The Eighth Chapter
Vector equations that are separated into 2 parts , one for each variable
x= xo + m1
y= yo + m2
r = [x,y] , ro = [xo,yo], m =[m1,m2]
[x,y] = [xo,yo] + t [m1,m2]
Example: Graph: -x+2y-10=0
Direction vector: m = [2,1]
known: ro = [2,6]
changing: r = [6,8]
Using the 2 position vectors a triangle is formed and the unknown line becomes
Since s is parallel to m,
S = tm
Substitute into: r = ro + tm
Therefore: r = ro + s
r = [x,y]
ro = [xo,yo]
Placed together: [x,y] = [xo,yo] + t [m1,m2]
Perpendicular vector to a line
If the line is : Ax +By +C=0, then n is considered [A,B]
Can be defined by vector or parametric equation
Cannot be defined by scalar equations
Scalar equations are used to define planes in a 3D space.
[x,y,z] = [xo,yo,zo] + t[m1,m2,m3]
x= xo +m1
z=zo + m3
[8.2] The Second Lesson: Equations of Planes
So firstly there is a point in the plane, which is point A (X,Y,Z).
Next, we have two vectors that are "in" the plane which are u and v
Technically, it is better of saying that there are parallel to the plane as when you dot product either vectors with the normal vector of the plane we'd get 0.
So what happens is that we also have a terminal point which is point P(X,Y,Z) that creates a vector with point A. So the new vector would be AP which is equal to
Then what we can do is add the vectors multiples of u and v to reach and reach Point P from Point A. Therefore:
AP = tu + sv
(The s and t are both scalar numbers)
<X-X0,Y-Y0,Z-Z0> = tu + sv
P(X,Y,Z) - A(X0,Y0,Z0) = tu +sv
P(X,Y,Z) = A(X0,Y0,Z0) + tu +sv
[x,y,z] = [X0,Y0,Z0] +t[u1,u2,u3]+s[v1,v2,v3]
r = r0 +tu + sv
And with that vector equation, if you solve for x, y, or z you'll get the parametric equation. For example with the Vector equation (X,Y,Z) = A(X0,Y0,Z0) + tu +sv...
The Parametric Equations will be:
x=x0 +t (u1) + s(v1)
y=y0 +t (u2) + s(v2)
z=z0 +t (u3) + s(v3)
The first number in the equations (x0,y0, and z0) are points on the plane, while the remaining two numbers are the components of the vectors parrallel to the plane.
So rather than having two vectors parallel to the plane, we can have one point in the plane and a vector perpendicular to the plane. Since no other plane will have that specific point and that perpendicular vector, we can define a plane that way too.
If you aren't given the normal vector and you want a Scalar equation and you have two parallel vectors to the plane, you can find the cross products of the two vectors parallel to the plane.
[8.3] The Third Lesson: Properties Of Planes
The properties of planes is fairly easy to understand. It is about determining the scalar equation from the normal vector to the plane, n, and a point on the plane, P0(X0, Y0, Z0) or vice versa.
Scalar Equation: Ax+By+Cz+d=0.
- Where vector n=(A,B,C)
- Any vector parallel to the normal of the plane is also a normal to that plane.
Therefore, if vector (J,K,L) is parallel to vector n=(A,B,C) then vector (J,K,L) can also be considered a normal plane.
- The coordinates of any point on the plane must satisfy the scalar equation
- A normal vector and a point can be used to define a plane
[8.4 ]Intersections of Lines in 2 Space and 3 Space
Case 1: No Solutions
Parallel Lines and Distinct
Case 2: One Solution
Case 3: Infinite solutions
Coinciding Lines (Parallel and Non-Distinct)
Although in three space you need to worry about 4 cases. Three of them being the three from 2 space and skew lines.
Case 4: Skew Lines