The notationa∧bis sometimes used.
This vector product is often represented in the form of the array shown below, in order
to aid memory.
a×b≡
∣
∣
∣
∣
∣
ijk
a 1 a 2 a 3
b 1 b 2 b 3
∣ ∣ ∣ ∣ ∣ ≡
∣
∣
∣
∣
a 2 a 3
b 2 b 3
∣
∣
∣
∣i−
∣
∣
∣
∣
a 1 a 3
b 1 b 3
∣
∣
∣
∣j+
∣
∣
∣
∣
a 1 a 2
b 1 b 2
∣
∣
∣
∣k
where we introduce the shorthand
∣
∣
∣∣ab
cd
∣
∣
∣∣=ad−bc
NB: This is just a mnemonic device for remembering the formula for the vector product
and for calculating it – you don’t need to know anything about determinants at this stage.
Problem 11.9
Find the vector product a×b between the vectors a=i−jY2k and b=
2iYk. What is b×a?
Using the component expression given we obtain (check these calculations against the
symbolic expressions given above)
a×b=
∣
∣
∣
∣∣
ijk
1 − 12
201
∣
∣
∣
∣∣
=[(− 1 )( 1 )− 2 ( 0 )]i−[1× 1 − 2 ×2]j+[1× 0 −(− 1 )( 2 )]k
=−i+ 3 j+ 2 k
You can also verify explicitly that
b×a=i− 3 j− 2 k=−a×b
From the above definition it can be shown that the vector (orcross) product has the
following properties:
(i) a×bis a vector
(ii) a×(b+c)=a×b+a×cand(a+b)×c=a×c+b×c
i.e. the vector product is distributive over vector addition
(iii) (ka)×b=k(a×b)=a×(kb)for any scalark
(iv) Ifa×b=0anda,bare not zero vectors, thenaandbare parallel
(v) b×a=−a×bas illustrated by Problem 11.9
(vi) a×a=0, which follows from (v)
(vii) The vector product is not associative, i.e.
a×(b×c)=(a×b)×c