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Current time:0:00Total duration:12:12

Video transcript

what I want to do in this video is start with some point that's not on the plane or maybe not necessarily on the plane so let me draw let me draw a point right over here and let's say the coordinates of that point are X naught X sub 0 Y sub 0 and Z sub 0 or it could be specified as a position vector I could draw the position vector like this so the position vector let me draw a better dotted lines the position vector for this could be X naught I plus y naught J plus Z naught K it specifies this coordinate right over here what I want to do is find the distance between this point and the plane obviously there could be a lot of distance I could find the distance between this point and that point and this point and this point at this point at this point now when I say I want to find the distance I want to find the minimum distance and you're actually going to get the minimum distance when you go when you go the the perpendicular distance to the plane or the normal distance to the plane and we'll hopefully see that visually as we try to figure out how to calculate the distance so the first thing we can do is let's just construct a vector between this point that's off the plane and some point that's on the plane and we already have a point from the last video that's on the plane this X sub P Y sub P Z sub P so let's construct a vector here let's construct let's construct this orange vector that starts on the plane its tail is on the plane and it goes off the plane I'm going to do that in orange it goes off the plane to this vector that's to this to this position X naught Y naught Z naught so what would be what would be how could we specify this vector right over here well that vector let me call that vector well I'll just call that vector what letters have I not used yet let me call that vector F vector F vector F is just going to be this yellow position vector minus this Green position vector so it's going to be this each of these chord this X component is going to be the difference of the x coordinates its y-coordinates squared be the difference of the y coordinate is going to be X naught X naught minus X sub P I subtracted the x-coordinates i plus y naught mine YP j+ go to the next line plus Z naught minus Z P minus Z P K so fair enough that's just some vector that comes off of the plane and on to this point but what we want to find out what we want to find out is this distance we want to find out this distance in yellow the distance that if I were to take a normal off of the plane and go straight to the point that's going to be the shortest distance and actually you can see it visually now because if you look at we can actually form a right triangle here so this this base of the right triangle is along the plane this side is normal to the plane so this is a right angle and you can see if I take any point any other point on the plane it will form a hypotenuse to on on a right triangle and obviously the shortest side here or the shortest way to get to the plane is going to be this distance right here as opposed to the hypotenuse this side will always be shorter than that side so given that we know this vector here how can we figure out how can we figure out this length here how can we figure out this length here in blue well we could figure out the magnitude of this vector so we could this the length of this side right here is going to be the magnitude of the vector so it's going to be the magnitude of the vector F that will just give us this length but we want this blue length well we could think about it if this was some angle and on the writing ceiling small if this was some angle theta we could use some pretty straight up pretty some pretty straightforward trigonometry if the distance under question is D you could say cosine of theta we could say that cosine of theta is equal to the adjacent side over the hypotenuse or is equal to D D is the adjacent side is equal to D over the hypotenuse well the hypotenuse is the magnitude of this vector it's the magnitude of the vector F or we could say the magnitude of the vector F times the cosine of theta I'm just multiplying both sides times the magnitude of the vector X F is equal to is equal to D but that's still you might say okay well Sal we know what Z is we could we could are sorry we know what f is we can figure that out we could figure out its magnitude but what we don't know what theta is how do we figure out what theta and to do that let's just think about it a little bit this angle this angle theta is the same angle so we don't know this distance here isn't necessarily the same as the length of the normal vector but it's definitely going in the same direction so this angle here is really the same thing as the angle between this vector and the normal vector and so you might remember from earlier linear algebra when we talk about the dot products of two vectors it involves something with the cosine of the angle between them and to make that fresh in your mind let's divide let's multiply and divide both sides let me multiply and divide the left side of this equation by the magnitude of the normal vector so I'm obviously not changing its value I'm multiplying and dividing by the same number so I'm going to multiply by the magnitude of the normal vector and I'm going to divide divide by the magnitude of the normal vector so I mean I'm I'm just essentially multiplying by one so I have not changed this but when you do it in this it might ring a bell this expression up here this expression right here is the dot product of the normal vector of the normal vector and this vector right here F so this right here is the dot product this is n this is n dot F up there it's equal to the product of their magnitudes times the cosine of the angle between them so the distance that shortest distance we care about is a dot product between this vector the normal vector divided by the magnitude divided by the magnitude of the normal divided by the norm magnitude of the normal vector so let's do that let's take the dot product between the normal and this and we already figured out in the last video the normal vector if you have the equation of a plane the normal vector is literally its components are just the coefficients on the x y&z term so this is a normal vector right over here so let's take let's literally take the dot product so n n dot F is going to be equal to to a times X not minus XP so it's going to be equal to I'll do that in pink so it'll be a X not minus a XP and then plus B times the y component here so plus B why not I'm just distributing the B minus B Y P and then plus do another color here and this C is too close of a color plus C times this come Z component so plus C Z naught minus C minus C Z P and all of that over the magnitude of the normal vector so what's the magnitude of the normal vector going to be it's just the square root of the normal vector dotted with itself so it's just each of these guys squared added to themselves and you're taking the square root so it's the square root the square root we I can do a nicer looking radical sign than that it's the square root of a squared plus B squared plus plus C squared now what does this up here simplify to let me just rewrite this so this is this is the distance in question this right here is equal to the distance but let's see if we can simplify it so first we can take all of the terms with the X naught these are involved the point that sits off the plane remember X not Y not Z not set off the plane so this is a X naught plus B Y naught plus C Z naught and then what are these terms equal to what are these terms over a X P a X negative a XP minus B YP minus C Z P well if you remember here D in the equation of a in the equation of a plane D when we started in the last video when we tried to figure out what the normal to a plane is D is if we this this point XP sits on the plane D is a XP plus B y plea plus B YP plus zc zp or another way you could say it is negative D negative D would be negative a and it's just the difference between lowercase and uppercase here right we say that lowercase a is the same as this uppercase a so it's negative a XP minus B YP minus c zp I'm just using what we got from the last video this is what D is so negative D will be this business and that's exactly what we have over here of negative a XP negative B yp- C Z P so all of this term this term and this term simplifies to a minus D and remember this negative capital B this is the D from the equation of the plane not the D not the distance D so this is the numerator of our distance and so and then the denominator or distance is just the square root of a squared plus B squared plus C squared and we're done this tells us the distance between any point and a plane this is a pretty intuitive formula here because all we're doing if we have some if I'm given you if I give you let me give you a give you an example let's say I have the plane if I have the plane 1 X minus 2y plus 3z is equal to 5 so that's some plane and let me pick some point that's not on the plane so let's say I have the point let's say I have the point I don't know let me say have the point 2 2 3 and let me make sure it's not on the plane so it's 2 minus 6 is negative negative yes this one let me just pick a random one so this definitely is not on the plane because we have 2 minus 6 plus 3 that gives us negative 1 which is not 5 so this is definitely not on the plane we can find the distance between this point and the plane using the formula we just derived we literally just evaluate it at so this will just be 1 times 1 times 2 1 times muses that same color one times 2 minus minus 2 times I'm going to fill it in plus 3 Plus 3 times something minus 5 all of that over and I haven't put these guys in let me do that right now so 1 times 2 minus 2 times 3 plus 3 times 1 this one minus 5 and kind of bring it over to the left-hand side all of that over all of that over the square root of 1 squared which is 1 plus negative 2 squared which is 4 plus 3 squared which is 9 so it's going to be equal to let's see this is 2 minus 6 or negative 6 and then you have plus 3 and then minus 5 so this is what this is 5 2 plus 3 is 5 minus 5 so those cancel out so this is negative 6 so it's equal to negative 6 over the square root of 5 plus 9 is 14 over the square root of 14 and you're done so hopefully you find that useful and hopefully we can apply this in other example problems