# Parametric Lines & Planes

• Jul 16th 2010, 05:58 AM
Parametric Lines & Planes
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

$y \ = \ \frac{b}{a} \cdot x \ + \ q$

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 that passes through some random point P by forming a located vector $\overline{PX}$, where X is the set of all points surrounding P, and then taking the dot product with some other located vector $\overline{ON}$ that is perpendicular to $\overline{PX}$.

For some located vector $\overline{ON}$ 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) = $\overline{ON}$ 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 :confused:
• Jul 16th 2010, 06:33 AM
HallsofIvy
First, you don't "dot product" variables with a point. The dot product is defined for two vectors. In this case, "ON" is a vector normal (perpendicular) to the plane which means its dot product with any vector in the plane is 0. If P and X are two points in the plane, with P a fixed point and X representing the coordinates, (x, y, z), of any point in the plane, then X- P is in the plane: the dot product of ON and X- P is 0. That's the "intuitive" idea behind this: N is perpendicular to the plane so its dot product with any vector in the plane is 0.

From that, as you write, N.(X- P)= N.X- N.P so N.X= N.P where "X" and "P" now are the vectors OX and OP.
• Jul 16th 2010, 06:44 AM
If so, if I want to find the plane that passes through point P but doesn't intersect the origin, it intersects point A, I don't use $\overline{ON}$ I use $\overline{AN}$ and