Point distance to plane Distance between a point and a plane in three dimensions
Point distance to plane
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- What I want to do with this video is start with some point that's
- not on the plane, or not necessarily on the plane.
- So let me draw a point right over here
- and let's say the coordinates of that point are x-naught
- x-sub-zero, y-sub-zero and z-sub-zero.
- 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 line
- so the position vector for this could be x-naught i
- plus y-naught-j, plus z-naught-k
- specifies this coordinate right over here.
- What I want to do is find the distance between this point
- and the plane. And obviously there could be a lot of distances
- I could find the distance between this point and that point
- And this point and this point, and this point and this point
- And 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 the perpendicular distance to the plane
- or the normal distance to the plane.
- We'll hopefully see that visually as we try to figure out
- how to calculate the distance.
- So the first thing we do, 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: x-sub-p, y-sub-p, z-sub-p.
- So let's construct a vector. Let's construct this orange vector
- that starts on the plane.
- its tail is on the plane, and it goes off the plane
- to this vector, to this position x-naught, y-naught, z-naught.
- So what would be, how could we specify this vector
- right over here?
- Well, that vector, let me call that vector, well
- let me call that vector--what letters have I not used yet?
- let me call that vector f.
- Vector f is just going to be this yellow position vector
- minus this green position vector.
- So it's going to be--the x component is going to be
- the difference in the x-coordinates.
- The y-coordinate is going to be the difference in the
- It's going to be x-naught minus x-sub-p.
- I subtracted the x coordinates i plus y-naught minus y-p j
- plus, I'll go to the next line, plus z-naught
- minus z p k.
- So fair enough. That's just some vector that comes
- off of the plane and onto this point.
- And 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 we can see it visually now.
- Because, if you look at--we can actually form
- a right triangle now.
- So 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 on a right triangle.
- And obviously, the shortest side here, 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 this length here in blue?
- Well, we could figure out the magnitude of this vector.
- So the length of this side right here is going
- to be the magnitude of this vector f.
- That will give us this length.
- But we want this blue length.
- Well, we could think about it.
- If this was some angle (I know the writing's getting small)
- if this was some angle theta, we could
- use some pretty straight up,
- pretty straightforward trigonometry.
- If the distance under question is d, you could say
- cosine of theta is equal to the adjacent side
- over the hypotenuse.
- or, is equal to d, d is the adjacent side, over the hypotenuse
- Well, the hypotenuse is the magnitude of
- this 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 vector f)
- is equal to d.
- So that's still, well you might say
- well, Sal, we know what f is.
- We could figure that out.
- We could figure out its magnitude.
- But we don't know what theta is.
- How do we figure out what theta is?
- And to do that, let's just think about it a little bit
- This angle, this angle theta, is the same angle
- so 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 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 by the magnitude of the normal vector.
- So I'm essentially multiplying by 1.
- 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
- and this vector right here, f.
- So this right here is the dot product.
- This is n dot f up here.
- 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 the dot product between this vector
- the normal vector divided by the 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, the components are just the
- coefficients on the x, y, and z terms
- So this is the normal vector right over here.
- So let's take
- let's literally take
- the dot product.
- So n dot f is going to be equal to
- A times x-naught minus x-p
- So it's going to be equal to, I'll do that in pink
- So it's going to be A x-naught minus A x-p
- and then, plus B times the y component here
- so plus B y-naught, I'm just distributing the B
- minus B y-p, and then plus
- (I'll get another color here)
- C (that's too close a color)
- plus C times the z component.
- So, plus C z-naught 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 this is just each of these guys squared, added to themselves
- and then take the square root.
- So the square root is (and I can make a nicer looking radical
- sign than that)
- So it's the square root of A-squared plus B-squared
- plus C-squared.
- Now what does this up here simplify to?
- Let me just re-write this.
- So this is the distance in question. This right here is
- equal to the distance.
- So let's see if we can simplify it.
- So first we can take all of the terms
- with the x-naught, these are involved
- with the point that sits off the plane
- and remember x-naught, y-naught, and z-naught sit off the plane
- so this is Ax-naught plus B y-naught plus C z-naught
- and then what are these terms equal to?
- A x-p
- negative A x-p minus B y-p minus C z-p
- and if you remember up here,
- D 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 this point x-p sits on the plane, D is
- A x-p plus B y-p plus C z-p
- or another way you could say it is
- negative D would be negative A (and it's just
- the difference between lower case and upper case
- here, right. We see that lower case a is the same as this upper case A)
- So it's negative A x-p minus B y-p minus C z-p
- I'm just using what we got from the last video.
- This is what D is, so negative D is just this business.
- And that's exactly what we have over here.
- negative A x-p, negative B y-p, negative C z-p.
- So all of this term, this term, and this term,
- simplifies to a minus D.
- And remember, this negative capital D, this is the
- D from the equation of the plane
- NOT the distance d.
- So this is the numerator of our distance
- and then the denominator of our 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
- And this is a pretty intuitive formula here.
- Because all we're doing--
- if we have some
- Let me give you an example.
- If I have the plane 1x minus 2y plus 3z is equal to five
- 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 2,3, (and let me make sure
- this is not on the plane
- so it's 2 minus six is negative, yes, let me just pick
- a random 1. So this is definitely not on the plane
- because we have 2 minus six 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 2 minus 2 times (I'm going to fill it in)
- 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
- minus 5 (kind of bringing it over to the left hand side)
- 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, minus 5
- so this is what, this is 5, 2 plus 3 is 5
- minus 5, so those two just cancel out
- so this is negative 6 over
- the square root of 5 plus 9 is 14
- over the square root of 14.
- And you're done.
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