Sec1Sess15

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equations of lines
We have seen: line =
intersection of 2 planes

another way to think of line:
trajectory of moving point
= "parametric equation"

example line through:
<math>Q_0 = (-1,2,2)</math>
<math>Q_1 = (1,3,-1)</math>

Q(t) = moving point
<math>Q(0) = Q_0</math>
moves at constant speed

position at time t is <math>\vec{Q_0}\ Q(t)</math>
= t * <math>(\vec{Q_0,Q_1})</math> due to traveling at constant time
<math>Q_0 = (-1,2,2)</math>
<math>Q_1 = (1,3,-1)</math>
<math>t * (\vec{Q_0,Q_1}) = t * \langle (1-\ -1),(3-2),(-1-2) \rangle = t * \langle 2, 1, -3 \rangle </math>

<math>Q_0 = (-1,2,2)</math>
Q(t) = (x(t),y(t),z(t))

x(t) = -1 + 2t
y(t) = 2 + t
z(t) = 2 - 3t
<math>Q(t) = Q_0 + t*Q_0*Q_1</math>
= (-1 + 2t, 2 + t, 2 - 3t)
= first parametric equation example in the class

in a (x,y, and z) parametric equation the independent variable is t and dependent variables are x,y,z

Example 1: Line through <math> P_0 = (1,2,3)</math> and parallel to vector <1,3,5>
<math> P_0 = (1,2,3)</math>
P = <x,y,z>
->P_0P = <x-1,y-2,z-3>
->P_0P being parallel to <1,3,5> makes it a scalar multiple of <1,3,5>
therefore <x-1,y-2,z-3> = t*<1,3,5>
t = x - 1, 3t = y - 2, 5t = z - 3
x = 1 + t, y = 2 + 3t, z = 3 + 5t
<1 + t, 2 + 3t, 3 + 5t>

In general, the line through P0 = (x0, y0, z0) in the direction of (i.e., parallel
to) v = <v1, v2, v3> has parametrization
<x, y, z> = <x0 + tv1, y0 + tv2, z0 + tv3>
= x = x0 + tv1, y = y0 + tv2 z = z0 + tv3.

Example 4: Find the line through the point <math>p_0 = (1,2,3)\ and\ p_1 = (2,5,8)</math>
<math>(\vec{p_0,p_1}) = <2-1,5-2,8-3> = <1,3,5> = v</math>
<math>(\vec{O,p_0}) = <1,2,3></math> where O is origin
line <math>= (\vec{O,p_0})+t*(\vec{p_0,p_1}) =</math> <1 + t, 2 + 3t, 3 + 5t>

Recitation 2:
L_1 <2-t, 1+t>
L_2 <2+t, 4 + 2t>

Set x in <math>L_2</math> to be equal to x in <math>L_2</math> and the same for y
create new variable "u"
L_2 x = 2 + u = 2 - t
y = 4 + 2u = 1 + t

-u = t

insert into t in y equations
4 + 2u = 1 - u
3u = -3

u = -1

t = 1

L_1:
x = 1
y = 2

L_2:
x = 2+u
y = 4+2u
x = 1
y = 2

Point of intersection is <1,2>