now if we take these two values so we have t and we have S if we plug these values into the third equation we get so if we look at here so 4 minus t and negative 3 plus 4S so we'll get uh 4 minus t like this that's a 4 minus t equals to negative three plus 4S so they have to be equal like that so then if we plug in t so we're going to get a 4 minus 11 over five equals two remove this Arrow further and then equals to negative 3 plus 4 and then s is eight over five Times by 8 over 5 like that and now what we'll do is yeah it just keeps solving it but just leaving it in place so here's the same condometers I'm going to multiply this by five over five uh it's going to be 20 minus 11. we're actually better yet instead of doing that let's just get rid of the 5 altogether so I'm going to multiply the entire thing by 5. so we're not changing uh anything uh we're just scaling up everything by five but then the the equations relative to each other don't change so this is going to be equals 2 4 times 5 is 20. it's 20 minus 11 equals 2 and then here we're going to get negative 3 times 5 is 15. like that and then this 5 cancels 4 times 8 is just going to be well 32. like that so then this part right here is going to be 20 minus 11 is 9. and then this part right here is negative 15. so 32 or 30 minus 15 is 15 plus another two so that's 17. so in other words this is not equal to 17. so the equation is wrong like yeah I mean they don't intercept each other so we can't so they don't have the same values there all right yeah so therefore there are no values of T and S that satisfies the three equations so L1 and L2 do not intersect thus L1 and L2 are skew lines so they're not uh parallel and they do not intercept so here is a graph of the lines using the geogebra 3D graphing calculator and there's line one one negative two four uh that's the point and there's the vector one three negative one and see that over here so 1 negative 2 4 1 3 negative one so again 1 2 negative four and then the next one is zero three negative three and then two one four is a vector zero three negative three two one four is that Vector Direction vector and if you graph it out it looks like this so This the block one is this one is L1 there's the red is L2 and the skew line so it looks like it's intersection but it's not if you turn it uh the geogebra you'll see that they don't actually intercept and they're not parallel see it's it's uh going out words like that yeah so it's the angle is getting bigger and bigger all right now let's go to the next section and that's on planes or equations of planes and the previous section was on lines now we're looking at planes so all the way line and space is determined by point and a direction a plane and space is more difficult to describe and that brings us to the vector equation of a plane so a single Vector parallel to a plane is not enough to convey the equal direction of a plane but a vector perpendicular to the plane does completely satisfy its direction yeah for example if you had a plane like this and then if you had a parallel Vector well it could have multiple of them so it's not specifying a Direction but you only have well one parallel uh I mean one perpendicular Vector to it yeah so this is specifying its direction if you shift it over it's still the same Vector there you're just shifting the position of the same thing whereas a parallel one you're going to have different actual directions there these are actual different directions but it's still parallel now it's going further so thus a plane in space is determined by a point p a p naught with the coordinates X naught y naught Z naught in the plane and a vector n that is orthogonal to the plane and this orthogonal or perpendicular Vector n is called a normal Vector so let P of uh with coordinates x y z be an arbitrary point in the plane and let R naught and R be the position vectors of P naught and P then the vector R minus r naught is represented by P naught to p as Vector as shown in the figure below so let's graph this whole thing out so what I'll do is first I'll draw the plane first draw a big one like this let's say we had a plane like that all right make this a bit better all right so we have that and now let's draw the z-axis it can go like this and I'll make a dashed line dash dash across and then here just to Signal signifies 3D and then it's going to go dash dash dash and goes across it like this so this is the Y there's our x coordinate yeah our x-axis so yeah so let's draw all these out let's draw first the orthogonal vector I'm going to draw actually let's do the point first let's go P right here so let's say we have a point p and has coordinates x y z so an arbitrary one like that and now let's draw a point here this one here we will call this uh this is going to be our po so let's call this p o and then this coordinates our X not y naught and Z naught all right so we have this to fix this up and now let's say we have a orthogonal vector n and let's say let's draw a big line like that this is going to be our n vector and it's going to be parallel I mean a perpendicular or orthogonal and let's say we have a vector here from here to here and this one right here is uh perpendicular to it and it's sticking out of the page all right towards the out of the page so and uh this right here let's say we had the position vectors now and this is the origin and this is going to go from here and then dot dot dot all the way across to this this one is going to be our R naught vector and then this one here goes across here and then goes all the way across here dot dot dot this is going to be our R vector just a regular R vector and then the difference between here and here is just a difference so we'll have R naught uh yeah R minus r naught because again you could use the triangle law so you'll have R naught plus this equals this so then move that around you'll get the difference so anyways it's going to be R naught like that and yes it's going to be r minus r naught all right now that we have this going further so the normal Vector n is orthogonal to every Vector in the plane so it doesn't matter where the vector is where you choose your uh P value there it could be here it could be here Etc it's all going to be perpendicular this n is going to be perpendicular to it so in particular n is orthogonal 2 this Vector R minus r naught and using the fact that the that the dot product zero for perpendicular vectors we have and here's a link to my earlier video so we have we're going to have just well the dot product of the normal vector to any Vector on the plane such as R minus r not this is going to equal to zero because it's perpendicular and so that's our plane equation and let's box this whole thing in and now are this above equation this can be Rewritten as we'll multiply this n times R and then minus n times R naught and move that over to the other side uh we'll just write it as uh n times I mean n dot so it's a DOT product equals two let's make it bigger so this is going to be n dot R vector and then this equals two against dot product not just a multiplication this is going to be now it's going to be and then you shift it off to the other side n dot or uh R naught vector so you could write the equation of a plane in Vector form and both of these formats and move this up and uh yeah so either equation above is called a vector equation of the plane and let's go further and now look at now the scalar equation of a plane so to obtain a scalar equation for the plane we write uh the normal Vector n with its components a b c and let's say R has one is x y z and the r naught vectors uh components X naught y naught Z naught then the vector equation B comes so let's write this out so we have n Vector dot this is going to be R minus r naught equals zero and then plug in everything into it this is going to be now it's going to be well the N is ABC so this is going to be a B and C uh Dot and this is going to be now well these are subtracting from each other yeah so just to subtract these vectors so it's going to be x minus X naught and then y minus y naught and then Z minus Z naught components for the triangle bracket equals zero and now when you do the dot product this is just a summation where this multiplies by this plus and then this multiplied by this and then this multiplied by this plus 2 add them all together so let's do that and that's going to be equals 2 a x minus X naught uh and yeah so A minus I mean a times x minus X naught Plus B times y minus y naught plus C and then Z minus Z naught now this whole thing equals a zero and there's our equation of a plane our scalar equation of a plane where there's no vectors involved yes uh so this above equation is again called the scalar equation of a plane through the point P naught X naught y naught Z naught with normal Vector n equals to A B C uh components or has ABC components all right going for this so example four so this one States find an equation of the plane through the point two four negative one with normal Vector 2 3 4 then we're asked to find intercepts and sketch the plane so let's go over here so putting in uh so this is the normal vectors n so n is equals A B C equals to two three four and P naught that goes through this point so p naught's x naught y naught Z naught that's gonna be two four one and so we plug that into the scalar equation of a plane we obtain so we should throw that in there so we obtain so a is two so two and then x minus X naught is going to be 2. plus and then it's going to be right here three that's three right here times it by y minus and then y naught is four and then plus uh this is going to be uh the next one is going to be four and then this is going to be right here as Z minus this negative 1 that's going to be plus one equals zero all right so we have this now we could just uh simplify this further multiply this inside so we're going to get 2x minus two times two two times negative 2 is negative 4. and this is going to be plus and 3 times y minus and then 3 times 4 is 12 so there's a minus 12 plus the next one 4z plus 4 equals to zero so let's do some simplifications well there's a minus four there's a plus four these can cancel and then when we're left as well there's a negative 12 right there so we just move that over to the other side and then put everything else neatly two it's going to be a 2X and plus 3y plus 4z equals to 12. those are equation of a plane yeah that's our simplified equation of a plan but technically we could have just stopped right here so we could have stopped right here and box this out or we could do or and then put it right here all right and uh not going further so the next step is well to find intercepts and sketch the plane so we find the equation of a plane and we're going to find intercepts so to find the x-intercepts we set the Y and Z equals to zero and then we could find uh yeah the point that crosses that x-intercept so if we set the Y is z equals zero in this equation and we can obtain uh yeah we'll obtain x equals six in other words just throw that in there we'll just just double check or work two plus I mean 2X plus well this is all zero plus zero plus zero this equals a 12 and again move the 2 over or divide it out so x equals to 6 like that and then similarly the Y intercept is 4 and the Z intercept is three and we can uh check this out yeah now we could double check this again using this and then do the same thing so to get the Y intercept we make everything else zero so we're going to get a 0 plus 3y plus zero equals a twelve like that and again divide by three uh this is going to be y equals to four and then similarly for the Z we make everything else zero so we're just going to get zero plus zero equal that plus uh four Z equals to 12 and then divide by 4 or 12 divided by 4 is going to be 3. so Z equals 2. 3. all right so now that we have the intercepts and this enables us to sketch the portion of the plane that lies in the first OCT end in other words that's all the um positive values and so again if we're asked to sketch the plane there so it's going to look something like this we have the uh the intercepts so if we draw this out you know what I'm doing this is the Z I'm going to move this down until a point where we get the Z equals three that's going to be the point zero zero and then three and then uh and then downwards is going to be like this and then going further uh now if we draw the other Corner the next one is the Y is four so that means it's going to go like this it goes across uh all the way until the four and this is going to be right here uh this point this point is going to be 0 is the x 4 is the Y and then zeros to Z and this is going to be our Y and then so then line like this this is just the first octant all right now the next one is six so it's going to be further out I mean I guess Dash cross all the way across let's say it goes something like this I guess it's here okay so that is going to be this point is going to be six zero zero all right and then this means that it's going to be a triangle like that triangles is basically the side that's on yeah there's a triangle that's in the first option for the triangle but then afterwards it's gonna this whole plane is going to go below this uh y uh X Y plane all right and then extend this out um this is going to be our X like that all right yeah so that is the graph and we can go further and let's graph this using the geogebra Gog bra 3D graphing calculator and uh here's inputs I did I just put in the plan 2x plus 3y plus Z equals to 12 and there's the points zero zero three zero four zero and then six zero zero and then I put the line segments from 0.1 to 0.2 and then also the point uh 1 to 0.3 segment and point two to point three anyways you get a graph like this it's so again there's a plane extending Way Beyond that triangle but then you can graph the the intercepts and then it forms exactly what we just graphed there there's Point uh p1003 the zero four zero and there's a six zero zero and I made it to the side so we can see this plane across like that so again you could play around with this link there all right now let's go further and now look at a linear equation of a plane so by collecting the terms in this scalar equation of the plane we could rewrite the equation of a plane as Falls and it also will look uh something like this how we simplified it here so if we go back to this plane equation so write this out then we could simplify it and then collect all of the constant terms such as a x naught B and then this B naught when they multiplied with each other is to separate all those and what we're going to get is also let's just start collecting the terms let's just write down the scalar equation first so x minus X naught plus b times y minus y naught plus c foreign this so let's see and then this is going to be Z minus Z naught this equals to zero and yeah multiply this all inside so what we're going to do is you know let's just multiply it all out it's going to be ax minus a X naught and then plus b y b y naught and then uh erase this all right it's going to be plus c times Z times c z naught this equals to zero so we're going to collect all the like terms and move all the positive stuff all the uh the variables uh Y is that all together so that's just going to be equals 2 a x plus b y plus c z and then the next one is going to be now we're just going to combine all these I'm going to take out the negative as well take out the negative and we're going to have a x naught plus b y naught plus c z not this equals zero and then specify this uh the this term right here this whole thing is negative including let's call this D so then that means we could write this as a linear equation we'll call this a linear equation but this equation for planes going to be ax plus b y plus c z plus d equals to zero there's our linear equation of a plane and this is going to be where again D for completeness is going to be equals to negative um a x naught plus b y naught plus c Z naught like that all right it's going further yeah so this equation this is called uh yeah the above equation is called a linear equation in X Y and Z so conversely it could be shown that if a B and C are all not zero then the linear equation represents a plane with normal Vector a b c and it says uh yes the exercise one at the end of the video and basically it's saying you know it's saying prove that this is equation of a plane starting from here or something yeah and uh it's pretty straightforward you just go backwards and yeah you can go backwards but without without needing to input this whole setup here it's with this D value here and I'll show that an exercise one at the end of the video anyway let's go further so let's take a look at example five and this one States find an equation of the plane that passes through the points p with coordinates one three two and Q with coordinates three negative one and six and R with coordinates five to zero and then and then uh yeah so let's go ahead and look at the solution right here so we need an equation of the plane to pass through these three points so the vectors A B uh yeah let's call the vectors A and B corresponding to P Q so we'll do the position Vector from yep p to Q and then PR so going from P this one and then this P to the R1 uh is and we again we just subtract uh the components of each one so a is going to equal 2 so Q terms the Q terms minus the P terms this equals two I'll just write this fully a equals to P Q vector or the position Vector parallel to this one and that's going to be PQ is going to be three minus one and then it's going to be negative one minus three and it's going to be six minus 2. like this and this equals 2 2 negative four and then four like that and then box that in and then uh continuing further the next one is going to be B this equals to p r this equals two um R is 5 minus 1. and then this can be 2 minus 3. and then it's going to be 0 minus two and this equals 2 4 negative 1 negative two I got all right so that's the two vectors like that all right so now since both A and B vectors lie in the plane because we're told that uh the equation passed through all these points so the vectors that associated with them are all on the plane so then that means they're cross product A cross B and again here's a link to my earlier on it is orthogonal or perpendicular to the plane and can be taken as the normal Vector yes the cross product gives a perpendicular Vector of two of two vectors so that means and then if we write this indeterminate form so we have uh n uh normal Vector n equals 2 a vector yeah a vector cross B Vector this equals to write this in determine forms can be uh let's write this out the big table there so I two four uh I mean yeah the a is going to be right here two two negative four or four like that and this is going to be J b k then this next one is four negative one two so put the vectors in in rows I mean in columns like this yeah so each of our each row is a vectors so this this vector and then this one here so negative one negative one negative two right here oops yeah negative two like that and then do this all right so if we take the cross product here if you remember what I'll do is cross out this cross out this to get the I terms or this is in standard basis Vector form so we're going to have well this negative four times negative two that's going to be positive eight then we're going to subtract 4 times negative one that becomes well uh minus becomes plus so let's be plus four and this is the I or the X component and now the next step is we'll cross this out I mean I mean erase that across this so now what we have is now we're gonna have a minus the middle one is a minus and we're going to have two times negative two that's going to be negative four and then four times four so we're going to subtract 4 times 4 that's 16. like that and this is going to be J like that now the next one is Plus so plus so that means we're gonna I mean the next one is the K1 so we'll cancel out the K so just remember this method so now we go 2 times negative 1 that's negative two and then minus negative four times four equals minus because plus because there's another negative there 4 times 4 16. and this is K uh Vector all right so now what this equals to let's just add it all up so 8 plus 4 is 12. 12i and then minus this becomes well it's me plus is negative 4 minus 16 is negative 20 then the negative plus 20 J like that next one's mean plus uh negative 2 plus 16 that's going to be plus 14. 14 k like that all right so then this is the normal vector and we can just box this whole thing out like that all right it's another way of the normal vector and we also have the point p is one three two so we have a point right here we could uh pick any point I believe it should all work out but again we did P then p q p r so we'll use the point P so one three two so with the point p is one three two and the normal vector and above here an equation of the of the plane is and we'll just write down remember it's going to be a which is 12. and then x minus XO x minus well XO is one and then we're going to have a plus and then it's going to be 20. and then y minus y o that's going to be 3. then plus uh this is going to be 14 is C so that's 14 is C and then it's going to be Z minus z o that's two like that this equals to zero all right so we could uh stat we could just stay with this so that that's our equation of the plane do a check mark Or we can continue further and simplified further what I'll do is I'm going to divide by 2 on the whole thing because you'll see there's a 12 there's a 20 there's a 14. there's a factor of 2 and then we'll get a 7 there so 7 should be a it can divide that anymore by two so let's go ahead this is going to be so yeah this is going to be 12 divided by two six so six x minus 1. plus 10 y minus 3 plus 7 Z minus two equals to zero and now we could multiply this inside get a linear equation and we could do yeah just go multiply it all out so what we're going to get is well let's multiply all this uh the variables so 6X 10 y 7z so we'll get a 6X Plus 10y plus seven Z 7 Z like that and the next one is going to be well 6 times negative one that's just going to be negative 6 like this yeah negative 6 and then this is going to be 10 times negative 3 that's going to be negative 30. the next one seven times negative 2 that's going to be negative 14 and this equals to zero like this all right and now let's just uh solve what this would be well um this is going to be right here this uh this one I mean this negative 6 negative 14 is negative 20 so then negative 30 negative 20 that becomes well negative 50. this whole thing is negative 50. negative 50. well that's it's negative 50 and now we'll just move it over to this side so we get a plus 50 on the right side so what we'll get is linear equation 6x plus 10y plus 7 Z equals to 50. so there's a linear equation of a plane more simplified than this equation of the plane but this is still correct as well all right so now going further uh let's graph This Plane using the Gog breath UD graphing a calculator so you could plug that into 6X 10y plus 7z equals 50 and then here's the points one three two three negative negative 1 6 then R is five two zero and also I did a double check actually this was just for my own sake so I did plenty of using the equation p q r so three points you just plug that in there and it has automatically solves the plane and the linear equation exactly what we had over here just quite remarkable all right and then I also did the segments uh from my this Q to p r to P and R to Q Etc and what we have is this so there's a plane I have it I shifted it around so that we can see the plane and we see the points on there's the p q r right there which is absolutely fascinating stuff and you can see it's right on the plane and you can even see that with this uh how this shaded area is at the bottom so it's showing that it's is on the plane there all these are so you see the circle that's a half light blue half dark blue so it's literally crossing the plane or just on the plane so that's the circles and cut in half there epic stuff all right let's make this actually a bit smaller Maybe all right just leave it yeah all right it's pretty cool stuff anyways now let's go further and look at example six an example six Six States find the point at which the line with parametric equations x equals two plus three t y equals negative four T and Z equals five plus T intersects to plane four X plus five y minus two Z equals 18. all right so uh let's take a look at the solution so uh to find this point so we want to point odd which the line has these equations intersex explains in other words we could just throw these X values for example inside here throw the Y inside here through the Z inside here and then find we're going to find a t value and then we're going to just solve for the XYZ so we substitute the expression for X Y and Z from the parametric equations into the equation of the plane so what we're going to get is yeah so we're going to get four so that's this 4X plus 5y minus 2z this is going to be now two plus three t and then plus five y y is going to be negative 4T and then the next one is z i mean is it going to be a negative 2z negative 2 and then Z is 5 plus t like that and then this is going to be equals to 18. now we've got to solve for the T value well let's uh multiply this inside inside so we're going to get a four times two is eight plus twelve T three times four then here's going to be minus 20t the next one is going to be right here negative 10 negative 2 t equals to 18. so would we get there and let's see what else we get is is right here so uh so we have this let's move this everything over to the other side and then combine all the like terms so I'll move this over to this side end this over to there now we're going to add up these like terms so we're going to get a uh 12. minus 20 minus 2 t equals to 18 plus 10 and then minus 8. so in other words we're going to get 18 minus 8 is going to be it's going to be 10. so for example this one's we'll just look at this one actually 10 minus 8 is 2. so then 18 plus 2 is well it's going to be this whole thing is equal to 20. yeah so it's going to be 20 there now so that's 20 and then over here let's see what we get here so 12 minus 2 is 10. or or we can just do it over here so 12 minus 20 is negative 8 and then minus uh minus uh 2 is going to be this is going to be negative 10. so in other words we get a negative 10 T equals to 20. solve for T divide this out so T is equal to 20 divided by 10 negative 10 that's going to be negative 2. all right so uh therefore the point of intersection occurs when the parameter value is t t of T equals negative two so plugging this value into the parametric equations of a line we get the point of intersection to be negative 4 8 and 3 and we could solve that here so x equals 2 plus 3T so we get x equals two plus three T equals two well two plus three negative two so this equals to two minus six equals to negative four so this is correct over there so that's negative four there the next one is going to be y is just uh equals negative fourteen so y equals two negative four T which equals two negative four negative two which equals to plus eight like this so that's plus eight that is correct I'll write this neater all right so this is this is correct that's an eight right there now the next one is uh Z is z is equal to 5 plus t so equals five plus T is equals to five minus two equals to three all right and that's three so that's the points all right and uh continuing further yeah we can graph this as always using the Gog bra 3D graphing calculator yeah I like it's pretty epic stuff you can double check your work and get some more insights so here's the plan 4X plus 5y minus 2z uh that is this right here 4X plus 5y minus 2z equals 18 and then what you do is draw a line and then here's a line with the initial point two zero five and uh yeah that's starting the starting one you could see that from the parametric equations uh there is the two and there's a zeros in there and then there's the five those are the starting ones and then it has a parallel Vector three negative four and one I mean uh yeah that's a parallel Vector right here three negative four and one and then it gets you this this uh vector equation and it draws it and there's the plane there and then you have the point four negative four eight and three negative four eight and three and it shows it across here so there's the line L and it intersects this plane and you can see right at this Edge there this plane there uh p is equal to negative four eight three CS epics epic stuff okay again you can now click the link and play around with this and see what else you can find anyways let's continue further all right let's take a look at parallel plane so two planes are parallel if they're normal vectors are parallel that is that's no Vector like this and it's a then you have another one like this as well pointing in the same direction then you have the same then you have parallel planes so for instance the planes X plus two y minus three Z equals four and two X plus four y minus six Z equals three are parallel because they're normal vectors right here is one two negative three so one two negative three and this N2 is uh that this is going to be 2 4 negative six uh yeah so because that's their normal vectors and N2 equals to two and one you can see these these are just uh multiples of it so all you're doing these are parallel because you're just scaling these up by two you're just multiplying this every component by two like that so it's a so it's exactly parallel so this one in this case you have one uh one vector that's pointing this way and then the other one is the same but double so this is going to be 2x so it's again it's parallel so then you can have parallel um planes and this term right here the three and four that just uh shows it um yeah that just shifts it in space but it's gonna be still parallel all right so if two planes are not parallel then they intersect in a straight line and the angles between the two planes is defined as the acute angle between their normal vectors yes and as shown in the figure below so yes if you have any non-parallel planes they're always going to intersect you because you're going to infinity and uh yeah just you might really speaking it will intersect so let's draw two planes so let's say you had a plane like this here I'll make this a straight line like this and then you had another one I mean this this is the same plane all right so there's one plane all right so let's say that's one plane now there's another one like this and there's that angle to it and I'm going to draw this angle as Theta so this angle and I'm going to go as dashed lines across dot dot dot and this dot dot dot dot all the way across here like that there's the line there all right so let's say that at the bottom here of this plane is gonna we'll call that plane one that's going to have a like this let's put a dashed line Dash um going dashed up like this Dash Dash Dash and then it goes through this uh through this point all the way across all the way on top here this is going to be our N1 Vector like that so now because uh yeah this one here is uh this is angle's Theta like that the next one is going to be and this is from the bottom plane the next one we're going to have let's say it's going to be on this plane on the top plane and that's going to be well this is going to be pointing out like this actually instead of going all the way there let's make it right next to it yeah it's made it right next to it and I'm going to do is I'm going to draw a line like this this is going to be perpendicular to the top plane and now we have an angle here this angle is going to be also Theta and yeah you can see this is a Theta this is exactly the same because you're you're turning it both by the exact same for example if you had it uh if you had both of these flat the angle is going to be zero and Theta is equal zero and then the difference between these are also going to be uh zero they're all they're pointing the exact same direction so then the theta equals zero on the top and then as you rotate uh either or yeah so if you had uh let's say you had here and you had a flat so you initially had zero theta equals zero and then you shift it you can shift by exactly the same amount Theta Theta so that's all you're doing is turning it so yeah so the angle is going to be the difference um between these and this is going to be our N2 all right so now let's continue further and take a look at an example on this so example seven so uh part A States find the angle between the planes X Plus y plus Z equals one and x minus two y plus three Z equals one and then Part B States find symmetric equations for the line of intersection L of these two planes now we got an interesting Little Nugget here so a solution to 7A so the normal vectors of these planes are and they go again the normal Vector is just going to be whatever variable is in front of these uh these XYZ variables so uh the first one is going to be well X Plus y plus Z that's going to be 1 1 1. so N1 N1 is equal to the components one one and then N2 is going to be 1 negative two and three one negative two and three like that all right so that is what we have all right so and so if Theta is the angle between the planes we could apply the equation for the geometric equation interpretation of the dot product and get and get the following equation because yeah so then basically we need to know the angle between the planes is just the angle between these uh Direction vectors so let's uh take a look at this and if we recall from my earlier video the geometric interpretation of the dot product is basically we get a cosine Theta angle between the two vectors is going to be equal to the dot product of two vectors divide this by the absolute value or the distance of the lengths of each one multiply by each other like that yes epic stuff so this equals two so the dot product uh this is yeah the top parts I mean the dot product this is going to be one one two vectors Dot one negative two negative three and now this is going to be divided by the lengths well the the bottom one is going to be length is going to be one squared plus one squared plus one squared and then multiplied by um this is going to be uh this one here one squared plus uh yeah plus uh yeah this is going to be negative two squared that's just going to be 2 squared and then plus 3 squared like that so let's see what this gets this one we know is just going to be three uh this part right here or the inside I'm just going to look at the inside so this part right here that's just going to be three this part inside is going to be well one plus four plus three squared is nine so this is going to be 14. so the inside is going to be 14 like that so put these all together we get uh this is going to be equal to and this equals two and then here's a DOT product so this one's going to be one times one plus yes we are multiply and add the components one negative two that's just going to be negative two those erases plus put a negative two and then next one is one times three so Plus 3. divide this by square root three times the Y square root fourteen all right so then what this is be becomes well one yeah one plus three is four the minus two is going to be two and then this one here is three we could put the square roots together three times fourteen is going to be well 14 times 2 is 28 and then uh and then uh Times by another I mean add another 14. so it's going to be 38 42. let's do that in our head that's going to be 4d2 like that so what this means is we this means that cosine theta equals to 2 over square root 42 so then the the angle Theta we could use the inverse of it so inverse um yeah inverse Theta I mean the inverse of this two uh 42 like that so the inverse of n yeah so this inverse or Arc cosine uh yeah of this two of the two divided by square root of 442. if you plug this into the calculator what we're going to get is well this equals 2 roughly 72 degrees just quite epic stuff here I'm going to box this in all right uh yes that's 72 degrees and now we can do a calculation check so we could write a cosine the same thing as R cosine or inverse cosine like that and then 2 divided by square root 42 and this is using the my built-in uh OneNote calculator so we go equals space and it goes automatically calculated 72 102 Etc so yes it rounds up to 72 degrees it rounds down to 72 degrees I pick stuff and now let's graph out the planes and double track our work using the geoge Bro 3D graphing calculator as always so here's the first plane uh P1 the plane One X Plus y plus Z equals one then P2 x minus two y plus three Z equals one and then we could do is it also calculate the angle directly using Gog bra so angle from of the planes P1 P2 72.02 absolutely remarkable uh stuff here and here's the graph of it so there's the plan Lane um and there's the other plane like that the green is the plane two so this is P2 P1 and there's the angle 72.02 all shown automatically in 3D epic stuff here or now let's take a look at solution to B and this is finding the the L the that intersects those two so find symmetric equations for the line of intersection L of these two planes so what we're asked to find is this line that's intersecting here like that that's uh intersecting these two planes across like that epic stuff there all right so we first need to find a point on L for instance we can find the point where the line intersects the X Y plane by setting Z equals to zero in the equations of both planes yeah so when we plug in Z equals zero into the equation both planes well again we're just kind of guessing that it would be crossing the exit line theoretically could it could never cross that X Y plane and yeah so that's this is the X this is this is the X this is the Y I mean the the red is the X there's the Y and there's a z up top here but looking at the graph here it looks like yeah it should cross it somewhere across somewhere across there it's going to cross uh the Z equals zero uh plane there or the X Y plane uh but yeah so you can again a trial and error but in this case it actually does so uh for instance we can find the point where the line under intersects X Y plane by setting Z equals zero in the equations of both planes this gives the equations X Plus y equals one and x minus two y equals one so if you look at the two uh planes so right here yeah you just get rid of this uh this set this goes vanishes so x minus two y equals one and X Plus y equals one and let's see if we get values for those and then whose solution is z equals uh I mean x equals one and Y equals zero and again you can see that so we'll have this X Plus y X Plus y equals to one and then the other one is going to be x minus two y equals to one and again it's double check there yes so then this X Plus y there's a z goes zero x minus two y plus three Z as in the three set goes zero so what we have here yeah so we have this we have to use two equations as an x and x line up so we just subtract this whole thing at the bottom and get a third equation so so that it cancels out so x minus X is zero y minus negative 2y is going to be plus 3y and then this 1 minus 1 is 0. so we get 3y equals to zero this means that Y is equal to zero so three just divided out plus 0. all right so Y is equal to zero so that's correct and then the last one we just plug that inside so this means now we'll get X plus that's this part right here X plus zero equals one so in other words uh x equals one yeah so we have these two points all right I mean these two values so now we have the point is going to be one zero zero all right uh yeah so the point uh one zero zero lies on L since this point satisfies the equations of both planes now we observe that since L is in both planes yeah so then it is perpendicular to both of the normal vectors yeah so this means that a vector pair yeah vector v parallel to L is given by the cross product since the cross product gives a perpendicular Vector so for example if you have these two planes so you had one plane like this uh it has a normal Vector let's call that and this is a yeah let's call it N1 or N2 actually if you look at that that's the green one that's N2 like this so that's N2 and then the other one is over here is going to be N1 uh perpendicular to this these are perpendicular um I mean it's not familiar to this one permeated to the plane like that and then this one is perpendicular to uh the green one Etc so let's ignore those I just erase that so yeah so we have those perpendicular um we'll make this a bit less looking like regular to it so this is like that it's gonna look something like yeah let's just call it like that all right so then the cross product is going to be parallel to this line right here this is going to be the cross product like that so let's make it parallel yeah to me perpendicular to these two like that but also it's going to be parallel to this line L so that we know the direction of it basically we know the direction and we have a point on that line L that means we can write the expression for that with the full line and uh yeah so we could just do the cross product of that so then the vector v parallel uh the direction Vector of the line L and equal to the cross product and N1 cross N2 vectors and now yeah the crossbar gives a perpendicular Vector to these vectors that are perpendicular to their respective planes this equals two as always I J k and then the components uh this is an N1 Vector is just going to be a one one one one and the next one is going to be um let's just go up here so that's one one one one negative two three so one negative two and three one negative two and three and now we'll do the cross product as always this is going to be equal to cross this out cross this out 1 times 1 is 3. and 1 times 3 is 3 and then minus 1 times negative 2 is negative two but we're minusing becomes plus two like that this is going to be I and then subtract is the middle one erase this erase no erase that across the middle now we're going to have a 1 times 3 can be 3 minus 1 times 1 is 1. J and then next one's going to be plus d k set up so erase this cross across this one one times negative 2 is negative two minus one times one like that and then I just solve this out or simplify it's a three plus uh two is five I and then minus three minus 1 is 2. like that and then this one's gonna be plus negative two minus 1 that's going to be a negative three so minus three okay like that and that is what we got so 5 negative two negative 3. and let's Circle this this is going to be our V Vector all right and uh so the symmetric equations of L can be written as and again we're asked for a symmetric equations of the line as opposed to just the vector equation so here find symmetric equations for the line of intersection L of these two planes and again remember the symmetric equation is just when you have the parameter uh set that all equal to each other so x minus X naught over a equals to Y minus y naught over b equals to Z minus Z naught over C where the um you know the ABC correspond to this vector v like that so that means that means or I'll continue furthest aware uh where A B C equals to V Vector equals to this component 5 negative 2 and negative three and also um yeah so and and the components uh the the point here x x 0 y 0 Z 0 this is just a point on l if we and remember we'll resolved it for both planes yeah this was just equals to one zero zero as per um the setup here as per here so we sold it here the points x uh is one and Y zero is that zero and so that one lies on the plane on the point L yeah so put this all together we get x minus X zeros one over a is five this equals to Y minus zero is a zero divided by uh B is negative two and and then this equals two Z minus zero zero uh yes n minus zero as you said and then divided by C is negative three but those are symmetric equations like that all right uh yeah so we have the symmetric equations uh let's just graph this out so let's graph out the line of intersection using again the amazing geogebra 3D graphing calculator so there's a link to it and uh there's the uh this I'm just using the same actual link as above so kttt Etc let's see this uh ktttfn and so on this is the same equation as added more terms uh just to add the line of intersection so we could solve this so there's a plane One X Plus y plus Z equals one there's the plane two x minus two y plus three Z equals one there's the angle 72 degrees and now what I did was I did a line just using the vector equation so there's the point line L and then the vector 5 negative two negative 3 from the vector equation I mean from the um yeah the vector one here from the cross product and then what happens is you'll see this is the plane this is the plane one this is uh plane one plane two and there's the line right there exactly on the intersection which is absolutely amazing yes it's a great way to double check your work all right now let's continue further and now look at node one so two linear equations can represent a line so since a linear equation in X Y and Z represent represents a plane and so linear equations of plane and two non-parallel planes intersect a line it follows that two linear equations can represent a line yeah so when you have two linear equations then The Intercept of that is is gonna be L can represent a line of those uh planes are not parallel which is epic so so the points x y z that satisfies both uh a1x plus B1 y plus Z One Z plus d equals one equals zero and A2 X Plus B2 y plus z c two Z Plus D2 equals zero these are linear equations yeah um yeah so the points x y z that satisfies both these linear equations these points rely on both of these planes and so the pair of linear equations represents the line of intersection of the planes if those planes are not parallel uh for instance in example seven the line L was given as the line of intersection of the plane's X Plus y plus Z equals one and x minus two y plus three is that equals to one and now the the symmetric equations that we found 4L could be written as yeah and then we could uh go and write these out first what I'll do is write down these symmetric equations from examples of seven so x minus 1 divided by five equals y divided by negative two and then equals z divided by negative three so let's write this out first X over 1 divided by five equals to Y over negative two equals to Z over negative three and uh yeah let's just double check that again so yeah x minus one over five y over negative two is that over negative three so we have this now what we'll do is we could take this and then write it write it separately and then we could write this as x minus 1 over 5 equals to y uh yeah y divided by negative two yeah so we could leave it like this this is actually a linear equation it's a box this out that's a linear equation and now the next one we could do is take this and then write it out as y as uh divided by negative two equals to Z over a negative three so we get this or uh just cancel all these negatives just write y over two let's make it simpler equals to Z over three now this is a linear equation both of them are exactly the same as our other linear equations just just written a bit differently and you can see that these are linear equations because uh we could just rearrange this or or not necessarily arrange it just look at this right here and then um it's separated out so that looks like our typical linear equation form so X over 5 minus one over five and then move this over to this other side we're going to have a plus we'll just shift this over to this side we're going to have a plus yeah plus y over 2 equals to zero so then the ABC uh yeah let's just shift this over actually now what I'll do is I'll move this out of the way and here I just move this out of the way uh move it on to the other side and then ship this over to this side and let's see all right let's do one more and then move this over to this side yes now we have that form that's exactly how we add a linear equation so X Plus y plus Z equals one but instead we have this uh one over five X plus one over two y That's So then the a is going to be one over five D is going to be one over two and then uh the Z One the C is going to be zero and then we'll have is negative five there so again these are linear equations or you can move this over to the right side it doesn't matter same thing uh and then likewise this one right here is going to be uh this one's already done for us y over two move this over the other side Z over three equals zero so again linear equations all right and uh these equations are yeah again are a pair of linear equations uh so they exhibit L as the line of intersection of the planes x minus one divided by five equals y divided by negative two and Y over two uh equals to Z over three so this part right here either or um and then here is uh yeah as in the figure below is for my calculus book so there is the line of intersection L and there's the two planes so this is this top vertical one is this x minus one divided by five equals y minus and a y divided by negative two and then this uh slanted one is this y divided by two equals z over three and then we have an equation of a line right there so the above figure shows that the line L and example seven can also be regarded as the line of intersection of planes from its symmetric equations which is quite epics up so these are different planes from those other ones from example seven so we can graph out the four equations of the planes two from example seven and two from the symmetric equations and show that they all intersect and intersected at the same almond Lane which is hashtag amazing so again from geogebra it's the same link there I just added more terms so we're going there's a plane one from I'm plane two of example seven and now here's a plane from symmetric equation one so x minus one divided by five equals y uh divided by negative two and there's the symmetric equation two y over two equals z over three so you could write it either or you could write it as this wait what is this this format and it'll just solve everything and there's the angle 72 degrees and I actually I I hit it so that's why it's not selected so I didn't select it just so get it out of the way anyways uh and now we have the line number the line is the same one from example seven um the point one zero zero and there's that cross product five negative two negative three that uh that parallel vector and now there's the uh four uh equation so there's the red and purple are the symmetric equations so this is yeah this red and purple are symmetric and they're different from these blue and green ones and notice how they're all uh exactly exactly intersect at the same line L which is absolutely uh remarkable yeah that's L absolutely amazing and now going further all right so uh yeah in general uh when we write the equations of a line in symmetric form such as this one right here x minus X not over a equals to Y minus y naught over B and again I like writing T equals just so that we always remember it's the parameter or set it all equals on for all of them and then you have Z equals z naught over c yeah so if we have the symmetrical equations as equation of a line in symmetric form then we could regard the line as the line of intersection of the two planes and again you just separate uh these across and you should also get another plane this one here and this equation equate those yeah I believe that should also work as well all right yeah so then anyways going further so it's going to be intersection of the two planes so we'll have X minus X naught over a equals y minus y naught over B and and then I'll just shoot this over all right shifted it over and we also have this one y minus y naught over b equals to Z minus Z naught over C all right uh now here's another note that actually alluded to it earlier is by using well this equation and this one you know those equating it as well and that is right here so no two line of intersection from two parametric planes again this is a Titan node one so note that another way to find the line of intersection is to solve the equation of the planes for the two of the variables in terms of the third which can be taken as the parameter so in our earlier example seven we had x minus one over five equals to Y over negative two this equals to Z over negative three yeah so what we could do is when we take this and then equate it to this and then also take uh this one and equate it to this one here so what we'll get is uh let's just write this out uh we'll get x minus one over five equals to Z over negative three and if we solve this I will multiply this 5 on top and then shift this over we're gonna get X is going to equal to well 5. I also put the plus one in front one plus uh and one it's gonna be one minus because there's a negative 3 there 1 negative five over three set like that so that's one equation and then the next one is going to be right here and that's going to be what we already used to that one y over negative 2 equals z over negative three I'll write this better and negative three and then multiply this inside we're going to get y equals two plus two over three Z so the parameter is z right here parameter equals that or a parameter yeah parameter equals to um is that and uh yeah but yeah basically the only difference is uh we have this equation now because this one here is the same thing as the other one we had uh that we graphed out it's gonna be yeah it's gonna be the same exact plane that's this one right here so y divided by negative two equals that divided by negative three or this equals to Y over two equals z over three but now we have another equation so all we're doing is adding another plane and that's this one right here because we equated this with the Z so that we get that plane so now I have another one here so I added a parametric equations uh I mean from the symmetric equation so s 1.2 so second one x equals one minus five Z over three then we have as 2.2 is this right here two over three Zed and these are the same exact colors because they're the exact same planes all you're doing is Shifting the two over to this side all right and now we have an another plan so we added all of them together so now we have one two three four five six and so again we go one two three four and then there's five yes five planes right here one two three four five and then this one is the same as that yeah so we have five planes all intersecting at this one line absolutely amazing which is uh yeah quite fascinating um yeah fascinating Little Nugget there all right and now let's continue further and look at example eight and this one's on distance from a point to a plane so we're asked to find a formula for a distance D from a point P1 with coordinates X1 y1 Z1 to the plane ax plus b y plus z z plus d equals zero so there's a linear equation of a plane yeah so let's take a look at the solution so let uh the p p naught X naught y naught Z not be any point in the given plane and let B be the vector corresponding to uh the uh the vector P naught P1 so P yeah P naught P one that'll be the vector B like that yeah so in other words what we get is well B Vector is equal to uh this is going to be um let's write this uh like this P naught p a uh P naught p one like that the vector this equals to again you just take a difference so we'll go X1 okay so to subtract the components X1 minus y y naught y1 minus y naught Z one minus Z naught like that all right so now that we have this so from the figure below you can see that the distance D from P1 to the plane is equal to the absolute value of the scalar projection of B onto the normal Vector n equals to A B C so let's make sense of all this and recall from our earlier video scalar projection I put a link to that one there yeah so let's just make sense of all this and let's graph stuff out so we'll have a plane like this all right uh planes can be like this all right so uh the point p is a point P of P1 to the plane so the point P naught is on the plane so let's say we had a point P naught here uh this one is on the plane so this is p naught is on the plane let's say we had a point not on the plane or just any any random point this is going to be our P1 like that so that's P1 and then the vector B is from P naught to P1 so this Vector right here this is B a vector like that and now uh so that's the plane so now we have a normal Vector here n and what we'll do is from the figure below we can see that the distance d uh the distance D is going to be well from let's say you had a point right here across here or uh before we get to that so let's look at this Vector B now if we project this on uh onto the normal Vector n so the absolute value is going to be the distance so in other words if you project it all the way across to this point right here also if we project it all the way across here to this it's going to be a right angle so we have this project this all the way on here and then this angle is going to be let's call this Theta and then this uh from here to here is going to be n so there's the normal Vector F so then the distance right here is going to be from here to here is d that's the distance from there to there it's just a scalar projection of this B onto it over there and first off recall and we're doing the absolute value recall right here if you look at just a basic trigonomic bunches if we take the absolute value of cosine here cosine is going to be equal to adjacent over hypotenuse and adjacent's D equals D over hypotenuse is the absolute value of this distance B like that yeah so then this is uh so then a d is going to be equal to so if this over it's going to be like this uh the distance uh the length of the the hypotenuse B Times by the absolute value of cosine we're doing absolute values just because we want the absolute values of all of these cosines and so on yeah since we're dealing with absolute values I'll recall that this is from our earlier video just a scalar projection of B so this is this scalar and it's and the way you wrote it right it is component of uh of this is a component of b onto n or the scalar projection that's how you write it but basically it's just this part right there and also recall that this angle right here of these two vectors uh B and N that's just a DOT product so we can also apply the dot product uh remember that cosine Theta is equal to B I was right recall this and cosine Theta is equal to the dot product of B Dot N the two vectors with the angle between them and then divided by the length B Times by length n like that so then put this all together what we end up getting is D equals to this right here com and the scale of projection is just the notation of how it's written so this is the component of b onto N I believe that's how it's pronounced all right so we have this and then we're taking absolute value and then it's going to be the absolute value of that it's going to be now um yeah it's going to be this B right there the length B then there's cosine we put that inside uh that's what we'll put this inside here so it goes down here this is going to be B of vector Dot N divide Us by there's a B on the bottom like this and there's the n all right yeah computer just froze for a bit anyway so we have this and this cancels this cancels so we're left with uh and this is again we're doing an absolute value so this whole thing is going to be absolute value we'll put that in there so we get uh B Dot N dot because we're doing absolute value of the distance and then divided by the length n like that that's what we have so anyways this equals two now this equals to the uh dot product absolute value of the dot product is going to be uh you have the B1 right here is uh just x minus yeah it's just x one minus X naught Etc this is right all for completeness so it's going to be X1 minus X naught y1 minus y naught Z one minus Z naught then dot then the vector n is going to be ABC yeah this uh n is just going to be the ABC yes per here so normal vectors A B C like that all right it's going to be a b C and then absolute value divided by uh the N the bottom is going to be the length it's going to be a squared plus b squared plus c squared of the normal background uh yeah so then now so then what we end up getting is we'll multiply the a inside keep going so we're going to get a um put the triangle bracket in front so we're gonna get a x actually we're not doing the triangle bracket because we're doing a DOT product so the dot product is uh is a scalar so then it's going to be a x one minus X naught plus b y1 minus y naught plus c Z 1 minus Z naught like that absolute value and then divided by and again it's frozen so divide this by scalar a squared plus b squared and plus c squared like that so that's what we have and then what's next is what we could simplify this even further so multiply it out just similar to how we made the linear equation for a plane all right so this equals to um this is going to be ax one multiply all these together in the other ones separately plus b y 1 plus c Z one and then we're going to have a minus uh this is going to be minus and then put a bracket ax not and then that next one here is going to be minus and there's a b yeah plus b y naught so we took the negatives out plus C is that not like that and again divide this by a squared plus b squared plus c squared like that all right so now that we have all this so since P naught lies in the plane its coordinate satisfies the equation of the plane so we have a x not plus b y naught plus c z naught plus d equals to zero yeah now we can see that above here so we again we were uh we let P not with coordinates X naught y naught Z naught B any point on the given plane right here so if you plug in these X naught y naught Z naught then then yeah that just equals to zero uh the plus d equals zero so in other words this part right here let's move this like this let's make some room so then what we'll get is if we write it like that we'll get if you make this whole thing over to the other side we're gonna get d equals to negative a x o plus b y o plus c z o like that all right so we have this so this is D so this whole thing is going to be D now yeah so this is d right there yeah so that's what we have all right so thus the formula for D can be written as well we're gonna have D equals to the absolute value of a x one plus b y 1 plus c z one and then plus d an absolute value and divided by the length of the normal Vector n which is a a square root a squared plus b squared let's see squared so let's box this whole thing in epic stuff all right let's continue further and take a look at example nine this utilizes this distance one so we were asked to find the distance between the parallel planes a 10x plus 2y minus 2z equals five and five X Plus y minus that equals one so let's take a look at the solution here first we know that the planes are parallel because they're normal vectors so here's a normal one here 10 2 negative 2 and 5 1 negative 1 are parallel that is there are multiples of each other they're just multiplied over here by two so two times five is ten two times one is two two times negative one is negative two or to find the distance between the planes we choose any point on one plane and then calculate its distance to the other plane so we just need to find one point as a result we have two planes like this two yeah parallel planes for example I just we just need to find any point here and then apply the distance formula this plane so we use a Point here and then plane here for example all right all right so in particular if we put y equals z equals zero in the equation of the first plane we get 10x equals five and so one half zero zero is a point on this plane and we could see this so if we look at the first plane right here 10x plus 2y minus 2z equals 5. so we'll go plugins that equals y equals zero so ten X plus 2y is going to be L plus zero plus zero equals to 5 like that so we get uh 10x equals five in other words and move this divide this out x equals two one half like that and then we're going to get the the point one half and zero and zero all right so that's so we could apply the distance formula so that's by the distance formula the distance between this point one half zero zero on the first plane and then that to the second plane and the second plane 5x plus y minus Z equals one uh equals this is minus minus Z equals one not equal zero equals zero no actually now this was a correct uh because what we did uh what a caucus book just did right here is it just took this negative one and is this one and moved it over so to negative one in other words we have this format like that so that's going to be d d is equal to negative one foreign so let's just plug this into the formula so we get D equals to let's write this whole thing out first ax1 plus b y 1 plus c z one plus d absolute value over a squared plus b squared plus c squared and it's going to be on this plane 5x plus 5 8 plus y minus plus uh I mean minus Z minus 1 equals zero so that means the a is just going to be the terms in front of the variables so it's going to be 5 and then X I mean uh yeah x one right here is going to be well uh the point yeah on the first plane this is the second place we're getting a point on the first plane that's going to be this right here one half zero zero what's going to be times it by one half and then it's going to be plus uh or one there b is 1 times 0. plus C is a minus C then it's going to be minus C which is one Times by zero and then plus d which is negative one minus one like that all right so we have all this and then that's absolute value and then divide this by the length of the normal Vector 5 squared plus one squared and there's a negative one that's going to be plus 1 squared like that yeah but for completeness I could put a negative 1 squared same thing negative 1 squared so what we end up having is a five over two plus zero plus zero I mean minus zero this will be minus one like that and then this absolute value sign but then five over two is bigger than one we can just remove that absolute value sign so it's going to be like this uh square root 5 is 25 plus 1 plus 1 27 so square root 27 and uh erase this yeah this equals two let's multiply this top and bottom by two so then same common denominator so 5 minus two is three over 2 like that and then this part right here is interesting so what I'll do is this 27 that's just the same thing as writing 3 times 9. and then we know that 9 square root of 9 is 3. so that means we could have a 3 over 2 over square root 3 times by three so this cancels like that and we're left with uh one yeah this is going to be 1 over 2 square root of three like that and here my caucus book actually simplifies this even further it doesn't like having a square root at the bottom I'll also this over here so it does is multiply the top and bottom by square root 3. so then basically what we'll get is this part of here will be square threes will become three I got so this will become square root 3 over 2 times 3 is equals to square root of three over six like that all right and now going further uh we can calculate this out so the distance between the planes is square root three over six which equals to square root uh you can just type this out in the calculator right here sqrt or square root three over six and using my again one note calculator space equals to 0.2886 Etc keeps going on it's pretty amazing I've built in a OneNote calculator anyways let's go for this let's graph the two planes and calculate its distance using the amazing Gog bro 3D graphing calculator as always it even has distances from a planes so here's the plane one 10x plus 2y minus two Z equals five equals five and here's plan two and green 5x plus y minus Z equals one and calculate the distance a I just type distance bit from between plane one plane two I just point two two nine here's rounded up two eight eight rounds up two nine and there's two planes there's the a plane one is red plane two is green so the notice that are parallel and if you turn it around you can see that there are indeed parallel and the distance from here to here's point uh not two nine absolutely amazing stuff all right all right now let's continue further and look at example 10. so in example three we show that the following lines are skew in other words they don't intercept and they're not parallel so we have uh right here line one is x equals one plus t y equals negative two plus three T and then Z equals four minus t and line two is uh x equals two s and Y equals three plus s and Z equals negative three plus four S so this is line two and there's line one so now we're asked to find a distance between them all right so solution since the two planes uh L1 and L2 are skew RSU they can be viewed as lying on two parallel uh planes P1 P2 for example it just means that the you'll have a line like this so they're not parallel but then so they never intersect and they're not parallel I mean yeah they never intercept uh so it's so imagine one lines on this plane one lines on that plane then we can find a distance between them all right let's go further so let's take a look at the solution so since the two lines L1 and L2 are skewed they can be viewed as lying on two parallel planes P1 and P2 the distance between L1 and L2 is the same as the distance between P1 and P2 which can be computed as an example and that's exact exact same thing here's find a point find equations of plane and solve so the common normal Vector to both planes must be orthogonal to uh to both we had to both V1 which is the direction angle of a direction Vector of L1 V1 is going to be the in front of the T So 1 3 negative one so one three negative one this direction of L1 and V2 which is 2 1 4 direction of alt is a 2 as a 2s 1 4 so that's the direction of that line two yeah so the common normal Vector must be orthogonal to both directions so if if it's in if it's on two planes like this and like this if this is One Direction uh let's say V2 this one is V1 then then the cross product is going to be a perpendicular and like that and it's going to be perpendicular to both like that I'll just visualize this and so a normal Vector is orthogonal to both and that is again that's the cross product so n a vector equals to V1 cross V2 which equals to our trusty I J okay like that this is going to be the first vectors 1 3 negative one one three negative one and the next one is two one four like that and now let's take the cross product as always is equals 2 cross the cell cross this out we get 3 times 4 is 12 minus negative 1 times 1 is is negative one so we have a plus one this is a negative there negatives becomes positive and then the next one is as always let's erase this one all right uh now we draw a line and the middle one and then do the same thing there but now we subtract so one times four is four and then minus is important negative already so it becomes positive one times two is two this can be our J and then the next one is plus and erase this next one is going to be erase the K I cancel out this so one times one is one and then minus three times two is six like that all right so now the next step is it's going to be 13. I minus 6 J then minus 5 K like that all right now I'm going to box this in like that all right so now we have this normal vector and the next thing that we could do is well if we put s equals zero in the equations of L2 just to get a point on it uh we get the point zero three minus 3 on L2 and so in equation for P2 is well let's just write this down first so if L2 so at s equals to zero we get well if we take a look at the equations um yeah so X is 2s X is 0. so x equals to zero the next one is y is equal to just to double check here the next one is um we have 3 plus s and the next one is negative 3 plus 4S so 3 plus s yeah s is 0. just leave it like that the next one is z is equal to negative three plus four S so that's going to be zero yeah so negative 3. so that is correct yeah so we have this point is going to be zero three negative 3. it's going to be on the line uh yeah line two and so an equation for P2 is so P2 let's write this down P2 planes the plane uh it's going to be well we're going to write the normal Vector it's a there's a normal Vector 13 negative 6 5 the components and there's the point on this uh on this line and we're going to say we're just going to form a parallel plane with this line on it so that's going to be over that point yeah so this point here on the line and that's going to be 13 it's going to be x minus zero and then minus 6 y minus 3 and then and then this is going to be yeah minus 5 is this value there that's C minus five Z minus the negative three plus three make that equals to zero all right so this is the equation of the plane but we could simplify it further this equals to 13x and zeros there's eliminated next next one is going to be negative six Y and then 5z and then move this over inside so negative 6 minus three and times that is going to be plus 18. all right this can be plus 18 and then uh the next one is going to be negative 5 times 3 negative 15 this equals to zero like that so uh this becomes just three that's 13x minus six y minus 5z plus three equals to zero so this is our equation of the plane 2p2 all right here I just box that and so now if we set T equals zero in the equations uh for L1 we get this point here so we have this plane right here for P2 so now we just need another point on P1 and then we get to solve for the using distance formula from point to the plane so if we set T equals zero in the equations for L1 we get the point one negative two at four uh on P1 and let's just double check L2 at T equals to zero we get and if we scroll up let's take a look here so if we plug in 0 we'll get x equals one this goes to zero so then y equals negative two and then Z equals to four yeah so that is for X of 1 negative two and four so x equals to one y equals to negative two and Z equals to four yeah so that's the point is going to be one negative two four on this is on L1 not L2 L1 all right so the distance between L1 and L2 is the same as the distance from this point right here on L1 1 negative two four to the plane right here on Parallel to L2 so it's 13x minus 6y minus 5z plus three equals zero so thus by the distance formula distance is D equals two this is going to be uh absolute value of a is 13. um yeah 13 now we're going to multiply the exponent of the point so it's one and then it's going to be plus but this would be minus right here is six uh the negative six there's a negative six and then it's going to be negative 2 is right here and then this is going to be minus 5 and then this is 4 right there and then plus d is three like that and then divide us all by the distance or the length of the normal Vector so 13 squared plus negative 6 squared plus negative 5 squared and and let's uh let's just do some uh algebra fun here so solve all of this this is going to be 13 times 1 is just 13 minus is going to be plus 12. 6 times 2 is the other negative become positive and then minus 20 plus 3 like that and then the bottom part here well what I'm going to do is I'm going to just put a side a bubble right here so we'll just just multiply on the side right here is 13 times 13. uh to get the squared is going to be well 3 times 3 is 9. 3 times 1 is 3. put a zero just doing long multiplication uh one times three is three one times one is is one and now we're going to add these up nine six one let's add so this is equals to one six nine plus six times six is 36 negative six times negative six is also 36. negative becomes positive negative five squared is just five squared equals plus 25. like that all right and uh let's do some more addition here so 13 plus 12 is 25 yeah so just add a 10 23 plus 2 is 25. and this right here is 17. negative 17. so negative 20 plus 3 is uh negative Seventeen and the difference 25 minus Seventeen that's just uh well three uh goes to 20. and then five yeah so uh 20 uh minus five this is M25 minus five is twenty minus another three Seventeen so that's five plus three is eight so that's going to be positive that's eight and now 169 plus 36 put it over here it's 36 so 6 Plus 9 is 15. put an add carry the one and then this is going to be 6 plus 1 I have 6 plus 3 is 9 plus the one to carry the 1 is going to be zero it's a ten and then this can be two two of five and then added 25 plus 25. 5 plus 5 is is ten carry the one this is going to be three this would be two this equals to 8 over square root twenty two thirty all right and I'm going to put the D right here and then this is going to be um approximately equal to when you solve it with a calculator 5.53 a box this whole thing in all right and going further let's do a calculation check right here uh so so we have right here 13 minus uh six times twelve that's this part right here 6 times negative 2 minus 5 times 4 plus 3 and then use a calculator delete press equals eight that's exactly right this next one is 13 squared um yeah so 13 squared plus negative six squared just plus six squared plus 5 squared and this equals two space 230 and then lastly 8 divided by square root 30 raise this equals space you'll you'll see this 0.527 round of the 0.53 so epic epic stuff there so now let's graph and double check our calculations using again the amazing Gog bra 3D graphing calculator here's the inputs I did and yeah here is the link right here you can play around with it and uh so there's the uh I hit these vectors away but anyways um going back here yeah I believe this the skew lines were from example three I believe but anyway so here's the line so what I have graphed here is line uh L L one right here with a point one negative two four and then it has that uh parallel vector or the direction Vector V1 that we had one three negative one and this other one was two one four and let's just scroll back to those Direction vectors there's V1 right here V1 is one three negative one V2 is one two one four and again that's you could see that from the equations from the parameters so one three negative one two one four all right now that we have that so that's the first line there's a second line and then the second line is 0 3 negative three there's two one four um the yeah the direction Vector there anyways and then there's a cross product I did a double check of the cross product V1 uh V2 so that's V1 V2 and we get 13 negative six negative five so you can even do a cross product check as well as 13 negative 6 negative five it's absolutely amazing all in one calculator right here and there's the plane one uh 13 x minus 6y minus five Z minus five uh equals zero this one I actually had to solve it also I'll solve it soon and then there's the other one that we had P2 13x minus six y minus five Z plus three because uh because the calculator here only could do distances from between planes not between point and plane for some reason the distance of P1 P2 equals 0.53 so you could actually solve the distance directly and here it is when you graph it you'll see the line two is parallel to this plane this one's line one's parallel to the outline and shift it around looks like that and there's the normal vector and you could also draw the normal Vector uh n right here with the cross product you can graph the cross product uh like that and again uh yeah so that's uh what it is and it's again it it confirms our calculations so note that in jujuba the distance from the recorder that I know the equation of both planes and in which I solved the first plane as follows so the first planet going back up here so this is 13x minus 6y minus 5z minus 5 and to solve that well we need a point on uh the plan so uh then that point is going to be 1 negative two four and we know and then we just use a nodal light so we can just solve that ourselves all right so let's go back here so I solved it as follows we had a point right here this is going to be 1 negative 2 4 is going to be on L1 and on the plane so then P1 is going to be 13 uh this is going to be at 13 negative 6 negative 5 is the um uh is the normal Vector so p x x minus 1 then minus six y then minus negative 2 this would be plus 2. and then minus five Z minus four equals zero solve this out this will be 13x and then 6y over here then minus 5z then move this over so plus 13 out of negative 13. all right this can be uh yeah negative 13 not plus 13. negative 6 times 2 is negative 12 and then negative 5 times negative 4 is Plus 20. equals zero and let's do some addition this is negative 13 minus uh negative 12. uh this is just negative 25 and then negative 25 plus 20 that's going to be negative 5. and there's our equation 13x minus 6y minus 5z minus 5 equals to zero and yeah and you could uh confirm that it's correct because the distance formula the distance gets it correct so we're going back up here scroll up there's that 13x negative 6y negative 5 is that negative five and then we plug in it to the calculator P1 P2 gets 0.53 exactly as we calculated with the point two distance uh near point two plane distance formula whereas this one just uses plane to play but it probably uses variation of the same formula all right and now finally let's take a look at the exercise one and this uh yeah last uh last example of this uh video so if a B and C are not all zero show that the equation ax plus b y plus c z plus d equals zero represents a plane and a b c is a normal Vector to the plane so we're just given this equation and we're given this D value so that it's uh actually a plane so hence suppose a is not equal to zero and rewrite the equation in the form a uh times X plus d over a plus b of times y minus zero and then plus Z times Z minus zero equals zero so solution so since a B and C are not all zero uh then we can assume that then one of them is not zero so let's assume a one is not a zero if it's not a we could do B C or whatever it doesn't matter so let's just pick a and the result will yeah the result will mean that at least one of them is true so hence the equation would be right if it is all true so anyways so what we have is let's say we have uh assume a is not um equal to zero so ax plus b y plus c Z so we have this and then plus d like this uh and then what I'm going to do is I'll multiply the top and bottom by a and this is equals zero so now we multiply top and bottom we're not changing anything and now we'll factor out the A out of there so that we get a X plus d over a and then do the exact same format over here but with the zero so y minus zero doesn't change anything plus c z minus zero equals zero all right and now we can rewrite this as a DOT product so we'll just uh yeah this is the same thing as writing it as um right here a b c Dot and then it's going to be X plus d over a and then y minus zero Z minus zero equals zero yeah because when you take the dot product you just multiply a into inside this and B inside this and and then see inside this one here and we get this top part right here which can be Rewritten back into this format right here so this one right here all right so now we have this uh this one here is our normal Vector because it's the dot product zero so this can be our normal Vector n all right so now we have that we can continue further and then separate this addition right here our subtraction uh remove those so what we're going to get is n Vector Dot and this one here is the same thing as writing um like this uh X I'll put a bracket here so uh X y z so move this x y z out and then this is going to be um plus the point d a well actually instead of using plus we'll use negative so then if we're taking a negative and then to get a plus there we need to also have this negative here just so that's it's familiar with the equation of a line that we had negative D over a and then 0 0 so we're subtracting everything like that equals to zero like that now this one here remember this is just our general uh position Vector at any point it's going to be r this one right here is going to be our R naught so the point on the plane I mean on the line all right so that's what we get is well this is going to be our n Dot then we can do R minus r not like that equals zero and then the other format is uh yeah so we can just we can stop it here or uh we could uh multiply this inside n r yeah so n dot R minus and Vector r dot like that equals to zero so we could do this or and then move this over uh put this order right here or yeah or finally move this over it's gonna be n dot R equals to n dot R naught put this like that and uh yeah these are all just the vector equations of uh of a plane so thus we have shown that the given equation can be written as a vector equation of a plane with normal Vector n equals A B C and R minus r naught representing a vector on the plane so there's this R minus r not there's fascinating fastening so anyways yeah that is all for today hopefully you learned and enjoyed another epic long math video and again as always I'm going to upload sections of these in uh yeah in a playlist format so you can watch uh yeah short clips of each segment and and so on and it's easier also to link back to them as well so I can just link back to a specific video as opposed to this giant long video for example this one here is a section that I uh that I took from the dot product and made it a separate video on the equation for the geometric interpretation of the dial product as opposed to the full like multi-hour dog product video and you have to sift through it but even in that video I put the um timestamps in it as well anyways that is all for if we learned and enjoyed and again the notes to all of these uh yeah the the link to all the notes will be in the description below in both PDF format and article format and timestamps will be there in the link will be to the uh sections playlist as well as the main uh 3D geometry playlist video or the geometry of space uh let's see what vectors in the geometry space play is all in the description below and also make sure to shout out to make sure follow my um telegram for Real Time research that I post everything on there conspiracy stuff science stuff math stuff a lot of stuff anyways thanks for watching and stay tuned for another math easy solution