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sponsoredwalk
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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.
(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
[tex] y \ = \ \frac{b}{a} \cdot x \ + \ q [/tex]
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.
(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) = [tex]\overline{ON} [/tex] 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
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
[tex] y \ = \ \frac{b}{a} \cdot x \ + \ q [/tex]
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 [tex] \overline{PX} [/tex], where X is the set of all points surrounding P, and then taking the dot product with some other located vector [tex] \overline{ON} [/tex] that is perpendicular to [tex] \overline{PX} [/tex].For some located vector [tex]\overline{ON} [/tex] 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) = [tex]\overline{ON} [/tex] 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