Mathematical Physics Vol 1

3.1 Vector algebra

59

Given that a is normal to b , the vector a × b is normal to the plane formed by vectors a and b and its magnitude is a · b sin90 ◦ = ab . The same vector is obtained when b is multiplied by a and then rotated for 90 ◦ , as in the picture above. Similarly, vector a × c is obtained when vector c is multiplied by a and rotated for 90 ◦ , as in Figure 3.6. We obtain the vector a × ( b + c ) in a similar way by a vector product of vectors b + c andvector a rotated for 90 ◦ , as depicted in Figure 3.6. Given that a × ( b + c ) is the diagonal of the parallelogram with sides a × b and a × c , it follows that a × ( b + c )= a × b + a × c . In the second part of the proof, let us as sume that vectors a , b and c can have arbi trary orientations in space (Fig. 3.7).

Figure 3.6: Figure in Problem 5.

Figure 3.7: Figure in Problem 5.

Let us decompose vectors b and c into two components, one that is parallel to vector a and the other that is normal to a . The vector product of components normal to a and the vector a is given in the first part of the proof, and the vector product of the components parallel to vector a and the vector a is equal to zero. c) Given that

a = a x i + a y j + a z k , b = b x i + b y j + b z k , c = c x i + c y j + c z k

it follows that

a × ( b × c )=( a x i + a y j + a z k ) × i

j k b x b y b z c x c y c z

=

=( a x i + a y j + a z k ) × [( b y c z − b z c y ) i +( b z c x − b x c z ) j +( b x c y − b x c z ) k ]=

i

j

k

a x a z ( b y c z − b z c y ) ( b z c x − b x c z ) ( b x c y − b y c z ) a y

=

=

=( a y b x c y − a y b y c x − a z b z c x + a z b x c z ) i + +( a z b y c z − a z b z c y − a x b x c y + a x b y c x ) j + +( a x b z c x − a x b x c z − a y b y c z + a y b z c y ) k .