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Current time:0:00Total duration:9:58

Video transcript

what I want to do in this video is make sure that we're good at picking out what the normal vector to a plane is if we are given the equation for plane so to understand that let's just start off with some plane here let's just start off so this is a plan drawing part of it obviously it keeps going in every direction so let's say that that is our plane and let's say that this is a normal vector to the plane so that is our normal vector of the plane it's given by a I plus V J plus C K so that is our normal vector to the plane and let's say that we have so it's perpendicular it's perpendicular to every other vector that's on the plane and let's say we have some point on the plane we have some point it's the point X sub P I'll say P for plane so it's a point on the plane XP YP ZP if we pick the origin so let's say that our axes are here so let me let me draw let me draw our coordinate axes so let's say our coordinate axes look like that this is our z-axis this is let's say that's a y-axis and let's say that this is our x-axis let's say this is our x-axis coming out like this this is our x-axis you could specify this as a position vector there is a position vector let me draw it like this let me draw it like this then it would be behind the plane right over there you have a position vector that position vector would be X P I plus YP j plus z PK it specifies this coordinate right here that sits on the plane let me just call that something let me call that position vector I don't know let me call that let me call that P let me call that p1 so this is a point on the plane so it's P it is p1 and it is equal to this now we could take another point on the plane let's say let's say we're taking this is a particular problem plane let's say we just say any other point on plane X Y Z but we're saying that XYZ sits on the plane so let's say we take this point right over here X Y Z that clearly same logic can be testified by another position vector we can have a position vector that looks like this and dotted line it's going under the plane right over here and this position vector I don't know let me just call it let me just call it P instead of that particular that p1 this would just be X I plus y J plus ZK now the whole reason why did this set up is because I want to find given given some particular point that I know is on the plane and any other XY Z that is on the plane I can find I can construct a vector that is definitely on the plane and we've done this before when we try to figure out what the equations of a plane are a vector that's definitely on the plane is going to be the difference of these two vectors and I'll do that in blue so if you take the yellow vector minus the green vector or take this position or you take you will get the vector that if you view it that way that connects this point in that point although you can shift the vector but you'll get a vector that definitely lies along the plane even if you so if you start one of these points it'll definitely lie along the plane so the vector will look like this and it would be lying along our plane so this vector lies along our plan that vector is P minus p1 this is the vector P minus p1 it's this position vector minus that position vector gives you this one or another way to view it is this Green position vector Plus this blue vector that sits on the plane will clearly equal this yellow vector right heads to tails it clearly equals it and the whole reason why I did that is we can now take the dot product between this blue thing and this magenta thing and we've done it before and they have to be equal to 0 because this lies on the plane this is perpendicular to everything that sits on the plane and equals 0 and so we will get the equation for the plane but before I do that let me make sure we know what the components of this blue vector are so P minus p1 that's the blue vector you're just going to subtract each of the components so it's going to be X minus XP it's going to be X minus XP I plus y minus y PJ minus YP j plus Z minus zp K and we just said this is in the plane and this is this right the normal vector is normal to the plane you take their dot product is going to be equal to zero so n dot n dot this vector is going to be equal to zero but it's also equal to this a times this expression I'll do it right over here so this these find some good colors so a times that which is a X minus a XP plus B times that so that is plus B Y minus B Y P and then let me make sure I have enough colors and then it's going to be plus that times that so that's plus C Z minus c zp and all of this is equal to zero now what I'm going to do is I'm going to rewrite this so we have we have all of these terms I'm looking for right color we have all of the X terms ax remember this is spit this is any X that's on the plane will satisfy this so ax B Y and C Z let me leave that on the right-hand side so we have ax plus B Y plus C is Z is equal to and what I want to do is I'm going to subtract each of these from both sides or in other ways I'm going to move them all over let me do it let me not do too many things I'm going to move them over to the left hand side so I'm going to add positive axp to both sides that's the equivalent of subtracting negative ax P so this is going to be positive a X P and then we're going to have positive B Y P Plus do that same green plus B Y P and then finally plus C zp + C Z P is going to be equal to that now the whole reason why I did this and I've done this in previous videos where we're trying to find the formula or trying to find the equation of a plane is now we said hey if you have a normal vector and if you're given a point on the plane where in this case it's XP YP ZP we now have a very quick way of figuring out the equation but I want to go the other way I want you to be able to if I were to give you if I were to give you a if I were to give you a equation for a plane where I were to say ax plus B y plus cz is equal to D so this is the general equation for plane if I were to give you this I want you to be able to figure out the normal vector very quickly so how could you do that well this ax plus B y plus cz is completely analogous to this part right up over here let me rewrite over this over here so it becomes clear this part is ax plus B y plus cz is equal to all of this stuff on the right-hand side I'm sorry on the left hand side so let me copy and paste it copy and paste it so I just I just essentially flipped this expression but now you see this all of this this a has to be this a has to be this a this B has to be this B the C has to be this thing and then the D is all of this and this is just going to be a number this is just going to be a number assuming that you knew what the normal vector is what you're a B's and C's are and you know particularly so this is what this is what D is so this is how you can get the equation for a plane now if I were to give you the equation or plane what is the normal vector well we just saw the normal vector this a corresponds to that a this B corresponds to that B that C corresponds to that C the normal vector to this plane we started off with is it has the components a B and C so if you're given if you're given equation for plane here the normal vector the normal vector to this plane right over here is going to be a I plus B j plus plus CK so it's a very easy thing to do if I were to give you the equation of a plane let me give you a particular example if I were to tell you that I haven't some plane in three dimensions let's say it's negative three although it'll work for more dimensions let's say I have negative 3x plus square root of 2 square root of 2y let me put it this way - or let's say plus 7z is equal to PI so you have this crazy it's not crazy it's just a plane in three dimensions and I say what is a normal vector to this plane you literally you literally can just pick out these coefficients and you say the normal uh normal vector to this plane is negative 3i plus a square root of 2 plus 2 square root of 2 J plus 7 K and you can ignore the D part there and the reason why you can ignore that is that'll just shift the plane but it won't fundamentally change how the plane is tilted so a normal vector to this normal vector will also be normal if this was if this was e or if this was if this was 100 it would be normal to all of those planes because all those planes are just shifted but they all have the same they have the same inclination so they would all kind of point in the same direction and so they're normal vectors would point in the same direction so hopefully you found that vaguely useful will now build on this to find the distance between any point in three dimensions and some plane the shortest distance that we can get to that plane