Mathematical Physics Vol 1

Chapter 3. Examples

60

By grouping the elements with b x , b y , b z , andwith c x , c y , b z we obtain: b x ( a y c y + a z c z ) i − c x ( a y b y + a z b z ) i + + b y ( a x c x + a z c z ) j − c y ( a x b x + a z b z ) j + + b z ( a x c x + a y c y ) k − c z ( a x b x + a y b y ) k If elements a x b x c x i , a y b y c y j and a z b z c z k are both added to and subtracted from the previous expression, we obtain

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

that is

a × ( b × c )= b x ( a · c ) i − c x ( a · b ) i + + b y ( a · c ) j − c y ( a · b ) j + b z ( a · c ) k − c z ( a · b ) k . By extracting the common elements and using (1.44), on p.31, we obtain the relation a × ( b × c )= b ( a · c ) − c ( a · b ) . (3.1) d) Given the anticommutativity of the vector product (1.23), it follows that ( a × b ) × c = − c × ( a × b )= a ( − c · b ) − b ( − c · a )= = b ( c · a ) − a ( c · b ) . (3.2) From (3.1) and (3.2) it follows that the associative law does not hold for the vector product a × ( b × c )̸=( a × b ) × c .

Problem6 Prove ( a × b ) · ( b × c ) × ( c × a )=( a · b × c ) 2 .

Solution By using the previous example we obtain

x × ( c × a )= c ( x · a ) − a ( x · c )

if x = b × c ( b × c ) × ( c × a )= c ( b × c · a ) − a ( b × c · c )= c ( a · b × c ) − a ( b · c × c )= c ( a · b × c ) ( a × b ) · ( b × c ) × ( c × a )=( a × b ) · c ( a · b × c )=( a × b · c )( a × b · c )=( a · b × c ) 2 .