Sec1Sess12

Equations of planes (2d plane graphing) with linear systems of equations (that matrix inverses can apply to).
 \mathrm { P_0 = } \langle \mathrm { x_0,y_0,z_0 } \rangle 
P = <x,y,z>
N = <a,b,c>
\vec{(P_0,P)} is in the plane when it is parallel to N.
A vector parallel to a plane:
When an equation of a plane is used the right hand side of the equation includes what
must be equaled for parallel to N to occur
\mathrm{E.g.\ 5x+2y+8z=13}
N = <5,2,8>
vector r must have \vec{r}\mathrm{\ .\ N\ =\ 13} for parallel position to be found.
A vector perpendicular to a plane:
The right hand side of the equation equaling 0 means it is orthogonal (perpendicular)
\vec{v}\mathrm{\ .\ N\ =\ 0}
** Find Normal when equation of plane is given:
Normal found as coefficients of the equation of plane given. A simple way to find the normal.
E.g. equation of plane 3x + 2y + z = 3
N = <3i, 2j, 1k>
** Find Normal of 2d points:
Find vector by subtracting the 2cnd point from the 1st point
With the resulting vector a vector perpendicular to <a, b> is <-b, a>
source: http://www.leadinglesson.com/problem-on-finding-a-line-between-two-points
Example
p_1 = <0,1>
p_2 = <3,0>
vector = <(3-0),(0-1)> = <3,-1>
N (perpendicular vector) = <1,3>
** Find Normal of 3d points: take cross product of 2 vectors
Example:
\mathrm{v_1\ =\ \langle\ x_1,\ y_1,\ z_1\ \rangle}
\mathrm{v_2\ =\ \langle\ x_2,\ y_2,\ z_2\ \rangle}
\mathrm{N\ =\ \langle\ x_3,\ y_3,\ z_3\ \rangle}
To find N:
\left \langle \begin{matrix} \langle \ i\ \ j\ \ k\ \rangle \\ \langle x_1\ y_1\ z_1 \rangle \\ \langle x_2\ y_2\ z_2 \rangle \end{matrix} \right \rangle=\langle x_3, y_3, z_3 \rangle
\mathrm{v_1\ and\ v_2} can be found by subtracting and 2 points to create the vector
for example points <3,2,5> and <-1,1,2> subtracted: <(3+1),(2-1),(5-2)> is a vector v_1
** For 2 vertex point (x,y,z) problems point being in a plane can be found by dot product:
\mathrm{N\ .\ }\vec{(P_0\ P)}\mathrm{\ =\ 0}
\mathrm{\langle a,b,c \rangle\ .\ \langle x-x_0,\ y-y_0,\ z-z_0 \rangle\ =\ 0}
equation described as "point normal form of plane": \mathrm{a(x\ -\ x_0)\ +\ b(y\ -\ y_0)\ +\ c(z\ -\ z_0)\ =\ 0}
** Find Intersection of 2 planes:
The cross product of the normals of the planes creates a vector that is the intersection of the planes.
\mathrm{v\ =\ n_1\ x\ n_2}
further references:
http://members.tripod.com/vector_applications/xtion_of_two_planes/
http://mathworld.wolfram.com/Plane-PlaneIntersection.html
Example:
Plane perpendicular to the normal vector <1,5,10>
P is in plane when
\vec{OP}\ .\ \vec{N}\ =\ 0
x + 5y + 10z = 0
** For 3 vertex problems the cross product can be used:
\mathrm{N\ =\ }\vec{(P_1*P_2)}\mathrm{\ x(cross\ prod.)\ }\vec{(P_1*P_3)}
Use point normal form (with P_1) to find the equation of the plane using cross product results:
\mathrm{P_1\ =\ \langle x_1,y_1,z_1 \rangle}
P = <x,y,z>
Cross Product results = <a,b,c>
\mathrm{a(x-x_1)+b(y-y_1)+c(z-z_1)=0=}"equation of the plane"
2)
Try to find Plane through P_0 (2,1,-1) and _|_*\vec{N} (normal vector N) = <1,5,10>
If P is in plane:
\vec{(P_0\ *\ P)}\mathrm{\ .\ }\vec{N}\mathrm{\ =\ 0}
<x-2,y-1,z+1> * <1,5,10> = 0
(x-2)+5(y-1)+10(z+1)=0
x+5y+10z=-3
In equation ax+by+cz=d
<a,b,c> = normal vector \vec{N}
For example: get \vec{N} by
cross-product of 2 vectors
in the plane
example 2:
vector <1,2,-1> and plane x+y+3z=5 are parallel, perpendicular, or neither?
step 1:
find normal vector
from the plane equation, normal \vec{N} = <1,1,3> is perpendicular to the plane
<1,2,-1> is not parallel to <1,1,3> and therefore is not perpendicular to the plane 
\vec{v} is perpendicular to -N?
\vec{v}\ .\ \vec{N} = 1+2-3 = 0 (dot product)
\mathrm{Yes,\ }\vec{v}\mathrm{\ is\ perpendicular\ to\ }\vec{N}
More examples of finding planes are in:
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/session-12-equations-of-planes-ii/MIT18_02SC_notes_5.pdf
If all variable intercepts (e.g. y-intercept) are known than a simpler formula can be used:
Plane with x,y, and z intercepts a,b, and c:
x/a + y/b + z/c = 1
* equation of a simple 2d line instead of a 3d plane can be found with a normal (perpendicular line):
\mathrm{P_0\ =\ \langle x_0,\ y_0 \rangle }
N = <a, b>
\mathrm{a(x-x_0)+b(y-y_0)=0}
* find equation of plane parallel to an equation and passing through a point (based on recitation 1 problem d) :
parallel equation: a*x + b*y + c*z = 0
point passed through: <x_2,\ y_2,\ z_2>
\mathrm{a*x_2\ +\ b*y_2\ +\ c*z_2} + i = 0 where i is variable needed to complete equation of plane
"distance between point and a plane" refers to the shortest distance from a point to a plane.
* Method to find distance of point from plane:
Find distance from a point given by a problem \mathrm{P_1} to a point on the plane \mathrm{P_2.\ \ P_2} is found by using the equation of the plane to find a point. Find vector from \mathrm{P_1\ to\ P_2} (e.g. <(x2-x1),(y2-y1),(z2-z1)>) Find the normal N to the equation of a plane (if equation of plane given use coefficients to find normal). Find length of N for equation \mathrm{(sqrt(x^2+y^2+z^2))}
Project the point onto the normal using this formula:
distance = | (\vec{PQ}\ .\ \vec{N}) / |\vec{N}| | where |\vec{N}| is length of N and other | | are absolute value signs
= |\vec{PQ}|*cos(angle to normal)
* Method to find distance of point from line (2d instead of 3d plane):
use the same formula from the plane method
distance = | (\vec{PQ}\ .\ \vec{N}) / |\vec{N}| | where |\vec{N}| is length of N and other | | are absolute value signs
* Distance between parallel lines:
Same as point to line method, find point on one line and find distance to other line
* Distance between skew lines:
Use point to line method. Find parallel planes by finding the normal N = v1 * v2 where v1 and v2 are the direction vectors of the lines.