Sec1Sess16

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An important approach to finding parallel vectors and directions is to look for the dot product with a
line's (cross product for plane) normal and another line (plane, etc.) to equal 0.

How to find the line intersection with a plane

Consider the plane x+2y+4y=7

Where does the line through <math>Q_0, Q_1</math> intersect the plane?

plane equation: x+2y+4z = P
<math>Q_0 = (-1,2,2)</math>
<math>Q_1 = (1,3,-1)</math>
<math>P\ .\ Q_0 = (-1)+2*2+4*2=11</math>
11 > 7
P . Q_1 = 1 + 2*3 + 4*-1=3
3 < 7
because no Q points resulted in 7 none were found to be in the plane

Due to <math>Q_0</math> passing through 7 as a result of the plane equation to get to <math>Q_1</math>
the 2 Q points are on opposite sides of the plane

Point of Intercept of a line with a plane:
use <math>Q_0</math> (point 1 of line) and <math>Q_1</math> (point 2 of line)
to find direction of line (x2-x1,y2-y1,z2-z1) to find parametric equation
x = (x2-x1)t, y = (y2-y1)t, z = (z2-z1)t
substitute the parametric equation into the equation of the plane:
plane = a + b + c = d
a(x)+b(y)+c(z)=d
solve t for d
apply t found to parametric equation to find x,y,z of point of intercept

Finding at what point the line intersects the plane:
what about Q(t)?
x(t)+2y(t)+4z(t)
<math> Q_1 \ -\ Q_0 = (1-\ -1,3-2,-1-2) = (2,1,-3) = Q_{1,0} = </math> change from time 0 to time 1
p = x+2y+4z
<math>Q(t)\ =\ p * Q_0 + t*(Q_1-Q_0)</math>
=(-1+2t)+2(2+t)+4(2-3t)
=-8t+11
is the result greater or equal to 7?
Q(t) is in the plane when -8t+11=7
when t = 1/2 the result is 7. time (t) is 1/2 when intersection occurs
Q(1/2) = (0,5/2,1/2) = intersection of the plane
if no solution for the equation of the plane's result (7 in this case) occurs than the line could be parallel to the plane or in the plane.

What is an equation for a line that lies on a plane?
Line should be perpendicular from normal of plane, therefore parallel to the plane.
Test for perpendicularity with dot product and normal. Normal of plane is given by
the equation of the plane.
E.g. plane: <math>3x+\ y+\ -5z=9</math>
N = <3i, j, -5k>
Perpendicularity is tested by <x,y,z> . N = 0 where <x,y,z> is the line.
E.g. x = at, y=bt, z=ct where 3x + y + -5z=0, a,b,c, not all 0
What is an equation for a line that lies on a plane and intersects a point?
plane: 3x + y + -5z=9
point: <4,8,10>
x = 4+at, y = 8+bt, z = 10+ct where 3x + y + -5z=0, a,b,c, not all 0

practice problem 1
plane = p = 2x + y -4z = 4
line = x = t, y = 2 + 3t, z = t
<math>Q_0\ =\ (0,2,0)</math>
<math>Q_1\ =\ (1,5,1)</math>
<math>Q_{1,0}\ =\ (1-0,5-2,1-0) = (1,3,1)</math>
<math>Q(t)\ =\ p\ *\ Q_0\ +\ t*(Q_1-Q_0)</math>
<math>t*Q_0*Q_1\ =\ (t,3t,t)</math>
<math>Q_0\ +\ t*(Q_1-Q_0)\ =\ (0+t,2+3t,0+t)\ =\ (t,2+3t,t)</math>
Q(t) = 2(t)+(2+3t)-4(t)=4
=5t-4t+2=t+2=4
when t+2=4 there is the intersection, therefore
t=2
since line = x = t, y = 2 + 3t, z = t
intersection point is at (2, 8, 2)

"parameterization of the plane" <math>=\ Q_0\ +\ t*(Q_1-Q_0)</math>