Parametric Lines & Planes

sponsoredwalk

Hi I'm reading Lang's Intro to Linear Algebra Page's 30 -36 and I'd just like some input on whether I understand this or not. Basically my problem is with Planes but the parametric eq. thing is just clarification.

Parametric Representation of a Line

Basically, if you have two n-tuples (n=2 for convenience!)

(a,b) = A and (p,q) = P

We form the equations:

X(t) = P + tA <==> (x,y) = (p,q) + t(a,b)

with;

x = p + ta
y = q + tb (times -(a/b))

---

x = p + ta
-(a/b)y = -(a/b)q - ta

---

x - (a/b)y = -(a/b)q

-(b/a)x + y = q

[latex] y \ = \ \frac{b}{a} \cdot x \ + \ q [/latex]

This is forming the eq. of the line going from point P in the direction of A.

If you want to go from point A in the direction of P the equation becomes

X = A + tP

if you want to go in the opposite direction in either of the equations you just give t a negative value.

If you want to go just from point A to point P let:

0 ≤ t ≤ 1

and let P = P - A

so that X(t) = A + tP = A + t(P - A)

i.e. "t" takes on a value less than 1 so you multiply, say, 0.04 by all the values (P - A) represents in the (x,y) dimensions before going on.

Planes

The idea is to find the equation of the plane in R³ that passes through some random point P by forming a located vector [latex] \overline{PX} [/latex], where X is the set of all points surrounding P, and then taking the dot product with some other located vector [latex] \overline{ON} [/latex] that is perpendicular to [latex] \overline{PX} [/latex].


For some located vector [latex]\overline{ON} [/latex] if we want to find the plane that passes through P we'll dot product it:

(X - P) • (N - O) = 0

(X - P) • N = 0

X•N - P•N = 0

X•N = P•N

Okay, obviously O is the origin and can be ignored.

I'm thinking that N = (N - O) = [latex]\overline{ON} [/latex] can be done all the time,
no matter where the plane is to simplify the algebra.

It's like point N determines the angle/direction of the plane with respect to the origin & point P determines the height up or down. Is that correct? On page 34 of the book the picture of the plane could go up and down the N arrow.

Still, X•N = P•N doesn't seem intuitive to me, is there a way to get it?

Like, you're dot producting the variables X = (x,y,z) with a point not on the plane and that's supposed to be equal to the original point that the plane passes through dot producted with this point not on the plane you're constructing

CJC86

The reasoning is that a 2d plane through 0 in 3 dimensions is determined by the direction of its normal. Thus, it's generally stated that X is in the plane if it is perpendicular to the normal N, ie. if N.X=0.

Now if we take a change of coordinates that takes 0 to the point P that we want the plane to be passing through, then X goes to X-P. So, we get that X is in the plane if N.(X-P)=0. You can say then X is in the plane if N.X = N.P, but it's not as clear as leaving it in the form N.(X-P)=0.

sponsoredwalk

So the plane passes through the origin, through P, and is perpendicular to the point N, & the set X is pretty much
any point you choose along the plane if you choose X : (x,y,z) i.e. variables!

If I were to create a plane passing through point P but not going through the origin I could just pick some point A and create the equation
(X - P) • (N - A) = 0 so that the plane passes through A & P and is still perpendicular to N.

If I call N - A = G

(X - P) • (G) = 0

X • G = P • G

Is that all sensible?

If I didn't create G I would get;

X•N - P•N - A•X + P•A = 0
X•N + P•A = P•N + •A•X

Which is pretty crazy, but doable - right?

CJC86

sponsoredwalk wrote: »

So the plane passes through the origin, through P, and is perpendicular to the point N, & the set X is pretty much
any point you choose along the plane if you choose X : (x,y,z) i.e. variables!

If I were to create a plane passing through point P but not going through the origin I could just pick some point A and create the equation
(X - P) • (N - A) = 0 so that the plane passes through A & P and is still perpendicular to N.

Sorry, I wasn't very clear in my last post. I initially pointed out that IF the plane was going through 0, then any point X such that N.X=0 would be in the plane, since the plane is completely determined by the direction of its normal.

However, if we choose another point P that we want the plane with normal N to pass through, then we look at the vector PX = (P-X) instead of OX = (X-0) = X, and we take the dot product of this with N to get the equation N.(P-X)=0.

So, the plane does not necessarily pass through the origin, no. It might, but generally will not if our determined point in the plane is not 0.

Basically, every plane is determined by a point P that it passes through and a normal direction N. If we were to choose 2 points ( P and the origin, say ) and a direction, then the system is overdetermined, and we may not have a solution for X.

sponsoredwalk

So in the initial example I gave, which is the one on page 34 of the book I linked to, the plane passing through P does go through the origin.

I think that is what is confusing me. Also, if I don't want the plane to go through the origin I don't choose the vector [latex] \overline{ON}[/latex] = (N - O) = N
I instead choose the vector [latex] \overline{AN}[/latex] = N - A and can call this vector G.

This way I can create a plane perpendicular to G using some point P and a set of variables X & some algebra.

(X - P) • (N - A) = 0
X•N - P•N - A•X + P•A = 0
X•N + P•A = P•N + •A•X

Example:

A : (3,2,4)
N : (1,1,3)
X : (x,y,z)
P : (p,q,r)

(X - P) • (N - A) = 0
X•N - P•N - A•X + P•A = 0
X•N + P•A = P•N + •A•X

(x,y,z)•(1,1,3) + (p,q,r)•(3,2,4) = (p,q,r)•(1,1,3) + (3,2,4)•(x,y,z)

x + y + 3z + 3p + 2q + 4r = p + q + 3r + 3x + 2y + 4z

2p + q + r = 2x + y + z

2x + y + z = 2p + q + r

so this is an overdetermined system with possibly no solution?

I'm a little confused, do you get to pick any numbers for p,q & r?
They have to lie along the plane, how would you find them?

CJC86

The initial example does not necessarily go through the origin, no. I must have confused you with the way I introduced a plane in my post, where I just pointed out that any plane through the origin is completely determined by its normal direction. However, if we choose a point away from the origin that the plane passes through, then you need to look at the direction that X is away from that point, ie. PX = X-P, and any point X that its direction away from P is perpendicular to N is in the plane.

You don't need to have this AN=N-A. What you are doing there is taking the normal direction to be AN rather than N. It's easiest to just think of N as a direction, so it doesn't matter that it is positioned at the origin. The plane you give as a example in your last post is not overdetermined, but its normal direction is N-A = (-2,-1,-1), rather than N = (1,1,3), so you have a completely different plane.


Example:
The plane through P= (2,3,5), with normal direction N=(1,0,1) is spanned by X=(x,y,z) such that:

N.(P-X)=0
=> (1,0,1).(2-x,3-y,5-z)=0
=> 2 - x + 5 - z = 0
=> x+z = 7

Edit:
note that the origin (x,y,z)=(0,0,0) does not satisfy this equation.

sponsoredwalk

Ahh... I get you now. When I was taking AN = N - A I was trying to include A as being part of the plane I was creating, with AN as being a vector starting on the plane & ending @ N.

I'll be back if I have any other problems Thanks a lot, I was pretty stuck there.

Michael Collins

sponsoredwalk wrote: »

Still, X•N = P•N doesn't seem intuitive to me, is there a way to get it?

I haven't read through everything in this thread so maybe this has been answered satisfactorially for you. But the answer to your question is definately yes!

Let's consider N to be a unit vector (if it isn't, just divide both sides by the size of N to get a lower case unit vector n).

X•N = P•N
X•n = P•n

Now n is a unit vector, perpendicular to the plane, from the origin.

P is a vector, from the origin, to the plane (not necessarily the perpendicular vector). So on the RHS we have, P•n, the projection of P in the direction of n (and so equals the perpendicular distance of the plane from the origin, but that's not too important at the minute) i.e. a constant.

X is a variable vector, from the origin to some point (x1, x2, x3), which satisfies the fact that the projection of this vector, in the direction of the normal (n) to the plane, is constant. This is the key. All the points that satisfy this clearly must form a plane. And since that constant is equal to the distance P•n, namely the perpendicular distance from the origin to the plane, it must be the plane through P.

A picture would really help here I know! You probably got it from the other posts anyway, but if not, let me know and I'll try and do up a picture.

sponsoredwalk

Michael Collins wrote: »

A picture would really help here I know! You probably got it from the other posts anyway, but if not, let me know and I'll try and do up a picture.

Lol, it really would have!

I haven't looked @ my book since last week & have let this whole concept grate on my mind
as some strange algebraic manipulation that "just works", but I wasn't satisfied all the same.
I kind of accepted it because the algebra worked but hadn't a clear picture of what was going on.
I just got it though & it's so simple, it's something I should of recognised or rederived for myself
to sidestep the apparent complicatedness I assumed was going on!

I'll just recreate the concept of a plane from scratch here to make sure I've got it,

[latex] \overline{OA} \ = \ \overline{X}[/latex] is a vector from the coordinate system origin to a set point on the plane.

[latex] \overline{OB} \ = \ \overline{Y}[/latex] is a vector from the origin to any point on the plane, i.e. this can be any point along the entire plane in R³.

[latex] \overline{ON} \ = \ \overline{N}[/latex] is a vector from the origin through the plane that is perpendicular to it.

The vector along the plane is [latex] \overline{Y} \ - \ \overline{X} [/latex] !

I wasn't visualising this at all for some reason...

So, basically the way to ensure you have a plane is to have [latex] ( \overline{Y} \ - \ \overline{X}) \cdot \ \overline{N} \ = \ 0 [/latex]

This seems to be the way to create a new plane with respect to some fixed coordinate system, as espoused in this video
(which visually showed me the simple (Y - X) vector that magically cleared all of the above up for me),

Am I right in thinking that you don't always need a reference coordinate system to create a new plane?
All you need is a set point X = (x₁, x₂, x₃) on the plane & an orthogonal vector N = < n₁, n₂, n₃ > where N = M - A, (A being a point on the plane & M being it's endpoint above it).
This way you can pick a variable point Y = ( y₁, y₂, y₃ ) lying on it & then do G • N = 0, where G = Y - X, to create the whole plane from, basically scratch.