# CHAPTER 8

### Lines and Planes

## Introduction

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

## 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.

## Example # 1

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

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

## 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

## Example # 1

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 . . .

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

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.

## Example #1

## Example #2

## 8.5 : Intersections of Lines ans Planes

## 3 POSSIBLE SITUATIONS

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.

## Situation # 1 Line intersecting plane one specific point. | ## Situation # 2 Line lies on the plane with an infinite amount of solutions. | ## Situation # 3 Line is parallel to the plane and there is no solutions meaning there is no intersection between the plane and line. |

## Intersection with the plane

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

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

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)

[ 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)

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)

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.

## Distance between Point P and a 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

- 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