For the basis vectorsi,j,kwe find (Reinforcement Exercise 24)
i×i=j×j=k×k= 0
i×j=k, j×k=i, k×i=jGeometrically the vector product can be expressed as:
a×b=absinθnwherenis a unit vector perpendicular to bothaandbsuch thata,b,nform a right-handed
set – see Figure 11.15.
q b
anFigure 11.15The vector product isa×b=absinθn.
The proof of this is an interesting exercise. Take thex-axis to be alongaandbto be
in thexyplane (we can always choose axes in this way). Then
a=aib=b 1 i+b 2 jso a×b=ab 2 i×j=ab 2 k
But also we have:
b 2 =bsinθso a×b=absinθk
which is equivalent to the above result.
The two vectorsa,bare parallel or opposite and parallel if and only ifa×b=0(then
θ=0 or 180°).
Exercises on 11.11
- Prove the properties (i)→(vi)
- Ifa=i+kb= 2 i−j+ 3 kc=i+ 2 j+ 3 k,evaluate
(i) a×b (ii) a×(b+c) (iii) a×c (iv) b×b
(v) b×a (vi) a×(b×c) (vii) (a×b)×cAnswers
- (i) i−j−k (ii) −i− 3 j+k (iii) − 2 i− 2 j+ 2 k (iv) 0
(v) −i+j+k (vi) 3i− 14 j− 3 k (vii) −i− 4 j+ 3 k