Lines and Planes
Two-Space and Three-Space
LINES IN TWO-SPACE AND THREE-SPACE
Any line in two-space (a space that has only two dimensions i.e. x and y) can be defined using its slope (m) and y-intercept (b). The slope of the lines represents the direction and the y-intercept represents the exact position in two-space.
The slope of the line is not defined in three space (a space that has three dimensions i.e. x, y and z), so line must be defined in other ways...
A line can be defined by an equation in Standard form (Ax+By+C=0), which is also known as the Scalar Equation, or in slope-y-intercept form (y=mx+b).
Do you know that a vector can also be used to define a line?
Scalar Equation -x+2y-10=0
A direction vector parallel to line is m = [2, 1]
A position vector that has its tip on the line is r0 = [2, 6]
Another position vector r = [6, 8] can be drawn to form a closed triangle.
The third side of the triangle is vector s
Therefore, the closed triangle can be shown using the vector sum.
r = r0 +s [vector r = vector r0 + vector s]
or
s = tm [ vector s = t x vector m] (since s is parallel to m)
Therefore, r = r0 + tm
which can also be written as [x, y] = [x0, y0]+t[m1, m2] in vector sum.
Therefore x = x0+tm1
&
y = y0 +tm2
This form of equation is known as Parametric Equation
A normal vector to a line 'l' is vector 'n', and vector n is perpendicular to the line
Fox Example:
If line 'l' is Ax+By+C = 0, then normal vector n=[A, B]
If a point (x ,y) is given along with a normal vector [A, B], then what will be the Scalar equation?
Lets recall that Scalar equation is in the form Ax+By+C=0
Therefore, substitute the given values A, B, x and y and find the unknown C
If, a vector equation is given, then write the equation in terms of parametric equations,then isolate "t" in both of the parametric equations and finally, set both of the t equations equations to each other and bring all terms on one side. The end result will be a scalar equation
Equation of Lines in Three-Space
Just for you knowledge, a plane is a two-dimensional flat surface extends infinitely in all directions
Therefore, we will add the variable z to the two-space vector equation and parametric equation, which will then give us the equation of line in three space.
Vector equation in three space will be...
r = r0 + tm or [x, y, z] = [x0, y0, z0]+t[m1, m2, m3]
and the Parametric equation will be...
x = x0 + tm1
y = y0 + tm2
z = z0 + tm3
In order to review Equation of Lines in Two-Space and Three-space in more detail, you can watch the following video...
Examples for Vector Equation of a Line in Two-Space
Example 1
Write a vector equation for a line given a direction vector m = [2, -1] and point P0 (4, 9)
First of all the position vector contains only ’x’ and ’y’ values.
Therefore, the vector equation for a line will be in two-space.
Given: A point and a position vector
Equation is... [x, y] = [x0, y0]+t[m1, m2]
[m1, m2] = [2, -1] (direction vector is given directly)
[x0, y0] = ???
We have to find the position vector.
if point (4, 9) is one the line, then the position vector [4, 9] has its tip on the line
A vector is also known as the difference between two points
Therefore, [x0, y0] = (4, 9) - (0 , 0)
= [4, 9]
Therefore [x, y] = [4, 9]+t[2, -1]
and we can also conclude that a given point on the line can also be used as a position vector
Example 2
If point (-2, -10) is on the line, then the position vector [-2, -10] has its tip on the line. Therefore, the only one "t" value exits that make the equation true.
[x, y] = [2, -3]+t[4, 7]
[-2, -10] = [2, -3]+t[4, 7]
Now, Evaluate the x-coordinates and the y-coordinates. (Hint: parametric equations can be used to evaluate the coordinates)
x-coordinate:
-2 = 2+4t
-2-2 = 4t
-4 = 4t
t = -1
y-coordinate:
-10 = -3+7t
-10+3 = 7t
-7 = 7t
t = -1
Since the t-values are equal, the point (-2, -10) lies on the line
Examples for Parametric and Scalar Equation of a Line in Two-Space
Example 1
x = 2-5t
y= 3+6t
From parametric equation, we can choose r0 = [2, 3] and m = [-5, 6]
Therefore, the vector equation is...
[x, y] = [2, 3]+t[-5, 6]
As told earlier, we can isolate "t" in both of the parametric equations.
x = 2-5t
x-2 = -5t
t = (-x+2)/5
y = 3+6t
y-3 = 6t
t = (y-3)/6
Equate both equations...
(-x+2)/5 = (y-3)/6
Therefore scalar equation is...
6x+5y-27=0
Example 2
Line one = [x, y] = [1, 3]+t[-3, 4]
Line two = [x, y] = [2, -5]+[-6, 8]
The direction vectors of parallel lines are scalar multiples of one another
Direction vector for line one = [-3, 4]
Direction vector for line two = [-6, 8]
If the lines are parallel, therefore [-3, 4] = k[-6, 8]
-3 = -6k
-3/-6 = k
k = 1/2
4 = 8k
4/8 = k
k = 1/2
Both of the k values are equal, therefore both of the lines are parallel to each other
Examples for Vector and Parametric Equation of a Line in Three-Space
Example 1
Write a vector and parametric equation for the line
For vector equation...
Find a direction vector first.
Direct vector is:
m = B - A
m = [3, 6, 2] - [2, 3, 5]
m = [1, 3, -3]
And as described earlier, we can use any point on the line as a position vector
Possible position vector is [2, 3, 5]
Therefore, the vector equation for the line is...
[x, y] = [2, 3, 5]+t[1, 3, -3]
For parametric equation...
Using the vector equation found above, we can conclude that...
x = 2+t
y = 3+3t
z = 5-3t
In conclusion (for a line in two and three space)...
For lines in two-space, a perpendicular vector and a point can be used to define a line
EQUATION OF PLANES
In three space, a plane can be defines by a vector equation, parametric equation or a scalar equation.
Equations of Planes
Starting from the origin. vector r0 goes to a known point on the plane, then to two vectos in the plane that are scalar multiples of the direction vectors a and b.
The resultant of these three vectors is the vector r
r = r0 + ta + sb or [x, y, z] = [x0, y0, z0]+t[a1, a2, a3]+s[b1, b2, b3]
r is the position vector for any point on the plane
r0 is a position vector for a known point on the plane
a and b are non-parallel direction vectors parallel to the planes
t and a are scalars
Parametric equations of a plane in three-space are:
x = x0 + ta1 + sb1
y = y0 + ta2 + sb2
z = z0 + ta3 + sb3
These equations are very similar in format to Parametric equations of a line as explained previously.
The scalar equation is form of Ax+By+Cz+D = 0 can also be used.
A, B and C represents the normal vector [A, B, C]
and
D can be found using a given point on the line
Additionally,
Y-int of a plane can be found by setting x=z=0
and
Z-int of a plane can be found by setting x=y=0
To find if a point lies on the plane (given a scalar equation), plug in the given point in the equation and check if LHS = RHS
To find if a point lies on the plane (given a vector or parametric equation), then there exists a single set of t- and s-values that satisfy the equations.
To find if two planes are parallel, their normals must be parallel.
Normals can also be found by using cross product of two directional vectors of the plane.
If the normals are scalar multiples of each other, then the both planes are parallel to each other.
In order to review Equation of Planes in more detail, you can watch the following video.
Examples of Equations of Planes
Example 1
X-int:
2x+4(0)-6(0)-24 = 0
2x-24 = 0
2x = 24
x = 24/2
x = 2
Therefore, the x-int is (2, 0, 0)
Y-int:
2(0)+4y-6(0)-24 = 0
4y-24 = 0
4y = 24
y = 24/4
y = 6
Therefore, the y-int is (0, 6, 0)
Z-int:
2(0)+4(0)-6z-24 = 0
-6z-24 = 0
-6z = 24
z = 24/-6
z = -4
Therefore, the z-int is (0, 0, -4)
Example 2
3(1)-2(3)+7 = 4
3-6+7 = 4
-3+7 = 4
4 = 4
LHS = RHS, Therefore the point A lies on the plane
Example 3
Write the vector and parametric equations for the plane
[x, y, x] = [x0, y0, z0]+t[a1, a2, a3]+s[b1, b2, b3]
[x0, y0, z0] = [3, 7, 1], a position vector for a point on the plane
t[a1, a2, a3] = [2, 3, 5], a direction vector
s[b1, b2, b3] = [1, 4, 6], also a direction vector
Therefore, the vector equation will be...
[x, y, x] = [3, 7, 1]+t[2, 3, 5]+s[1, 4, 6]
The parametric equation will be...
x = 3+2t+s
y = 7+3t+4s
z = 1+5t+6s
Example 4
Plane one: [x, y, z] = [10, 3, 6]+t[2, 3, 5]+s[1, 2, 3]
and
Planetwo: [x, y, z] = [1, 5, 9]+t[4, 6, 10]+s[2, 4, 6]
First of all, find the normals of the both planes by taking cross product of directional vectors.
The Normal of plane one is...
[-1, -1, 1]
and
The Normal of plane two is...
[-4, -4, 4]
Therefore, [-4, -4, 1] = k[-1, -1, 1]
-4 = -k; k = 4
-4 = -k; k = 4
4= k ; k = 4
The normal of plane two is the scalar multiple of the normal of plane one.
Therefore, both of the planes are parallel to each other
PROPERTIES OF PLANES
Equations of properties of planes
Ax + By +Cz +D = 0, where n = [A, B, C] is a normal vector to the plane
The scalar equation is form of Ax+By+Cz+D = 0 is used.
A, B and C represents the normal vector [A, B, C]
and
D can be found using a given point on the line
In order to review the properties of plane in more detail, you can watch the following video...
Example 1
Find the scalar equation of the plane.
Scalar equation is int he form Ax+By+Cz+D=0
Since A, B and C are the values of the normal vector,
[A, B, C] = [2, 1, 3]
Therefore, substitute the given normal vector and the point into the Scalar equation.
Plug in normal vector: 2x+y+3z+D = 0
Plug in the given point to find D:
2(4)+5+3(7)+D = 0
34+D = 0
D = -34
Therefore, the scalar equation is 2x+y+3z-34 = 0
Example 2
If dot product of vector a and the normal vector of the plane is equal to zero, then vector a is parallel to the plane
First of all find the normal vector.
We know that [A, B, C] = normal vector.
In the given plane, A = 3, B = 2 and C = 1
[A, B, C] = [3, 2, 1]
Therefore, [A, B, C]*[5, 4, -23] = 0
[3, 2, 1]*[5, 4, -23] = 0
(3*5)+(2*4)+(1*-23)=0
0 = 0
Therefore, vector a is parallel to the given plane.
Example 3
[x, y, z] = [2, -5, 4]+t[7, -3, 5]+s[2, -3, -2]
Since scalar equation is in the form Ax+By+Cz+D = 0 and A, B, C represents the normal vector, we need to find the normal vector first.
Normal vector is any vector perpendicular to the plane.
Therefore, cross product can of two given vectors, t and s can be used to find the normal vector.
[7, -3, 5]x[2, -3, -2]
The cross product comes out to be [21, 24, -15], which can we further simplified to [7, 8, -5].
Therefore, n = [7, 8, -5] = [A, B, C]
From our previous lesson, we know that vector [2, -5, 4] also represents the point (2, -5, 4)
Now we have normal vector and a point, we can plug in the value of normal vector and the point into scalar equation to find the value of D.
Ax+By+Cz+D = 0
7x+8y-5z+D = 0
7(2)+8(-5)-5(4)+D = 0
-46+D = 0
D = 46
Therefore, the scalar equation is 7x+8y-5z+46 = 0
INTERSECTION OF PLANES
Intersection of two planes in three-space
In the above example,
process of elimination is used to eliminate one of the variable from two of the equations.Multiply one of the equation by a number which will help us eliminate one of the variable from both of the equations.
After subtracting/ adding both of the equations, a third equation is formed.
In the third equation, you can set one of the variable from two left over variables to be equal to t.
"t" will help us find the parametric equations of the line.
Now solve the third equation in terms of non t variable.
Now, you will have two variables in terms of t.
Plug in both of them in the either of the equations of plane to find the third variable.
Now you will have all of the variables as parametric equations which can be re-written in form of vector equation of the line.
Intersection of three planes in three-space
There are three possible geometric models of three planes which are consistent.
The normals are parallel and not coplanar.
The normals are coplanar, but not parallel.
The normals are parallel.
There are three possible geometric models of three planes which are inconsistent.
The normals are parallel.
Two of the normals are parallel.
The normals are coplanar, but not parallel.
To check if the normals are coplanar, triple scalar product can be used. If the triple scalar product is equal to zero, then the normals are coplanar.
Two more equations can be drived using by subtracting/ adding first two equations and last two equations.
Now, solve in terms of parametric equations and re-write them in terms of vector equation of a line.