Distance between planes 2010 IIT JEE Paper 1 Problem 51 Distance Between Planes
Distance between planes
- 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.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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