8.4 Lines

Back

DEFINITION 8.4.1 A line in E3 is the set L(P0; X) consisting of all

points P satisfying

P = P0 + tX; t 2 R or equivalently

-

P0P= tX; (8.6)

for some ¯xed point P0 and ¯xed non{zero vector X. (See Figure 8.8.)

Equivalently, in terms of coordinates, equation 8.6 becomes

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

where not all of a; b; c are zero.

The following familiar property of straight lines is easily veri¯ed.

THEOREM 8.4.1 If A and B are distinct points, there is one and only

one line containing A and B, namely L(A;

-

AB) or more explicitly the line

de¯ned by

-

AP= t

-

AB, or equivalently, in terms of position vectors:

P = (1 ¡ t)A + tB or P = A + t

-

AB : (8.7)

Equations 8.7 may be expressed in terms of coordinates: if A = (x1; y1; z1)

and B = (x2; y2; z2), then

x = (1 ¡ t)x1 + tx2; y = (1 ¡ t)y1 + ty2; z = (1 ¡ t)z1 + tz2:

162 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

¢

¢

¢

¢

¢

¢®

-

6

y

z

x

O

P

@

P0

@

@

R

@

@

@

@

@R

C

D

@

@

@

@

@

@

@

@

@

- @

P0P= t

-

CD

Figure 8.8: Representation of a line.

¢

¢

¢

¢

¢

¢®

-

6

y

z

x

O©©©©©©* B

¢

¢

¢

¢

¢

¢¸

¡

¡

¡

¡¡µ

A

P

@

@

@

@

@

@

@

@

@

@

@

@

@

P = A + t

-

AB; 0 < t < 1

Figure 8.9: The line segment AB.

8.4. LINES 163

There is an important geometric signi¯cance in the number t of the above

equation of the line through A and B. The proof is left as an exercise:

THEOREM 8.4.2 (Joachimsthal's ratio formulae)

If t is the parameter occurring in theorem 8.4.1, then

(i) jtj =

AP

AB

; (ii)

¯¯¯¯

t

1 ¡ t

¯¯¯¯

=

AP

PB

if P 6= B:

Also

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

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

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

(See Figure 8.9.)

For example, t = 1

2 gives the mid{point P of the segment AB:

P =

1

2

(A + B):

EXAMPLE 8.4.1 L is the line AB, where A = (¡4; 3; 1); B = (1; 1; 0);

M is the line CD, where C = (2; 0; 2); D = (¡1; 3; ¡2); N is the line EF,

where E = (1; 4; 7); F = (¡4; ¡3; ¡13). Find which pairs of lines intersect

and also the points of intersection.

Solution. In fact only L and N intersect, in the point (¡2

3 ; 5

3 ; 1

3 ). For

example, to determine if L and N meet, we start with vector equations for

L and N:

P = A + t

-

AB; Q = E + s

-

EF;

equate P and Q and solve for s and t:

(¡4i + 3j + k) + t(5i ¡ 2j ¡ k) = (i + 4j + 7k) + s(¡5i ¡ 7j ¡ 20k);

which on simplifying, gives

5t + 5s = 5

¡2t + 7s = 1

¡t + 20s = 6

This system has the unique solution t = 2

3 ; s = 1

3 and this determines a

corresponding point P where the lines meet, namely P = (¡2

3 ; 5

3 ; 1

3 ).

The same method yields inconsistent systems when applied to the other

pairs of lines.

164 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

EXAMPLE 8.4.2 If A = (5; 0; 7) and B = (2; ¡3; 6), ¯nd the points P

on the line AB which satisfy AP=PB = 3.

Solution. Use the formulae

P = A + t

-

AB and

¯¯¯¯

t

1 ¡ t

¯¯¯¯

=

AP

PB

= 3:

Then

t

1 ¡ t

= 3 or ¡ 3;

so t = 3

4 or t = 3

2 . The corresponding points are ( 11

4 ; 9

4 ; 25

4 ) and ( 1

2 ; 9

2 ; 11

2 ).

DEFINITION 8.4.2 Let X and Y be non{zero vectors. Then X is parallel

or proportional to Y if X = tY for some t 2 R. We write XkY if X is parallel

to Y . If X = tY , we say that X and Y have the same or opposite direction,

according as t > 0 or t < 0.

DEFINITION 8.4.3 if A and B are distinct points on a line L, the non{

zero vector

-

AB is called a direction vector for L.

It is easy to prove that any two direction vectors for a line are parallel.

DEFINITION 8.4.4 Let L and M be lines having direction vectors X

and Y , respectively. Then L is parallel to M if X is parallel to Y . Clearly

any line is parallel to itself.

It is easy to prove that the line through a given point A and parallel to a

given line CD has an equation P = A + t

-

CD.

THEOREM 8.4.3 Let X = a1i + b1j + c1k and Y = a2i + b2j + c2k be

non{zero vectors. Then X is parallel to Y if and only if

¯¯¯¯

a1 b1

a2 b2

¯¯¯¯

=

¯¯¯¯

b1 c1

b2 c2

¯¯¯¯

=

¯¯¯¯

a1 c1

a2 c2

¯¯¯¯

= 0: (8.8)

Proof. The case of equality in the Cauchy{Schwarz inequality (theorem 8.3.1)

shows that X and Y are parallel if and only if

jX ¢ Y j = jjXjj ¢ jjY jj:

Squaring gives the equivalent equality

(a1a2 + b1b2 + c1c2)2 = (a2

1 + b2

1 + c2

1 )(a2

2 + b2

2 + c2

2 );

8.4. LINES 165

which simpli¯es to

(a1b2 ¡ a2b1)2 + (b1c2 ¡ b2c1)2 + (a1c2 ¡ a2c1)2 = 0;

which is equivalent to

a1b2 ¡ a2b1 = 0; b1c2 ¡ b2c1 = 0; a1c2 ¡ a2c1 = 0;

which is equation 8.8.

Equality of geometrical vectors has a fundamental geometrical interpre-

tation:

THEOREM 8.4.4 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.)

Proof. If

-

AB=

-

CD then

B ¡ A = D ¡ C;

C ¡ A = D ¡ B

and so

-

AC=

-

BD. Hence

-

AB k

-

CD and

-

AC k

-

BD.

Conversely, suppose that

-

AB k

-

CD and

-

AC k

-

BD. Then

-

AB= s

-

CD and

-

AC= t

-

BD;

or

B ¡ A = s(D ¡ C) and C ¡ A = tD ¡ B:

We have to prove s = 1 or equivalently, t = 1.

Now subtracting the second equation above from the ¯rst, gives

B ¡ C = s(D ¡ C) ¡ t(D ¡ B);

so

(1 ¡ t)B = (1 ¡ s)C + (s ¡ t)D:

If t 6= 1, then

B =

1 ¡ s

1 ¡ t

C +

s ¡ t

1 ¡ t

D

and B would lie on the line CD. Hence t = 1.

166 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

Навигация