Vector dot and cross products
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Vector Dot Product and Vector Length
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Proving Vector Dot Product Properties
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Proof of the Cauchy-Schwarz Inequality
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Vector Triangle Inequality
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Defining the angle between vectors
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Defining a plane in R3 with a point and normal vector
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Cross Product Introduction
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Proof: Relationship between cross product and sin of angle
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Dot and Cross Product Comparison/Intuition
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Vector Triple Product Expansion (very optional)
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Normal vector from plane equation
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Point distance to plane
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Distance Between Planes
Normal vector from plane equation Figuring out a normal vector to a plane from its equation
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- What I want to do in this video is make sure that
- we are good at picking out what
- the normal vector to a plane is
- if we are given the equation for a plane.
- So, to understand, let's start off with some plane here.
- Let's start off！！
- So, this is a plane and I have write part of it,
- obviously it keeps going in every direction.
- So let's say that that is our plane.
- And let's say this is a normal vector to the plane.
- So, that is our normal vector to the plane.
- It's given by ai plus bj plus ck.
- So, that is our normal vector to the plane.
- And let's say that we have！！ So it's perpendicular
- to every other vector that's on the plane.
- Let's say we have some point on the plane.
- We have some point.
- It's the ponit x sub p.
- And I will say p for plane.
- So it's the point on the plane, xp, yp, zp.
- If we pick the origin,
- So let's say that our axes are here.
- So let me draw, let me draw a cord in axis.
- So let's say that our cord in axes look like that.
- This is our z-axis. That is y-axis.
- And let's say this is our x-axis.
- Let's say this is our x-axis coming out like this.
- This is our x-axis.
- You can specify this as a position vector.
- There is a position vector. Let me draw it like this.
- Let me draw it like this.
- And then it would be behind the plane right over there.
- You have a position vector.
- That vector would be xpi plus ypj plus zpk.
- It specifies this coordinate right here
- that sits on the plane.
- Let me call that something.
- Let me call that position vector. I don't know.
- Let me call that,P, let me call that, p1.
- This is a point on the plane.
- So it is p1.
- It is equal to this.
- Now, we could take another point on the plane.
- This is a particular for.
- We just say any other point on the plane x, y, z.
- But we say that x, y, z sits on the plane.
- So let's say we take this point right over here, x, y, z.
- That clearly saying logic can be specified
- by another position vector.
- We could have a position vector that looks like this,
- In dotted line,
- It's going under the plane right over here.
- And this position vector.
- I don't know. Let me just call it.
- Let me just call it p, instead of that particular p1.
- This would just be xi plus yj plus zk.
- That the whole reason why I did this setup is
- because I want to find, given some particular point
- that I know on the plane and any other xyz
- that is on the plane,
- I can find, I can construct a vector
- that is definitely on the plane.
- And we have learned this before.
- We try to figure out what the equations for a plane are.
- That the vector that is definitely on the plane
- is going to be the difference of this two vectors.
- So, I will do that in blue.
- So if you take the yellow vector, minus the green vector.
- We will take this position.
- You will get the vector.
- If you view it that way,
- that connects this point to that point.
- Although you can shift the vector.
- But you will get a vector
- that definitely lies along the plane.
- Even if you！！
- If you start with one of these points.
- It will definitely lie along the plane.
- So, the vector will look like this.
- And it would be lying along our plane.
- So this vector lies along our plane.
- That vector is p minus p1.
- This is the vector p minus p1.
- It is this position vector minus
- that position vector gives you this one.
- Another way to view this is this green position vector
- plus that blue position vector
- that sits on the plane
- will clearly equal this yellow vector.
- Right! Heads to tails, clearly equals it.
- And the whole reason why that I did that is
- now we can take the dot product
- between this blue thing and this magenta thing.
- And we have done this before.
- And they have to be equal to zero,
- because this lies on the plane.
- This is perpendicular to everything that sits the plane.
- And it equals to zero,
- and we will get the equation for the plane.
- But before I do that,
- let me make sure that we know
- what the components of the blue vector are.
- So p minus p1, that's the blue vector.
- You are just going to subtract each of the components.
- So it's going to be
- x minus xpi, plus y, minus ypj, plus z, minus zpk.
- And we just said, this is in the plane.
- And this right, the normal vector,
- is normal to the plane.
- You take their dot product.
- is going to be equal to zero.
- So n dot, this vector,
- is going to be equal to zero.
- But it is also equal to this a times this expression.
- I will do it right over here.
- So this. I will find some good colors.
- So, a times that.
- Which is ax minus axp, plus b times that.
- So, that is plus by, minus byp.
- And then, I have to make sure that I have enough colors.
- And then it's going to be plus that times that.
- So that's plus cz minus czp.
- And all of these is equal to zero.
- Now, what I am going to is: I am going to rewrite this,
- so we have all of these terms
- that I am looking for right color.
- We have all the x terms ax.
- Remember, this is any x that sits on the plane
- that will satisfy this.
- So ax, by, and cz.
- Let me leave that one the right hand side.
- So we have ax plus by plus cz is equal to！！
- And what I want to do is:
- I am going to subtract each of these from both sides.
- Another way is:
- I am going to move them all over to the, let me do it.
- Not doing too many things.
- I'm going to move them over
- to the left hand side. .
- So I'm going to add positive axp to both sides
- as equivalent to subtracting negative axp.
- So, this is going to be positive axp, and then,
- we are going to have positive byp, plus,
- we are going to do that in the same green.
- Plus byp and then finally plus czp,
- plus czp is going to be equal to that.
- Now, the whole reason that why I do this！！
- I have done this in previous video
- where we were trying to find the formula
- or we were trying to find the equations of the plane.
- And now you said that hey!
- If you have a normal vector,
- and you are given a point on the plane.
- We say xp yp zp.
- We now have a very quick way to figure out the equation.
- But I want to go the other way.
- I want you to be able to！！
- If I were to give you.
- If I were to give you.
- If I were to give you an equation for a plane.
- Where I would to say ax plus by plus cz, is equal to D
- So, this is a general equation for the plane.
- If I want to give you this.
- I want you to be able to figure out
- the normal vector very quickly.
- So, how could you do that?
- Well, this ax plus by plus cz is completely
- analogous to this part right up over here.
- Let me rewrite this over here.
- so that this part becomes clear.
- This part is ax plus by plus cz is equal
- to all of these stuff on the right hand side.
- Sorry, on the left hand side.
- So, let me copy and paste it.
- Copy and paste it.
- So I just essentially flip this expression.
- But now you see, this all of this.
- This a has to be this A, this b has to be this B
- and this c has to be this thing.
- And D is going to be all this.
- And this is just going to be a number.
- This is just going to be a number.
- assuming you mean that you knew
- what the normal vector is.
- What you a,b,and cs are.
- You know the particular values.
- So, this is what, this is what D is.
- So, this is how you can get the equation for a plane.
- Now, what if I were to give you the equation for a plane.
- What is a normal vector?
- Well. We just saw.
- The normal vector, this a corresponds that A.
- This b corresponds to that B.
- That c corresponds to that C.
- The normal vector to this plane we started off with is,
- and has the components of a, b, and c.
- So, if you are given an equation for a plane here.
- The normal vector.
- The normal vector to this plane right over here,
- is going to be Ai plus Bj plus Ck.
- So it's very easy thing to do
- if I were to give you an equation of a plane.
- Let me give you a particular example.
- If I were to tell you
- that I have some plane, in three dimensions.
- Let's say it's negative 3.
- although we could work for more dimensions.
- Let's say I have negative 3x,
- plus the square root of 2, y, minus,
- So it's crazy, I mean it's not crazy.
- It should be the plane in three dimensions.
- And what's the normal vector to this plane?
- You literally,
- literally can just pick out these coefficients.
- You say a normal vector to this plane
- is negative 3i plus square root of 2 j plus 7 k,
- and you can ignore the D part there.
- And the reason why you can just ignore that
- is that can just shift the plane.
- but won't fundamentally change how the plane is tilted.
- So, this normal vector will also normal
- if this was e, if this was a hundred.
- It will be normal to all of those planes
- because all of those planes are shifted
- but they all have the same inclinations.
- So it would all come to point at the same directions.
- So, it would have normal vectors would point
- in the same direction.
- So, hopefully you found that vaguely useful.
- So now build on this,
- define the distance between any point
- in three dimensions and some plane,
- the shortest distance we can get to that plane.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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