Part 1: Determinants and the Cross Product

In this section, we introduce the cross product of two vectors. However, the cross product is based on the theory of determinants, so we begin with a review of the properties of determinants.

To begin with, the determinant of a 2×2 array of numbers is defined
u1
u2
v1
v2
 = u1v2-u2v1
(1)
Subsequently, the cross product of the vectors u = á u1,u2,u3 ñ and v = áv1,v2,v3 ñ is defined to be a vector of determinants:
u×v =

u2
u3
v2
v3
 ,
u3
u1
v3
v1
 ,
u1
u2
v1
v2

(2)
We have constructed (2) so that the direction of u×v satisfies the right-hand rule, which says that as the fingers of the right hand sweep from u to v through an angle of less than 180°, the thumb points in the direction of u×v.
 

       

EXAMPLE 1    Compute u×v for u = á 2,3,5 ñ and v = á6,7,9 ñ .   

Solution: To do so, we construct the vector of determinants in (2),
u×v =

3
5
7
9
 ,
5
2
9
6
 ,
2
3
6
7

and then we use (1) to evaluate the determinants:
u×v = á 3·9-7·5,5·6-9·2,2·7-6·3 ñ = á -8,12,-4 ñ
(3)
Notice however that v×u is
v×u =  

7
9
3
5
 ,
9
6
5
2
 ,
6
7
2
3

= á 8,-12,4 ñ

In example, v×u = -( u×v) , which can be shown to be true in general.  Theorem 3.1 lists similar properties of the cross product that follow from direct calculation (for example, u×u = -( u×u) implies that u × u = 0, where 0 = á 0,0,0 ñ ).       

Theorem 3.1: The cross product is defined only for 3 dimensional vectors u and v. Moreover, the following hold:
i)
v×u = -( u×v)
ii)
u×u = 0
iii)
( ku) ×v = k( u×v) = u×kv
iv)
a×( u+v) = a×u+a×v

           

The determinant of a 3×3 array of numbers is defined
r1
r2
r3
u1
u2
u3
v1
v2
v3
   =  r1
u2
u3
v2
v3
  + r2
u3
u1
v3
v1
  + r3
u1
u2
v1
v2
(3)
Alternatively, we can calculate (3) by repeating the first two columns after the determinant, and then summing the products along the 3 main diagonals and the 3 off diagonals. The determinant is the difference of the two sums:
(4)
Formally, if the first row of (3) is i, j, and k, then
i
j
k
u1
u2
u3
v1
v2
v3
  = i  
u2
u3
v2
v3
  + j  
u3
u1
v3
v1
  + k 
u1
u2
v1
v2
Thus, we can also calculate u×v using a 3-dimensional determinant.

EXAMPLE 2    Use a 3-dimensional determinant to compute u×v when u = á 2,1,2 ñ and v =   á 3,4,5 ñ           

Solution: We write u×v as a 3 dimensional determinant and then evaluate the determinant.

This simplifies to u×v = -3i-4j+5k.          

Check Your Reading: What is v×u when u = á 2,1,2 ñ and v = á 3,4,5 ñ ?