Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 10, Problem 32P
To determine

To Compute: A.(B×C)=C.(A×B)=B.(C×A)

Expert Solution & Answer
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Explanation of Solution

Given:

  A=axi^+ayj^+azk^B=bxi^+byj^+bzk^C=cxi^+cyj^+czk^

Formula used:

The property of vector cross product:

  · i^ ×j^=k^,  j^ ×k^=i^,  k^ ×i^=j^· j^ ×i^=k^,  k^ ×j^=i^,  i^ ×k^=j^· i^ ×i^=0,  j^ ×j^=0,  k^ ×k^=0

The properties of vector dot product:

  · i^.j^=0,  j^.k^=0,  k^.i^=0, · j^.i^=0,  k^.j^=0,  i^.k^=0· i^ i^=1,  j^ j^=1, k^ k^=1

Calculation:

Calculation of A.(B×C) :

A .( B × C )=( a x i ^ + a y j ^ + a z k ^ ).[ ( b x i ^ + b y j ^ + b z k ^ )×( c x i ^ + c y j ^ + c z k ^ ) ] =( a x i ^ + a y j ^ + a z k ^ ).[ b x c x ( i ^ × i ^ )+ b x c y ( i ^ × j ^ )+ b x c z ( i ^ × k ^ )+ b y c x ( j ^ × i ^ )+ b y c y ( j ^ × j ^ )+ b y c z ( j ^ × k ^ )                                                           + b z c x ( k ^ × i ^ )+ b z c y ( k ^ × j ^ )+ b z c z ( k ^ × k ^ ) ]

   =( a x i ^ + a y j ^ + a z k ^ ).[ b x c x ( 0 )+ b x c y ( k ^ )+ b x c z ( j ^ )+ b y c x ( k ^ )+ b y c y ( 0 )+ b y c z ( i ^ )                                                           + b z c x ( j ^ )+ b z c y ( i ^ )+ b z c z ( 0 ) ] =( a x i ^ + a y j ^ + a z k ^ ).[ b x c y ( k ^ )+ b x c z ( j ^ )+ b y c x ( k ^ )+ b y c z ( i ^ )+ b z c x ( j ^ )+ b z c y ( i ^ ) ]

   =( a x i ^ + a y j ^ + a z k ^ ).[ ( b y c z b z c y )( i ^ )( b x c z + b z c x )( j ^ )+( b x c y b y c x )( k ^ ) ] = a x ( b y c z b z c y )( i ^ . i ^ ) a y ( b x c z + b z c x )( j ^ . j ^ )+ a z ( b x c y b y c x )( k ^ . k ^ )

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

   C .( A × B )=( c x i ^ + c y j ^ + c z k ^ ).[ ( a x i ^ + a y j ^ + a z k ^ )×( b x i ^ + b y j ^ + b z k ^ ) ] =( c x i ^ + c y j ^ + c z k ^ ).[ a x b x ( i ^ × i ^ )+ a x b y ( i ^ × j ^ )+ a x b z ( i ^ × k ^ )+ a y b x ( j ^ × i ^ )+ a y b y ( j ^ × j ^ )+ a y b z ( j ^ × k ^ )                                                           + a z b x ( k ^ × i ^ )+ a z b y ( k ^ × j ^ )+ a z b z ( k ^ × k ^ ) ]

   =( c x i ^ + c y j ^ + c z k ^ ).[ a x b x ( 0 )+ a x b y ( k ^ )+ a x b z ( j ^ )+ a y b x ( k ^ )+ a y b y ( 0 )+ a y b z ( i ^ )                                                           + a z b x ( j ^ )+ a z b y ( i ^ )+ a z b z ( 0 ) ] =( c x i ^ + c y j ^ + c z k ^ ).[ a x b y ( k ^ )+ a x b z ( j ^ )+ a y b x ( k ^ )+ a y b z ( i ^ )+ a z b x ( j ^ )+ a z b y ( i ^ ) ]

   =( c x i ^ + c y j ^ + c z k ^ ).[ ( a y b z a z b y )( i ^ )( a x b z + a z b x )( j ^ )+( a x b y a y b x )( k ^ ) ] = c x ( a y b z a z b y )( i ^ . i ^ ) c y ( a x b z + a z b x )( j ^ . j ^ )+ c z ( a x b y a y b x )( k ^ . k ^ ) = c x ( a y b z a z b y ) c y ( a x b z + a z b x )+ c z ( a x b y a y b x )

   = c x a y b z c x a z b y c y a x b z c y a z b x + c z a x b y c z a y b x = a x b y c z a x b z c y a y b x c z a y b z c x + a z b x c y a z b y c x       ...............(ii)

   B .( C × A )=( b x i ^ + b y j ^ + b z k ^ ).[ ( c x i ^ + c y j ^ + c z k ^ )×( a x i ^ + a y j ^ + a z k ^ ) ] =( b x i ^ + b y j ^ + b z k ^ ).[ c x a x ( i ^ × i ^ )+ c x a y ( i ^ × j ^ )+ c x a z ( i ^ × k ^ )+ c y a x ( j ^ × i ^ )+ c y a y ( j ^ × j ^ )+ c y a z ( j ^ × k ^ )                                                           + c z a x ( k ^ × i ^ )+ c z a y ( k ^ × j ^ )+ c z a z ( k ^ × k ^ ) ]

=( b x i ^ + b y j ^ + b z k ^ ).[ c x a x ( 0 )+ c x a y ( k ^ )+ c x a z ( j ^ )+ c y a x ( k ^ )+ c y a y ( 0 )+ c y a z ( i ^ )                                                           + c z a x ( j ^ )+ c z a y ( i ^ )+ c z a z ( 0 ) ]

   =( b x i ^ + b y j ^ + b z k ^ ).[ c x a y ( k ^ )+ c x a z ( j ^ )+ c y a x ( k ^ )+ c y a z ( i ^ )+ c z a x ( j ^ )+ c z a y ( i ^ ) ] =( b x i ^ + b y j ^ + b z k ^ ).[ ( c y a z c z a y )( i ^ )( c x a z + c z a x )( j ^ )+( c x a y c y a x )( k ^ ) ]

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

Equations (i), (ii) and (iii) are equal.

Therefore,

  A.(B×C)=C.(A×B)=B.(C×A)

Conclusion:

Hence, A.(B×C)=C.(A×B)=B.(C×A)

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Physics for Scientists and Engineers

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