CHAPTER 8

Lines and Planes

Introduction

Chapter 8 introduces the concepts of lines and planes, by utilizing information from previous units such as vectors (magnitude/size and direction) and scalars (magnitude/size) you are able to determine various things. For example you can determine if a line lies on a plane, determining the normal vectors, and for what values of "x" is it parallel and for what values is it perpendicular with respect to other lines.

8.1 : Equations of Lines in Two-Space and Three-Space

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KEY INFORMATION


  • A normal vector is a line that is perpendicular to another line.
  • In order to define a line in either two or three space you need to know two points on the line , or a point and direction vector.
  • For lines in two space just a point and perpendicular line can be used to help define a line.
  • Skew Lines: they do not intersect nor are they parallel.
8.1 Equations of Lines in 2-space and 3-space

Example # 1

Find the vector, parametric , and scalar equation of the line passing through A(-5,1) and B(-1,-2).


Directions Vector:

[B - A] = [-1- (-5) , -2 - 1]

= [4, -3]


Vector Equation:

r = ro + tm

[x,y] = [Xo, Yo] + t[m1 + m2]

[x,y] = [-1, -2] + t[4, -3]


Parametric Equation:

X = xo = tm1

Y = yo + tm2


X = -5 + 4t

Y = 1 - 3t


X = -1 + 4t

Y = -2 -3t


Scalar Equation:

x + 5 = 4t

y - 1 = 3t


(x + 5)/4 = t

(y - 1)/3 = t


(x + 5)/4 = (y - 1)/3

3x - 4y + 11 = 0

Example # 2

Given A (-7, 9) and B (8, 3), which points are on the line [ x , y ] = [ 5 , 1 ] + t[ -3 , 2] ?


Step 1: Change the vector equation into a parametric equation.


x = 5 - 3t

y = 1 + 2t


Step 2: Substitute in the X values of the points and solve in terms of "t".


x = 5 - 3t

-7 = 5 - 3t

-12 = -3t

4 = t


x = 5 - 3t

8 = 5 -3y

3 = -3t

-1 = t


Step 3: Substitute in the "t" values into the parametric equation written in terms of "y" and solve for "y"


y = 1 + 2t

y = 1 + 2(4)

y = 9


y = 1 + 2t

y = 1 + 2(-1)

y = -1


Therefore because y = 9 for point A, the point lies on the line. However, for point B the y value does not equal 1 it equals -1, this means point B does not lie on the line.

8.2 : Equations of Planes

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KEY INFORMATION



  • n, a normal vector, is orthogonal to the plane
  • Any line on the plane is also orthogonal to the normal vector
  • A plane can be defined by
    1. three non-collinear points
    2. A point and 2 non-parallel direction vectors
  • Both t and s are scalars
  • In three space scalar equations define a plane, where as in two space they define a line
8.2 Equations of Planes

Example # 1

Write the coordinates of three points on the plane 8x + 4z = 16.


Step 1: Substitue in 3 random values for X or Z and solve in terms of the other variable.


8x + 4z = 16.


8(1) + 4z = 16

8(2) + 4z = 16

8(3) + 4z = 16


4z = 8

4z = 0

4z = -8


z = 2

z = 0

z = -2


Step 2: Write the coordinates for each diffrent value of X or Z


( 1 , 0 , 2 )

( 2 , 0 , 0)

( 3 , 0 , -2)

Example # 1 Continued . . .

Determine if the the point is on the plane 4x - 3y + 2z = 8


Step 1: Substitiute in one of the points you had solved for


4x - 3y + 2z = 8

4(1) - 3(0) + 2(2) = 8

8 = 8


Because LS is equal to RS the point does lie on the plane

Example # 2

Determine if the point J ( 3 , -6 , -1 ) is contained in the plane [ x , y , z ] = [ 2 , 1 , -3 ] + s[ 5 , 3 , 1] + t[ 6 , -4 , 3]


Step 1: Write the given vector equation as a parametric equation.


X = 2 + 5s + 6t

Y = 1 + 3s - 4t

Z = -3 + s - 3t


Step 2: Substitute in the point into any two of the parmetric equations.


3 = 2 + 5s + 6t

5s + 6t =1


-6 = 1 + 3s - 4t

-7 = 3s - 4t


Step 3: Make either the S values or the T values the same.


(5s + 6t = 1) 3

15s + 18t = 3

15s + 18t - 3 = 0


(3s - 4t = -7) 5

15s - 20t = - 35

15s - 20t + 35


Step 4: Solve for T and S


15s + 18t - 3 = 15s - 20t + 35

18t + 20t = 38

38t = 38

t = 1


5s + 6t = 1

5s + 6(1) = 1

5s = -5

S = -1


Step 5: Solve for Z by subing in the values for S and T


Z = -3 + (-1) + 3(1)

= -3 - 1 + 3

= -1

8.3 : PROPERTIES OF PLANES

KEY INFORMATION

  • The scalar equation of a plane is Ax + By + Cz + D = 0. The normal vector to a plane is represented as n = [A, B, C] where A, B, and C are taken from the scalar equation of the target plane.
  • Any vectors parallel to the normal vector of a given plane are also perpendicular to the plane. If vector "A" is parallel to the normal vector "n" of the plane "Pl", then "A" is also a normal vector to "Pl".
  • Any points on a plane will satisfy its scalar equation. This means any point on the plane will make the equation equal zero.
  • The scalar equation of a plane can be derived using a point and a normal vector.
Lesson 3-3: Scalar Equations of Planes

Example #1

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Example #2

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8.5 : Intersections of Lines ans Planes

3 POSSIBLE SITUATIONS

1. Line intersects the plane at a specific point, this results in the system having a unique solution.


2. The line lies on the plane in any way. This results in the system having a infinite amount of solutions.


3. The line is parallel to the plane. This means that the system has no solutions.

Intersection with the plane

From the diagram above, you can see that when a line is not parallel to the plane, it will intersect at only one point

An Example of this can be done with a pencil and sheet of paper. If you lay your sheet of paper flat on the desk, and angle your pencil towards any direction (other then parallel to the sheet) It will intersect through one point. This means that there will only be one solution (Point) in which the line intersects with the plane.

The Line lies on the plane

The second possibility is that the line lies on the plane

An example of this can also be done using a sheet of paper and pencil. If you lay a pencil on a sheet, it interacts with the sheet at many points, where as in situation one the pencil only interacts with one spot. This means in situation 2, there are an infinite amount of solutions (If you imagine the sheet and pencil are infinitely extended).

No Intersection

This happens when you line is parralel to the plane and does not lie on the plane.

Hold a sheet of paper out and hold a pencil parallel to it, there is not intersection

Example # 1 ( One Unique Solution between line and plane)

Determine the point of intersection between the plane 2x - 3y + z + 5 = 0 and the line

[ x, y, z ] = [7, 14 , -97] + t[3 , 6 , -4]


Step 1: Dot product


[2 , -3 , 1] . [3 , 6 , -4] = [6 - 18 - 4]

= -16


* Since it is a non 0 answer, it means that it is not perpendicular *


Step 2: Parametric equations


x = 7 + 3t

y = 14 + 6t

z = -9 - 4t


Step 3: Substitute


2x - 3y + z + 5 = 0

2(7 + 3t) - 3( 14 + 6t) + (-9 - 4t) + 5 = 0

14 + 6t - 42 - 18t - 9 - 4t + 5 = 0

-16t = 32

t = -2


Step 4: Solve for X, Y, and Z



x = 7 + 3t

x = 7 + 3(-2)

x = 1


y = 14 + 6t

y = 14 + 6(-2)

y = 2


z = -9 - 4t


z = -9 - 4(-2)

z = -1


The line intersects the plane at the point ( 1 , 2 , -1 )

Example #2 ( Infinite number of solutions with a line and plane)

Find the intersection of the plane x + 4y - 2z + 3 = 0 and the line [x,y,z]= [-3,1,2] + t[2,3,7].


Step 1: Parametric Equations


x= -3 + 2t

y= 1+ 3t

z= 2+ 7t


Step 2: Substitute the "x,y and z" values into the equation of the plane


1. x + 4y - 2z + 3 = 0

2. (-3 + 2t) + 4(1+ 3t) - 2(2+ 7t) + 3 = 0

3. -3 + 2t + 4 + 12t -4 -14t + 3 = 0

4. 0t=0


Therefore, t is "x" equals to all real numbers

Therefore, the line lies on the plane and all points are intersecting on the plane and has an infinite amount of solutions.

For example, (1,7,16) is a point that is both on the line and the plane if you out the t value as 2 in the parametric equations.

Example #3 ( No Solutions with a line and a plane)

Find the intersection of the plane 5x - y + 2z + 7 = 0 and the line [x,y,z] = [-4,6,-1] + t[1,5,0].


Step 1: Parametric Equations

x= -4 + 1t

y= 6 + 5t

z= -1


Step 2: Substitute the "x,y and z" values into the equation of the plane

1. 5x - y + 2z + 7 = 0

2. 5(-4 + 1t) - (6 + 5t) + 2(-1) + 7 = 0

3. -20 + 5t - 6 - 5t - 2 + 7 = 0

4. 0t = 21


Therefore, t is undetermined meaning there are no solutions.

Therefore, the line is parallel and distinct from the plane; the line and the plan share no common points.

Point of intersection of a line and a plane
Intersection of a Line and Plane : ExamSolutions Maths Revision

Distance between Point P and a plane

- To find the distance between point and a plane requires the parametric equations of the point and the plane.

- Once you have those, you plug in the " x,y,z" values into the plane's scalar equations and solve for t.

- Once solved for the value of t, take that value and substitute it back in the parametric equation to get the intersection of both the plane and the line.

- Then to find the distance you would use the pythogrieam therom in 3D space

The shortest distance from a line to the plane

- To find the shortest distance from a line to the plane, you would need to find a line that is perpendicular to both the line and the plane

- First, you would just need to find a random point on the plane

- Then, you pick any two variables from ' x,y,z' and solve for the remaining variable in the plane's scalar equation.

- Once you solve for a variable, you have found one point on the plane. So yu draw a vector from that point to the line.

- So you would use the concept of projection to find the distance

- Project the vector to the normal vector and the vector projection would be from the plane to the point on the line which helps to find the distance

- Once you get the normal vector's coordinates and the vector from the point to the line you use the distance formula to find the shortest distance.

- Formula: l dot product of the point to line vector and normal vector l / (magnitude of normal vector) = shortest distance

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