Chapter 10 Basic Concepts of Descriptive Vector Geometry

Section 10.2 Lines and Planes

10.2.3 Planes in Space

Starting from a vector

\vec{u}

in space one obtains all vectors that are collinear to

\vec{u}

(see Info Box 10.2.1) by taking all multiples

λ \vec{u}

λ \in ℝ

of this vector. Interpreted as position vectors, all these collinear vectors in combination with an arbitrary reference vector constitute the parametric equation of a line as discussed in the previous Subsection 10.2.2. With this in mind, one may ask which object we obtain starting from two fixed (but non-collinear) vectors

\vec{u}

and

\vec{v}

and considering all their coplanar vectors (all vectors that result from

λ \vec{u} + μ \vec{v}

;

λ, μ \in ℝ

- see Info Box 10.2.1). This, in combination with an arbitrary reference vector, generalises the concept of the parametric equation of a line resulting in the parametric equation of a plane in space which is outlined in the Info Box below.
Planes are usually denoted by uppercase Latin letters (

E

F

G

\dots

). Of course, the concept of a plane is only meaningful in

ℝ^{3}

Info 10.2.8

A plane

E

in space is given in vector form or parametric form as the set of position vectors

E = {\vec{r} = \vec{a} + λ \vec{u} + μ \vec{v} : λ, μ \in ℝ},

often written

E : \vec{r} = \vec{a} + λ \vec{u} + μ \vec{v}; λ, μ \in ℝ .

Here,

λ

and

μ

are called parameters,

\vec{a}

is called the reference vector, and

\vec{u}, \vec{v} \neq \vec{O}

is called the direction vector of the plane. Here, the direction vectors

\vec{u}

and

\vec{v}

are non-collinear. The position vectors point to individual points in the plane. The reference vector

\vec{a}

is the position vector of a fixed point in the plane, called the reference point.
(This figure will be released shortly.)

Just as two points in space uniquely define a line (see Section 10.2.2), three given points in space uniquely define a plane. From these three given points, the parametric form of the equation of the corresponding plane can be determined rather easily. The vector form of the equation of a given plane plane is, as for a line, not unique. An infinite number of equivalent equations in vector form exists to represent a given plane. The example below lists a few typical applications.

Example 10.2.9

The reference vector $\vec{a} = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})$ and the direction vectors $\vec{u} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})$ , $\vec{v} = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})$ define an equation in parametric form

$E : \vec{r} = \vec{a} + λ \vec{u} + μ \vec{v} = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + μ (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}); λ, μ \in ℝ$
of a plane that lies at an altitude of $1$ parallel to the $x z$ -plane in the coordinate system (see figure below).
(This figure will be released shortly.)
The parametric equation of the plane $E$ given above is not the only possible one. Each point in the plane $E$ can be used as a reference point. For example, the point defined by the position vector ${\vec{a}}^{'} = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix})$ lies in $E$ since for $λ = μ = 1$ we have:

$(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) + 1 \cdot (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + 1 \cdot (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) .$
Thus, this point can be used as a reference vector. All vectors that are coplanar to $\vec{u}$ and $\vec{v}$ but not collinear to each other can be used as alternative direction vectors. Examples are the vectors ${\vec{u}}^{'} = (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) = 1 \cdot (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + 1 \cdot (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})$ and ${\vec{v}}^{'} = (\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}) = 1 \cdot (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) - 1 \cdot (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})$ . Then, another representation of $E$ in parametric form is given by the equation

$E : \vec{r} = {\vec{a}}^{'} + s {\vec{u}}^{'} + t {\vec{v}}^{'} = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) + s (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) + t (\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}); s, t \in ℝ .$
Consider three points $A = (1; 0; - 2)$ , $B = (4; 1; 2)$ , and $C = (0; 2; 1)$ . Find the equation of the plane $F$ that is specified by these three points, in parametric form.
One of these three points, for example the point $A$ , is used as the reference point. $\vec{A} = (\begin{matrix} 1 \\ 0 \\ - 2 \end{matrix})$ is the corresponding reference vector. The connecting vectors from the reference point to the two other points are used as the direction vectors:

$\vec{A B} = \vec{B} - \vec{A} = (\begin{matrix} 4 \\ 1 \\ 2 \end{matrix}) - (\begin{matrix} 1 \\ 0 \\ - 2 \end{matrix}) = (\begin{matrix} 3 \\ 1 \\ 4 \end{matrix}),$

$\vec{A C} = \vec{C} - \vec{A} = (\begin{matrix} 0 \\ 2 \\ 1 \end{matrix}) - (\begin{matrix} 1 \\ 0 \\ - 2 \end{matrix}) = (\begin{matrix} - 1 \\ 2 \\ 3 \end{matrix}) .$
Hence, the equation

$F : \vec{r} = (\begin{matrix} 1 \\ 0 \\ - 2 \end{matrix}) + ρ (\begin{matrix} 3 \\ 1 \\ 4 \end{matrix}) + σ (\begin{matrix} - 1 \\ 2 \\ 3 \end{matrix}); ρ, σ \in ℝ$
is a correct representation of the plane $F$ in parametric form.
(This figure will be released shortly.)
Consider the two points $P = (1; 2; 3)$ and $Q = (2; 6; 6)$ . Verify whether they lie in the plane $G$ given by the equation

$G : \vec{r} = (\begin{matrix} 0 \\ 3 \\ 2 \end{matrix}) + μ (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) + ν (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}); μ, ν \in ℝ$
in parametric form.
The points $P$ or $Q$ lie in the plane $G$ if their position vectors arise for specific parameter values of $μ$ and $ν$ as position vectors from the equation of $G$ , i.e. $\vec{P} = \vec{r}$ or $\vec{Q} = \vec{r}$ for appropriate values of $μ$ and $ν$ . For the point $P$ , we have:

$\vec{P} = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) = (\begin{matrix} 0 \\ 3 \\ 2 \end{matrix}) + μ (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) + ν (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}) = (\begin{matrix} μ \\ 3 + 2 μ + ν \\ 2 + 3 μ + 2 ν \end{matrix}) .$
From the first component of this vector equation we get $μ = 1$ . Substituting this parameter value into the second and third component provides two contradicting equations in the parameter $ν$ :

$2 = 3 + 2 \cdot 1 + ν \Leftrightarrow ν = - 3$
and

$3 = 2 + 3 \cdot 1 + 2 ν \Leftrightarrow ν = - 1 .$
There are no parameter values $μ$ and $ν$ providing in the parametric equation of the plane $G$ the position vector $\vec{P}$ , so the point $P$ does not lie in the plane $G$ . For $Q$ , however, we have:

$\vec{Q} = (\begin{matrix} 2 \\ 6 \\ 6 \end{matrix}) = (\begin{matrix} 0 \\ 3 \\ 2 \end{matrix}) + μ (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) + ν (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}) = (\begin{matrix} μ \\ 3 + 2 μ + ν \\ 2 + 3 μ + 2 ν \end{matrix}) .$
From the first component we get $μ = 2$ . Substituting this parameter value into the second and third component results in

$6 = 3 + 2 \cdot 2 + ν \Leftrightarrow ν = - 1$
and

$6 = 2 + 3 \cdot 2 + 2 ν \Leftrightarrow ν = - 1 .$
This is not a contradiction. We see that the parameter values $μ = 2$ and $ν = - 1$ provide the position vector $\vec{Q}$ . Hence, the point $Q$ lies in the plane $G$ .
(This figure will be released shortly.)

As well as by three points, a plane can also be defined by a line and a point that does not lie on the line. The example below shows how this can be reduced to the case of three given points.

Example 10.2.10

Let a point

P = (2; 1; - 3)

be given. In addition, let a line

g

be given in parametric form by the equation

g : \vec{r} = (\begin{matrix} 0 \\ - 1 \\ 0 \end{matrix}) + t (\begin{matrix} 2 \\ 0 \\ - 1 \end{matrix}), t \in ℝ .

The point

P

does not lie on the line

g

since there is no value of the parameter

t \in ℝ

such that

\vec{P} = (\begin{matrix} 2 \\ 1 \\ - 3 \end{matrix}) = (\begin{matrix} 0 \\ - 1 \\ 0 \end{matrix}) + t (\begin{matrix} 2 \\ 0 \\ - 1 \end{matrix}) = (\begin{matrix} 2 t \\ - 1 \\ - t \end{matrix}) .

The second component of this vector equation results in the contradiction

1 = - 1

. The point

P

and the line

g

uniquely define a plane

E

that contains both

P

and

g

. A parametric equation of this plane can by found by choosing two additional points on

g

besides the given point

P

that can be used as a reference point and then proceeding as in the example above for three given points. Hence, the reference vector is in this case

\vec{P} = (\begin{matrix} 2 \\ 1 \\ - 3 \end{matrix}),

and the two additional points

Q_{1}

and

Q_{2}

g

result from the equation of the line for two different values of the parameter

t

, for example,

t = 0

and

t = 1

. Choosing

t = 0

results in the reference point of the line as position vector:

{\vec{Q}}_{1} = (\begin{matrix} 0 \\ - 1 \\ 0 \end{matrix}) + 0 \cdot (\begin{matrix} 2 \\ 0 \\ - 1 \end{matrix}) = (\begin{matrix} 0 \\ - 1 \\ 0 \end{matrix}) .

Choosing

t = 1

results in

{\vec{Q}}_{2} = (\begin{matrix} 0 \\ - 1 \\ 0 \end{matrix}) + 1 \cdot (\begin{matrix} 2 \\ 0 \\ - 1 \end{matrix}) = (\begin{matrix} 2 \\ - 1 \\ - 1 \end{matrix}) .

Thus, the direction vectors are

{\vec{P Q}}_{1} = {\vec{Q}}_{1} - \vec{P} = (\begin{matrix} 0 \\ - 1 \\ 0 \end{matrix}) - (\begin{matrix} 2 \\ 1 \\ - 3 \end{matrix}) = (\begin{matrix} - 2 \\ - 2 \\ 3 \end{matrix})

and

{\vec{P Q}}_{2} = {\vec{Q}}_{2} - \vec{P} = (\begin{matrix} 2 \\ - 1 \\ - 1 \end{matrix}) - (\begin{matrix} 2 \\ 1 \\ - 3 \end{matrix}) = (\begin{matrix} 0 \\ - 2 \\ 2 \end{matrix}) .

Hence, the plane

E

is given by the vector equation

E : \vec{r} = (\begin{matrix} 2 \\ 1 \\ - 3 \end{matrix}) + v (\begin{matrix} - 2 \\ - 2 \\ 3 \end{matrix}) + w (\begin{matrix} 0 \\ - 2 \\ 2 \end{matrix}); v, w \in ℝ .

(This figure will be released shortly.)

In the following Section 10.2.4 we will further discuss the relative positions of planes and lines, as well as other data that can be used to define a plane uniquely.

Exercise 10.2.11

The plane

E

uniquely defined by the three points

A = (0; 0; 8)

B = (3; - 1; 10)

, and

C = (- 1; - 2; 11)

has the parametric equation

E : \vec{r} = (\begin{matrix} 2 \\ - 3 \\ x \end{matrix}) + s (\begin{matrix} y \\ 1 \\ - 1 \end{matrix}) + t (\begin{matrix} 5 \\ z \\ - 4 \end{matrix}); s, t \in ℝ .

Find the missing components

x

y

, and

z

x =

y =

z =

The reference vector

\vec{A} = (\begin{matrix} 0 \\ 0 \\ 8 \end{matrix})

and the direction vectors

\vec{A B} = \vec{B} - \vec{A} = (\begin{matrix} 3 \\ - 1 \\ 10 \end{matrix}) - (\begin{matrix} 0 \\ 0 \\ 8 \end{matrix}) = (\begin{matrix} 3 \\ - 1 \\ 2 \end{matrix}),

\vec{A C} = \vec{C} - \vec{A} = (\begin{matrix} - 1 \\ - 2 \\ 11 \end{matrix}) - (\begin{matrix} 0 \\ 0 \\ 8 \end{matrix}) = (\begin{matrix} - 1 \\ - 2 \\ 3 \end{matrix})

define the following parametric equation:

E : \vec{r} = (\begin{matrix} 0 \\ 0 \\ 8 \end{matrix}) + μ (\begin{matrix} 3 \\ - 1 \\ 2 \end{matrix}) + ν (\begin{matrix} - 1 \\ - 2 \\ 3 \end{matrix}); μ, ν \in ℝ .

The reference point

(\begin{matrix} 2 \\ - 3 \\ x \end{matrix})

lies in the plane

E

(\begin{matrix} 2 \\ - 3 \\ x \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 8 \end{matrix}) + μ (\begin{matrix} 3 \\ - 1 \\ 2 \end{matrix}) + ν (\begin{matrix} - 1 \\ - 2 \\ 3 \end{matrix}) = (\begin{matrix} 3 μ - ν \\ - μ - 2 ν \\ 8 + 2 μ + 3 ν \end{matrix}) .

This is a system of linear equations in the three variables

μ

ν

, and

x

that can be solved using the methods described in Section 4.3. Considering the first and second components results in the two equations

2 = 3 μ - ν and - 3 = - μ - 2 ν

with the solution

μ = ν = 1

. Substituting these values into the third component results in

x = 8 + 2 + 3 = 13 .

The two vectors

(\begin{matrix} y \\ 1 \\ - 1 \end{matrix})

and

(\begin{matrix} 5 \\ z \\ - 4 \end{matrix})

are direction vectors of the plane

E

if they are coplanar to both

\vec{A B}

and

\vec{A C}

. For the first vector, we have

(\begin{matrix} y \\ 1 \\ - 1 \end{matrix}) = a (\begin{matrix} 3 \\ - 1 \\ 2 \end{matrix}) + b (\begin{matrix} - 1 \\ - 2 \\ 3 \end{matrix}) = (\begin{matrix} 3 a - b \\ - a - 2 b \\ 2 a + 3 b \end{matrix}),

which is again a system of linear equations in three variables. Considering the second and the third component results in the two equations

1 = - a - 2 b and - 1 = 2 a + 3 b

with the solution

a = 1

b = - 1

. Substituting this values into the first equation results in

y = 3 + (- 1) \cdot (- 1) = 4 .

For the second vector we have

(\begin{matrix} 5 \\ z \\ - 4 \end{matrix}) = a (\begin{matrix} 3 \\ - 1 \\ 2 \end{matrix}) + b (\begin{matrix} - 1 \\ - 2 \\ 3 \end{matrix}) = (\begin{matrix} 3 a - b \\ - a - 2 b \\ 2 a + 3 b \end{matrix}) .

Considering the first and the third component results in the two equations

5 = 3 a - b and - 4 = 2 a + 3 b

with the solution

a = 1

b = - 2

which finally results in

z = - 1 - 2 \cdot (- 2) = 3 .

Exercise 10.2.12

Consider the points

P = (h; 2; - 2)

Q = (1; i; 6)

R = (- 3; 2; j)

and the plane

E

be given by an equation

E : \vec{r} = (\begin{matrix} 3 \\ 0 \\ 2 \end{matrix}) + s (\begin{matrix} 2 \\ 1 \\ 7 \end{matrix}) + t (\begin{matrix} 3 \\ 2 \\ 5 \end{matrix}); s, t \in ℝ

in parametric form. Find the missing components

h

i

, and

j

such that the points

P

Q

, and

R

lie in the plane

E

h =

i =

j =

For the plane

E

, we have the equation

\vec{r} = (\begin{matrix} 3 \\ 0 \\ 2 \end{matrix}) + s (\begin{matrix} 2 \\ 1 \\ 7 \end{matrix}) + t (\begin{matrix} 3 \\ 2 \\ 5 \end{matrix}) = (\begin{matrix} 3 + 2 s + 3 t \\ s + 2 t \\ 2 + 7 s + 5 t \end{matrix}) .

The conditions

\vec{P} = \vec{r} \Leftrightarrow (\begin{matrix} h \\ 2 \\ - 2 \end{matrix}) = (\begin{matrix} 3 + 2 s + 3 t \\ s + 2 t \\ 2 + 7 s + 5 t \end{matrix}),

\vec{Q} = \vec{r} \Leftrightarrow (\begin{matrix} 1 \\ i \\ 6 \end{matrix}) = (\begin{matrix} 3 + 2 s + 3 t \\ s + 2 t \\ 2 + 7 s + 5 t \end{matrix}),

and

\vec{R} = \vec{r} \Leftrightarrow (\begin{matrix} - 3 \\ 2 \\ j \end{matrix}) = (\begin{matrix} 3 + 2 s + 3 t \\ s + 2 t \\ 2 + 7 s + 5 t \end{matrix})

each result in a system of linear equations in the variables

s

t

, and

h

i

j

that can be solved using the methods described in Section 4.3.
For the point

P

, considering the second and the third component results in a system of two linear equations in the variables

s

and

t

2 = s + 2 t and - 4 = 7 s + 5 t

with the solution

s = - 2

t = 2

. Substituting the solution into the first component results in

h = 3 + 2 \cdot (- 2) + 3 \cdot 2 = 5 .

For the point

Q

, considering the first and the third component results in a system of two linear equations in the two variables

s

and

t

- 2 = 2 s + 3 t and 4 = 7 s + 5 t

with the solution

s = 2

t = - 2

. Substituting the solution into the second component results in

i = 2 + 2 \cdot (- 2) = - 2 .

For the point

R

, considering the first and the second component results in a system of two linear equations in the two variables

s

and

t

- 6 = 2 s + 3 t and 2 = s + 2 t

with the solution

s = - 18

t = 10

. Substituting the solution into the third component results in

j = 2 + 7 \cdot (- 18) + 5 \cdot 10 = - 74 .

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics

Chapter 10 Basic Concepts of Descriptive Vector Geometry

10.2.3 Planes in Space

Info 10.2.8

Example 10.2.9

Example 10.2.10

Exercise 10.2.11

Exercise 10.2.12