8.1 Introduction

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In this chapter we present a vector{algebra approach to three{dimensional

geometry. The aim is to present standard properties of lines and planes,

with minimum use of complicated three{dimensional diagrams such as those

involving similar triangles. We summarize the chapter:

Points are de¯ned as ordered triples of real numbers and the distance

between points P1 = (x1; y1; z1) and P2 = (x2; y2; z2) is de¯ned by the

formula

P1P2 =

p

(x2 ¡ x1)2 + (y2 ¡ y1)2 + (z2 ¡ z1)2:

Directed line segments

-

AB are introduced as three{dimensional column

vectors: If A = (x1; y1; z1) and B = (x2; y2; z2), then

-

AB=

2

4

x2 ¡ x1

y2 ¡ y1

z2 ¡ z1

3

5 :

If P is a point, we let P =

-

OP and call P the position vector of P.

With suitable de¯nitions of lines, parallel lines, there are important ge-

ometrical interpretations of equality, addition and scalar multiplication of

vectors.

(i) Equality of vectors: Suppose A; B; C; D are distinct points such that

no three are collinear. Then

-

AB=

-

CD if and only if

-

AB k

-

CD and

-

AC k

-

BD (See Figure 8.1.)

149

150 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

¢

¢

¢

¢

¢

¢®

-

6

y

z

x

O

¡

¡

¡

¡µ

-

A

¡

¡

¡

¡µ

C

Q

Q

Q

Q

Q

Q

D

Q

Q

Q

Q

Q

Q

B

-

AB=

-

CD;

-

AC=

-

- BD

AB +

-

AC=

-

AD

Figure 8.1: Equality and addition of vectors.

(ii) Addition of vectors obeys the parallelogram law: Let A; B; C be non{

collinear. Then

-

AB +

-

AC=

-

AD;

where D is the point such that

-

AB k

-

CD and

-

AC k

-

BD. (See Fig-

ure 8.1.)

(iii) Scalar multiplication of vectors: Let

-

AP= t

-

AB, where A and B are

distinct points. Then P is on the line AB,

AP

AB

= jtj

and

(a) P = A if t = 0, P = B if t = 1;

(b) P is between A and B if 0 < t < 1;

(c) B is between A and P if 1 < t;

(d) A is between P and B if t < 0.

(See Figure 8.2.)

8.1. INTRODUCTION 151

¢

¢

¢

¢

¢

¢®

-

6

y

z

x

O

@

@

@

@

@R

R P

A

B

-

AP= t

-

AB; 0 < t < 1

Figure 8.2: Scalar multiplication of vectors.

The dot product X¢Y of vectors X =

2

4

a1

b1

c1

3

5 and Y =

2

4

a2

b2

c2

3

5, is de¯ned

by

X ¢ Y = a1a2 + b1b2 + c1c2:

The length jjXjj of a vector X is de¯ned by

jjXjj = (X ¢ X)1=2

and the Cauchy{Schwarz inequality holds:

jX ¢ Y j · jjXjj ¢ jjY jj:

The triangle inequality for vector length now follows as a simple deduction:

jjX + Y jj · jjXjj + jjY jj:

Using the equation

AB = jj

-

AB jj;

we deduce the corresponding familiar triangle inequality for distance:

AB · AC + CB:

152 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

The angle µ between two non{zero vectors X and Y is then de¯ned by

cos µ =

X ¢ Y

jjXjj ¢ jjY jj

; 0 · µ · ¼:

This de¯nition makes sense. For by the Cauchy{Schwarz inequality,

¡1 ·

X ¢ Y

jjXjj ¢ jjY jj · 1:

Vectors X and Y are said to be perpendicular or orthogonal if X ¢ Y = 0.

Vectors of unit length are called unit vectors. The vectors

i =

2

4

1

0

0

3

5 ; j =

2

4

0

1

0

3

5 ; k =

2

4

0

0

1

3

5

are unit vectors and every vector is a linear combination of i; j and k:

2

4

a

b

c

3

5 = ai + bj + ck:

Non{zero vectors X and Y are parallel or proportional if the angle be-

tween X and Y equals 0 or ¼; equivalently if X = tY for some real number

t. Vectors X and Y are then said to have the same or opposite direction,

according as t > 0 or t < 0.

We are then led to study straight lines. If A and B are distinct points,

it is easy to show that AP + PB = AB holds if and only if

-

AP= t

-

AB; where 0 · t · 1:

A line is de¯ned as a set consisting of all points P satisfying

P = P0 + tX; t 2 R or equivalently

-

P0P= tX;

for some ¯xed point P0 and ¯xed non{zero vector X called a direction vector

for the line.

Equivalently, in terms of coordinates,

x = x0 + ta; y = y0 + tb; z = z0 + tc;

where P0 = (x0; y0; z0) and not all of a; b; c are zero.

8.1. INTRODUCTION 153

There is then one and only one line passing passing through two distinct

points A and B. It consists of the points P satisfying

-

AP= t

-

AB;

where t is a real number.

The cross{product X£Y provides us with a vector which is perpendicular

to both X and Y . It is de¯ned in terms of the components of X and Y :

Let X = a1i + b1j + c1k and Y = a2i + b2j + c2k. Then

X £ Y = ai + bj + ck;

where

a =

¯¯¯¯

b1 c1

b2 c2

¯¯¯¯

; b = ¡

¯¯¯¯

a1 c1

a2 c2

¯¯¯¯

; c =

¯¯¯¯

a1 b1

a2 b2

¯¯¯¯

:

The cross{product enables us to derive elegant formulae for the distance

from a point to a line, the area of a triangle and the distance between two

skew lines.

Finally we turn to the geometrical concept of a plane in three{dimensional

space.

A plane is a set of points P satisfying an equation of the form

P = P0 + sX + tY; s; t 2 R; (8.1)

where X and Y are non{zero, non{parallel vectors.

In terms of coordinates, equation 8.1 takes the form

x = x0 + sa1 + ta2

y = y0 + sb1 + tb2

z = z0 + sc1 + tc2;

where P0 = (x0; y0; z0).

There is then one and only one plane passing passing through three

non{collinear points A; B; C. It consists of the points P satisfying

-

AP= s

-

AB +t

-

AC;

where s and t are real numbers.

The cross{product enables us to derive a concise equation for the plane

through three non{collinear points A; B; C, namely

-

AP ¢(

-

AB £

-

AC) = 0:

154 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

When expanded, this equation has the form

ax + by + cz = d;

where ai + bj + ck is a non{zero vector which is perpendicular to

-

P1P2 for

all points P1; P2 lying in the plane. Any vector with this property is said to

be a normal to the plane.

It is then easy to prove that two planes with non{parallel normal vectors

must intersect in a line.

We conclude the chapter by deriving a formula for the distance from a

point to a plane.

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