8.6 The cross{product of two vectors

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DEFINITION 8.6.1 Let X = a1i + b1j + c1k and Y = a2i + b2j + c2k.

Then X £ Y , the cross{product of X and Y , is de¯ned by

X £ Y = ai + bj + ck;

where

a =

¯¯¯¯

b1 c1

b2 c2

¯¯¯¯

; b = ¡

¯¯¯¯

a1 c1

a2 c2

¯¯¯¯

; c =

¯¯¯¯                                      

a1 b1

a2 b2

¯¯¯¯

:

The vector cross{product has the following properties which follow from

properties of 2 £ 2 and 3 £ 3 determinants:

(i) i £ j = k; j £ k = i; k £ i = j;

8.6. THE CROSS{PRODUCT OF TWO VECTORS 173

(ii) X £ X = 0;

(iii) Y £ X = ¡X £ Y ;

(iv) X £ (Y + Z) = X £ Y + X £ Z;

(v) (tX) £ Y = t(X £ Y );

(vi) (Scalar triple product formula) if Z = a3i + b3j + c3k, then

X ¢ (Y £ Z) =

¯¯¯¯¯¯

a1 b1 c1

a2 b2 c2

a3 b3 c3

¯¯¯¯¯¯

= (X £ Y ) ¢ Z;

(vii) X ¢ (X £ Y ) = 0 = Y ¢ (X £ Y );

(viii) jjX £ Y jj =

p

jjXjj2jjY jj2 ¡ (X ¢ Y )2;

(ix) if X and Y are non{zero vectors and µ is the angle between X and Y ,

then

jjX £ Y jj = jjXjj ¢ jjY jj sin µ:

(See Figure 8.14.)

From theorem 8.4.3 and the de¯nition of cross{product, it follows that

non{zero vectors X and Y are parallel if and only if X £ Y = 0; hence by

(vii), the cross{product of two non{parallel, non{zero vectors X and Y , is

a non{zero vector perpendicular to both X and Y .

LEMMA 8.6.1 Let X and Y be non{zero, non{parallel vectors.

(i) Z is a linear combination of X and Y , if and only if Z is perpendicular

to X £ Y ;

(ii) Z is perpendicular to X and Y , if and only if Z is parallel to X £ Y .

Proof. Let X and Y be non{zero, non{parallel vectors. Then

X £ Y 6= 0:

Then if X £ Y = ai + bj + ck, we have

det [X £ Y jXjY ]t =

¯¯¯¯¯¯

a b c

a1 b1 c1

a2 b2 c2

¯¯¯¯¯¯

= (X £ Y ) ¢ (X £ Y ) > 0:

174 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

¢

¢

¢

¢

¢

¢®

-

6

y

z

x

O

@

@

@

@I

¡

¡

¡ª

-

µ

¡ @

@

X

Y

X £ Y

Figure 8.14: The vector cross{product.

Hence the matrix [X £ Y jXjY ] is non{singular. Consequently the linear

system

r(X £ Y ) + sX + tY = Z (8.15)

has a unique solution r; s; t.

(i) Suppose Z = sX + tY . Then

Z ¢ (X £ Y ) = sX ¢ (X £ Y ) + tY ¢ (X £ Y ) = s0 + t0 = 0:

Conversely, suppose that

Z ¢ (X £ Y ) = 0: (8.16)

Now from equation 8.15, r; s; t exist satisfying

Z = r(X £ Y ) + sX + tY:

Then equation 8.16 gives

0 = (r(X £ Y ) + sX + tY ) ¢ (X £ Y )

= rjjX £ Y jj2 + sX ¢ (X £ Y ) + tY ¢ (Y £ X)

= rjjX £ Y jj2:

Hence r = 0 and Z = sX + tY , as required.

(ii) Suppose Z = ¸(X £ Y ). Then clearly Z is perpendicular to X and Y .

8.6. THE CROSS{PRODUCT OF TWO VECTORS 175

Conversely suppose that Z is perpendicular to X and Y .

Now from equation 8.15, r; s; t exist satisfying

Z = r(X £ Y ) + sX + tY:

Then

sX ¢ X + tX ¢ Y = X ¢ Z = 0

sY ¢ X + tY ¢ Y = Y ¢ Z = 0;

from which it follows that

(sX + tY ) ¢ (sX + tY ) = 0:

Hence sX + tY = 0 and so s = 0; t = 0. Consequently Z = r(X £ Y ), as

required.

The cross{product gives a compact formula for the distance from a point

to a line, as well as the area of a triangle.

THEOREM 8.6.1 (Area of a triangle)

If A; B; C are distinct non{collinear points, then

(i) the distance d from C to the line AB is given by

d = jj

-

AB £

-

AC jj

AB

; (8.17)

(ii) the area of the triangle ABC equals

jj

-

AB £

-

AC jj

2

= jjA £ B + B £ C + C £ Ajj

2

: (8.18)

Proof. The area ¢ of triangle ABC is given by

¢ =

AB ¢ CP

2

;

where P is the foot of the perpendicular from C to the line AB. Now by

formula 8.13, we have

CP =

q

AC2 ¢ AB2 ¡ (

-

AC ¢

-

AB)2

AB

= jj

-

AB £

-

AC jj

AB

;

176 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

which, by property (viii) of the cross{product, gives formula 8.17. The

second formula of equation 8.18 follows from the equations

-

AB £

-

AC = (B ¡ A) £ (C ¡ A)

= f(B ¡ A) £ Cg ¡ f(C ¡ A) £ Ag

= f(B £ C ¡ A £ C)g ¡ f(B £ A ¡ A £ A)g

= B £ C ¡ A £ C ¡ B £ A

= B £ C + C £ A + A £ B;

as required.

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