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if the distance between the plane a X minus 2y plus Z equals D and the plane containing the lines and they give us two lines here in three dimensions if that distance is square root of six then the absolute value of D is so let's think about it a little bit they're talking about the distance between this plane between this plane and some plane that contains these two lines so in order to talk realistically about distances between planes those planes will have to be parallel because if they're not parallel if they intersect with each other the distance is clearly zero and they're telling us here that the distance is the square root of six so we have a situation so that the planes can't intersect they must be parallel so you have this plane up here you have this plane up here we could call this the equation here is ax minus 2y plus Z is equal to D and then you're going to have another plane that's going to be parallel to it maybe it looks something like this you have the other plane that is parallel and it's going to contain both of these lines so maybe it has this line so this line is in I'll say green maybe this line looks something like this so on that blue plane and then this line maybe in magenta is also is also going to be on the blue plane so how can we figure out the distances well a good starting point would be to try to figure out try to figure out the equation for this blue plane here and since these planes are parallel this equation should look very much like this orange equation at least on the left hand side it might just have a different D value and that's because it has the exact same inclination and then once we figure out the equation for this plane over here then we could actually probably figure out what a is that we could find some point then we could find some point on the blue plane and then use our knowledge of finding the distance between points and planes to figure out the actual distance from any point to this orange plane so let's figure out the equation of this blue plane first and a good place to start is just to try to figure out two vectors try to figure out two vectors on this blue plane then we can take the cross product of those two vectors to find out a normal to the blue plane and then use that information to actually figure out the equation for the blue plane so let's figure out let's figure out first some points that sit on the blue plane so on this green line right over here you have see if I want all of these to be equal to zero you'd have the point X is equal to one y is equal to Z is equal to three so you'd have the point 1 1 2 3 that definitely sits on the blue plane let's come up with another point let's see if I want all of these to be equal to 1 I could make this if I want this to evaluate to 2 I would have the point 3 I would have the point if I want this to evaluate to 1 I would want 5 minus 2 over 3 so 3 5 and then I would want this to be 7 7 minus 3 over 4 is also 1 so that's another point actually both of these points sit on this line right over here 1 2 3 and then 3 5 & 7 and then let's do the same thing for this let's find 2 points on this plane actually we just have to find one point on this plane because if you have three points that's enough to figure out two different vectors vectors that I had that aren't scalar multiples of each other which would be enough to figure out the normal to this plane so let's just figure out one more point over here and one more point would be if we want all of these three to be equal to 0 would be the point 2 3 4 2 2 3 4 because this would be 0 0 0 so that's also sitting on the play and that sits on the magenta line right over there so let's use these three points to figure out two vectors on the plane that aren't multiples of each other then we could take their cross product to actually to actually figure out a normal a normal vector to the blue plane so let's say the first vector a that sits on this plane let's say it's the difference in the position vectors that specify these two points and we know that'll be on the plane and so that will be 3 minus 1 is 2 I plus 5 minus 2 is 3 j plus 7 minus 3 is is 4 k so vector a vector a is actually going to be is actually going to sit on this green line because both of these points are on this line so it's going to sit on that line if we put it on the plane or if we were just started one of those points it'll sit on that line and then we could do another vector it's essentially going between a point on the Green Line and a point on the purple line but that's definitely going to be a vector on our blue plane and let's go between these two points that looks pretty straightforward so let's call vector B let me do another color so we don't confuse ourselves with that purple let's call vector B let's call that let's see 2 minus 1 is I 3 minus 2 is J and then 4 minus 3 that's just 1 K so this also this vector right here is also sitting on the plane so if I take the cross product of a and B I am going to get a vector that is perpendicular to the plane or a normal vector to the plane so let's do that so let's find what a cross B is a cross B is equal to and this is how I find it easiest I just write i j k this is really the definition of the cross product or i guess one of them and we write our first vector we have two three four and then we have our second vector which is just 1 1 1 and then this is going to be equal to first we look at the I component so across that row that column out 3 times 1 minus 1 times 4 so that's just 3 minus 4 so it's negative I and then minus we're gonna have the J so let me write a minus here minus we just swap signs we have positive negative positive so J get rid of that column that row 2 times 1 which is 2 minus 1 times 4 so that's minus 4 is negative 2 so we could write a negative 2 here but the negatives cancel out so it becomes plus 2 J and then finally for the K get rid of that row that column 2 times 1 is 2 minus 1 times 3 is 2 minus 3 which is negative one so it's negative K so this right here is a normal vector this right here is a normal vector is a normal vector to the plane so if we want to find the equation the equation for that plane we've done it multiple times we just have to take we just have to take the dot product of that normal vector we just have to take the dot product of that normal vector and any arbitrary any arbitrary vector on that that's specified that we can specify with with an arbitrary x y&z and we've done this multiple times in multiple videos if this is any point X Y Z that sits on the plane so this is any point XYZ that sits on a plane then the point then the vector let me draw the vector let's say we go to this point right over here so this vector right over here is going to be let me draw it the other way actually so this vector right over here let's say we're going between this point this point and XYZ this vector right here is going to be X minus 3i plus y minus 5 J plus Z minus 7 K that's what this vector is it sits on the plane assuming XY and z sit on the plane so if we take the dot product of this and the normal vector that has got to be equal to 0 because it sits on the plane and then we'll have our equation so let's call so let's take n dot that over there so n dot X minus 3i plus y minus 5 J plus Z minus 7 K if any of this is confusing to you I've gone into a little bit more depth in previous videos especially in the linear algebra playlist where I talked about constructing the equation of a plane given a point on the plane and the normal vector and even how do you find that normal vector so you might want to watch those if you want some review there but these are going to be equal to 0 so when you take the dot product and our normal vector is this so we just take the X term which is negative 1 times this X term right over here so negative 1 times this is just 3 minus X and then Plus this y-component times this y component so it's two times this so it's plus 2y minus ten and then finally the Z component negative one times this so this is plus seven plus seven minus Z is equal to zero and the what do we get so we have our negative x negative x plus two I plus two I minus Z and then is equal to let's subtract three from both sides so if we take it out there it'll be minus three if we subtract if we add 10 to both sides so then you have a plus 10 over here and then we subtract 7 from both sides this becomes a minus seven so then on the right hand side negative 3 plus 10 minus 7 that's just going to be zero that's just going to be zero and just like that we have we have the equation we have the equation for this blue plane over here the plane that contains that contains these two lines now remember what we said at the beginning of the video these two planes are parallel so the ratio of the coefficients on the X terms the Y term and the Z term has got to be the same and so this one has a positive 2 that has a negative 2 this is a just to simplify it so it looks very similar to each other let's multiply this equation right here both sides by negative one and then we're going to get X minus 2 y plus Z is equal to 0 so this is a completely valid another alternate way of expressing the same plane and what I like about this is it looks very similar to this at least the ratios of the X YS and Z's negative 2y negative 2y 1 z 1z and remember the ratios have to be the same so here we have a one to one ratio between the Z coefficient of the Z coefficient the Y coefficient and the Y coefficient so it's also going to be for the X coefficient so here we know if this is going to be parallel to the blue plane we know that a has got to be equal to 1 so this is X minus 2y plus Z is equal to is equal to D so now let's figure out let's figure out the actual distance let's actually figure out the actual distance between these two planes so what we can do is we can take a point on this blue plane and we have several examples of points on the blue plane and find the distance between that point and this plane over here and actually I just finished doing some videos on how do you find the distance between a point and a plane so I'm just going to use that formula if you want it to be proved go watch that video it's actually a pretty interesting proof I think but the distance the distance between let's say this prove this point one two three and this plane over here and this plane so this distance right here it's going to be in the direction of the normal the distance is going to be and I want to be careful not using that so all right I'll actually write out the word don't want to overload variables the distance is going to be you literally just evaluate you literally just evaluate this let me do it this way you you literally put in this point for the XY and Z and then you subtract the D in the numerator and we saw that as the formula for finding the distance so we're going to be 1 1 I'm actually using this point right over here it's going to be 1 1 because we just have one X so it's just going to be 1 minus 2 times 2 1 minus 4 that's 2 times 2 plus 3 plus 3 minus D well here the D is just D so we're just going to write minus D just like that all of that over what is essentially the magnitude of the normal vector and we saw in several videos that's just the square of the coefficients on each of these terms right here and taking this the sum of those then the square root so it's going to be equal to 1 squared plus negative 2 squared which is 4 plus 1 squared which is 1 so this is going to simplify to the distance is equal to 1 minus 4 plus 3 is 0 so in the numerator we have negative D all of that over all of that over the square root of 1 plus 4 plus one so all over the square root of six so they say the distance if the distance between the plane a X minus 2y plus Z is equal between this plane and this plane over here is square root of 6 so they're saying the distance is equal to is equal to the square root of 6 that's what this information right over here is maybe I should do that in another color so this distance right and that's not another color the distance between the two planes is going to be the square root of 6 and so then if we solve for D multiply both sides of this equation times the square root of 6 you get 6 is equal to negative D or D is equal to D is equal to negative 6 now what they care about is the absolute value of D or the absolute distance so that's actually this would be kind of the signed distance it kind of specifies whether we're above or below the plane since we're below the plane we got a negative number I just happen to draw it right if we were above the plane we would get a positive number so this distance is negative 6 the absolute value of it the absolute value of D which is the same thing as the absolute value of negative 6 is equal to 6 so and the take any point any point on this blue plane and you look for the closest point on the orange plane and they will be they will be exactly 6 apart anyway hopefully you found that interesting