Mathematical Physics Vol 1

3.1 Vector algebra

67

Using the property of the triple product (example 5d, p. 58) we obtain:

1 V 3 ( b × c ) ·{ [( c × a ) · b ] a − [( c × a ) · a ] c } = = 1 V 3 ( b × c ) · [( b × c ) · a ] a = 1 V 3 [( b × c ) · a ] 2 = 1 V 3

a ′ · ( b ′ × c ′ )=

1 V

V 2 =

.

d)Vectors a , b and c are not coplanar, given that their mixed product is not equal to zero, and therefore its reciprocal is also different from zero. Given that, according to example (20c, p. 66), the value of this reciprocal is equal to the mixed product of vectors a ′ , b ′ and c ′ , it follows that these vectors are not coplanar (see example 19, p. 65).

Problem21 Prove that the vectors of one system can be uniquely determined by the vectors of the reciprocal system of vectors and that these relations are reciprocal.

Proof Let a , b and c be linearly independent vectors in a 3-D space. Thus, an arbitrary vector in this space can be represented as their linear combination d = α a + β b + γ c . Coefficients α , β i γ can be determined as follows. By multiplying both sides of the previous relation by the vector b × c we obtain d · b × c = α a · b × c , given that b · b × c and c · b × c are both equal to zero, being mixed products of coplanar vectors. By introducing V = a · b × c , we obtain α = d · b × c V . It can analogously be proved that

d · c × a V

d · a × b V

β =

, γ =

.

Finally, we obtain

( d · b × c ) V

( d · c × a ) V

( d · a × b ) V

d =

a +

b +

c .

According to the previous example, by introducing reciprocal vectors using the same relations, we obtain d = d · a ′ a + d · b ′ b + d · c ′ c .