## 2.10. The Cross Product Of Two Vectors In R3

### Definition: Cross Product

The cross product of 2 vectors A and B in R 3  is defined as another vector C=A×B  with components

( A×B ) i = c i = ε ijk a j b k

where the Einstein notation of implied summation over repeated indices is used and ε ijk  is the totally anti-symmetric Levi-Civita symbol defined by

ε ijk ={ +1 1 0 if      i,j,k  is          even  permutation  of  123 odd  permutation  of  123 otherwise

i.e., ε 123 = ε 231 ==1 , ε 213 = ε 321 ==1  and ε 113 = ε 221 ==0 .

Carrying out the summations, we have

A×B=( a 2 b 3 a 3 b 2 , a 3 b 1 a 1 b 3 , a 1 b 2 a 2 b 1 )

### Theorem 2.12: Properties

A,B,C R 3  and αR ,

(a)     A×B=( B×A )

(b)     A×( B+C )=A×B+A×C

(c)     α( A×B )=( αA )×B=A×( αB )

(d)     A( A×B )=0

(e)     B( A×B )=0

(f)      A×B 2 =     A 2 B 2 ( AB ) 2

(g)     A×B=0         Û            A and B are L.I.

#### Proof

Porperties (a-c) follow immediately from definition.

For (d), we have

A( A×B )= ε ijk a i a j b k = ε jik a i a j b k          [ ε ijk = ε jik  ]

= ε ijk a j a i b k                              [ ij  ]

=0                            [ x=x        x=0  ]

For (f),

A×B 2 =( A×B )( A×B )   = ε ijk a j b k ε ilm a l b m

Using the identity

ε ijk ε ilm = δ jl δ km δ jm δ kl

we have

A×B 2 =( δ jl δ km δ jm δ kl ) a j b k a l b m   = a j b k a j b k a j b k a k b j

=( AA )( BB ) ( AB ) 2   = A 2 B 2 ( AB ) 2

Using AB= A B cosθ  where q is the angle between A and B, we have

A×B = A B sinθ  = Area of parallelogram with sides A and B.

[see Fig.2.5]

For (g):  using (f), we have

A×B=O         Û            A 2 B 2 = ( AB ) 2

By the Cauchy-Schwarz inequality, this means A and B are linearly dependent.

### Examples

From property (a), we have

A×A=A×A=0

Also, writing the jth component of the ith coordinate unit vector as

( e i ) j = δ ij

we have

( e i × e j ) k   = ε klm ( e i ) l ( e j ) m   = ε klm δ il δ jm = ε kij = ε ijk

Þ            e i × e j = ε ijk e k            (1)

Note that geometrically, (1) can be implemented in 2 ways; thus giving rise to the right and left handed coordinate systems (see Fig.2.4)

### Theorem 2.13

Let A and B be L.I. in R3, then

(a)     A, B, and A×B  are L.I.

(b)     Every vector in R3 orthogonal to both A and B are scalar multiples of A×B .

#### Proof

(a)

Let C=A×B .  Since A and B are L.I., C0 .  Consider the equation

αA+βB+γC=0

Taking C  on both sides reduces it to γCC=0  so that γ=0 .  Since A and B are L.I., we must have α=β=0 .  QED

(b)

Let N be orthogonal to both A and B.  By part (a), A, B, and C=A×B  are L.I. so that they form a basis in R3.  Hence, we can write

N=αA+βB+γC                   (2)

Taking N  on (2) gives NN=γNC .

Taking C  on (2) gives CN=γCC .

Hence, ( NC ) 2 =( NN )( CC )  so that by the Cauchy-Schwarz inequality, the theorem is proved.