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### Course: Linear algebra > Unit 1

Lesson 5: Vector dot and cross products- Vector dot product and vector length
- Proving vector dot product properties
- Proof of the Cauchy-Schwarz inequality
- Vector triangle inequality
- Defining the angle between vectors
- Defining a plane in R3 with a point and normal vector
- Cross product introduction
- Proof: Relationship between cross product and sin of angle
- Dot and cross product comparison/intuition
- Vector triple product expansion (very optional)
- Normal vector from plane equation
- Point distance to plane
- Distance between planes

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# Distance between planes

2010 IIT JEE Paper 1 Problem 51 Distance Between Planes. Created by Sal Khan.

## Want to join the conversation?

- At14:30, Sal claims that given any point in the first plane, the "closest point" on the second plane will be exactly 6 units away. However, wouldn't that mean that the distance between the two planes is 6? And doesn't the original question state that the distance between the two points is equal to the
*square root*of 6?

Unless I'm mistaken, the "d" in the video is merely the right-hand side of the 1st equation for the plane, which is*not*the same as the distance between the two planes. Am I right or am I missing something??(101 votes)- I think that you are right and that Sal messed up in that last part. The equation for the red plane is x-2y+z=-6 and the equation for the blue plane is x-2y+z=0. This means that the planes are parallel with the red one is shifted down. If we calculate the distance between the two planes with those equations we get: (1-4+3-(-6))/sqrt(1+4+1) and that is equal to 6/sqrt(6), if you multiply by sqrt(6)/sqrt(6) you get that the distance between the two planes is sqrt(6), which is what was stated originally.(77 votes)

- How does (x-1)/2 = (y-2)/3 = (z-3)/4 represent a line? Is there any tutorial on this? I can't find one.(24 votes)
- Any equation with highest power of 1 is not a line but rather a plane.

For instance ax + by +cz = D is a plane. A line needs to be defined by intersection of two planes or parametrically using vectors

See Parametric representation of a line video

In this question (x-1)/2 = (y-2)/3 = (z-3)/4 = t represents a line because it is actually giving you in effect three parametric equations, where t is any real number

if you want to define a line in R3 through <x1, y1, z1> and < x2, y2, z2> parametrically, then it will have three equations

x = (x1-X2) + x1 t

y = (y1-y2) + y1 t

z = (z1-z2) + z1 t

the above equations can be arranged as

{(x - (x1-x2)}/x1 = { y- (y1-y2)}/ y1 = {z- (z1-z2)}/ z1 = t

where t is user set parameter(16 votes)

- At7:00, since we just found the normal vector of the blue plane to be (-1, 2, -1), can't we just go ahead and say the equation for the blue plane is -x+2y-z=D. (From the Normal vector from plane equation video).

Plug in a point on the plane (1, 2,3) we get -1+4-3=D=0. So the equation for the blue plane is -x+2y-z=0?(27 votes)- Yes, we can. 😊

The equation -x+2y-z=D applies to all planes parallel to (and including) the blue plane. Simply plugging a point into this formula lets us specify a particular plane.(15 votes)

- whats with all the i, j, and k's?(6 votes)
- i, j and k are the unit vectors. "i" is the unit vector in the x direction, "j" in the y direction and k in the z direction. For example,

let v be a 3D vector.

v=3i+5j+6k

This means the vector has components 3 in the "x" 5 in the "y" and 6 in the "z"

Hope this helps.(20 votes)

- At0:45Sal says that the planes must be parallel since the distance between them is sqrt(6), but couldn't the planes be skew? Skew planes still have a distance from each other, but the distance is different at each point on the planes.(5 votes)
- The reason he says they must be parallel is because it doesn't say the distance between them is sqrt(6) at a specific point, it just says that that's the distance between them period, meaning at every point.(18 votes)

- I was wondering why 2 planes can't intersect at only one point. What if the tip of one of the planes touched the other?(1 vote)
- theres no tip when you're talking about planes. theyre not actually rectangles, theyre infinitely large. theyre just visualized as rectangles.(18 votes)

- why sal is taking point x and multiply with n whereas points are already available in plane(6 votes)
- Because you are aiming to result an equation , whereas x,y,z contained.

If you use the give two lines, their dot product will all be scalar, useless then.

Hope you see that.(7 votes)

- Why dont we just get the point of intersection of the 2 given lines,

and then use the "point distance to plane" formula to get the distance??(4 votes)- Note that the 'A' (coefficient of 'x') in the equation of the first plane is an unknown number. We need to first find that out (by obtaining the equation of the second plane). But, if it were a known number, then you're right - we could have taken
**any**point on either of the two given lines (not just the point of intersection), and obtained the distance to the first plane.(4 votes)

- @2:33Why sal set the lines equation to zero, and one?

(x-1)/2 = (y-2)/3 = (z-3)/4 = 0 ?(5 votes) - why sal didn't dot producted normal vector and vector a or vector b for getting the equation of plane ..he said we do this with any arbitrary vector on plane..then why not a vector or b vector ? at8:00(4 votes)
- A plane is a locus of points, not a dot product of two vectors. You can't get an equation for a plane without a vector with a variable head. It might help to back to "https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/dot-cross-products/v/defining-a-plane-in-r3-with-a-point-and-normal-vector" and learn this, then either work forward or come back here.(2 votes)

## Video transcript

If the distance between the plane Ax-2y+z = d and the plane containing the lines, and they give us two lines here in three-dimensions, if that distance is square-root of 6, then the absolute value of d is... So let's think about it for a little bit. They're talking about the distance between this plane and some plane that contains these two line. So in order to talk realistically about the distance between the 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 square-root of 6. 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, you could call this the equation here is ax - 2y + z = 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 green, maybe this line looks something like this, it's on that blue plane. And then this line, maybe in magenta, 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 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, then we could find some point on the blue plane and then use our knowledge of finding the distance 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 on this blue plane, then we can take the cross-product of those two vectors to find out a normal to this blue plane, and then use that information to actually figure out the equation for the blue plane. So let's figure out 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 1, y is equal to 2, z is equal to 3, so you have the point (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 one, I could make this, if I want this to evaluate to two, I'd have the point three, I would have the point, if I want this to evaluate to one, I would want five minus two over three, so 3, 5, and then I would want this to be seven. Seven minus three over four 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 two points on this plane, and actually we just need to find one point on this plane, because if you have three points that's enough to figure out two different vectors -vectors 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 zero, would be the point (2,3,4). Because this would be zero, zero, zero. So that's also sitting the plane 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 can take their cross-product to actually figure out 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 then we know that will be on the plane. And so that will be three minus one is 2i plus five minus two is 3j, plus seven minus three is 4k. So vector 'a' 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 to start it at one of those points it will sit on that line. And then we can do another vector, and 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. Let's go between these two points - that looks pretty straight-forward, so let's call vector b, so let's call vector b, let's call that, let's see, two minus one is 'i', three minus two is 'j', and then four minus three, that's just 1k. So this vector 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 2, 3, 4, and then we have our second vector, which is just 1,1,1, and then this is going to be equal to, first we'll look at the 'i' component, so cross that row, that column out. three times one minus one times four, so that's just three minus four, so it's negative 'i', and then minus, we're going to have the 'j', so let we write up minus here, minus , let me just swap signs, we have positive, negative, positive, so 'j' , get rid of that column, that row, two times one which is two, minus one times four, so that's minus four, is negative 2, so we could have written negative 2 here, but the negatives cancel out, so it becomes plus 2j, and finally for the 'k', get rid of that row that column, two times one is two, minus one times three, is two minus three, which is negative one, so it's negative k. So this right here is a normal vector to the plane. So if we want to find the equation for that plane, we've done it multiple times, we'd just have to take the dot product of that normal vector and any arbitrary vector on that that specified that we can specify with with an arbitrary x y and z, and we've done this multiple times in multiple videos. If this is any point, x, y, z, that sits on the plane, 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 and x,y,z, this vector right here is going to be x minus three i, plus y minus five j, plus z minus 7 k. That's what this vector is, it sits on the plane, assuming x,y 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 zero, because it sits on the plane, and then we'll have our equation. So let's take 'n' dot that over there, so n dot x minus three i, plus y minus five j, plus z minus seven 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 talk about constructing the equation of a plane given a point on the plane and a normal vector, and even how 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 zero, so when you take the dot product, n, our normal vector is this, so we just take the x term, which is negative one, times this x term right over here, so negative one times this is just three minus x, and then plus this y component times this y component, so it's two times this, so it's plus two y minus ten, and then finally the z component - negative one times this. So this is plus seven minus z is equal to zero, and what do we get? So we have our negative x, plus two y, minus z, and then is equal to, let's subtract three from both sides So if we take it out there it will be minus three. If we add ten to both sides, so then you have a plus ten over here, and then we subtract seven from both sides, this becomes a minus seven So then on the right-hand-side, negative three plus ten minus seven, well that's just going to be zero! And just like that we have the equation for this blue plane over here - the plane 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. So this one has a positive 2, that has a negative 2, this is just to simplify it, so it looks very similar to each other. Let's multiple this equation right here, both sides, by negative one. And then we're going to get x minus two y plus z is equal to zero; so this is a completely valid, another alternate way of expressing the same plane. And what like about this about this is that it looks very similar to this - at least the ratio of the x, y's and z's, negative two y, negative two y; one z, one z, and remember the ratios have to be the same. So here we have a one-to-one ratio between the z-coefficient and 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 one. So this is x minus two y plus z is equal to d. So now let's 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 to 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 pretty interesting proof I think) but the distance between let's say this point, (1,2,3), and this plane over here, so this distance right here is going to be in the direction of the normal, the distance is going to be, you literally just evaluate this, let me do it this way: you literally just put in this point for the x, y and z, and then you subtract the d in the numerator, and we saw that as the formula for finding the distance. So it's literally going to be one, one, (I'm actually using this point right over here) So it's going to be one, one, because we just have one x so it's just going to be one minus two times two, one minus four (that's two times two), plus three, minus d, over 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 the sum of those, taking the square root, so it's going to be equal to: 1 squared, plus negative 2 squared which is four, plus one squared which is one. So this is going to simplify to the distance is equal to one minus four plus three is zero. So in the numerator we have negative d, all of that over the square root of one plus four plus one, so all over the square root of six. So they say the distance between this plane and this plane over here is square root of six. So they're saying the distance is equal to the square root of six, that's what this information right over here is, maybe I should do that in another colour. The distance between the two planes is going to be the square root of six, and so then if we solve for d, multiple both sides of this equation times the square root of six, you get six is equal to negative d, or d is equal to negative six. Now, what they care about is the absolute value of d, or the absolute distance, so this would be kinda the signed distance - it specifies whether we're above or below the plane. Since we're below the plane we got a negative number - I just happened to draw it right, if we above the plane we would get a positive number. So this distance is negative six, the absolute value of it, the absolute value of d which is the same as the absolute value of negative six, is equal to six. So take any point on this blue plane and you look for the closest point on the orange plane, and they will be exactly six apart. Anyway, hopefully you found that interesting.