Begin2.DVI

In a similar fashion one can show that the vectors (A
×B
)×C,
A
and B
are coplanar

so that there exists constants γand δsuch that

(A
×B
)×C
 =γA
+δB
 (6 .36)

The equations (6.35) and (6.36) show that in general

A
×(B
×C
)
= (A
×B
)×C

Example 6-9. Show that the triple vector product satisfies

A
×(B
×C
) = (A
·C
)B
−(A
·B
)C

Solution Use the results from the previous example showing there exists scalars α

and βsuch that

A
×(B
×C
) = αB
+βC
 (6 .37)

Let B
×C
=D
and write

A
×D
 =αB
+βC
 (6 .38)

Take the dot product of both sides of equation (6.38) with the vector A
to obtain

the triple scalar product

A
·(A
×D
) = α(A
·B
) + β(A
·C
)

By the permutation properties of the triple scalar product one can write

A
·(A
×D
) = A
·(D
×A
) = D
·(A
×A
) = 0 = α(A
·B
) + β(A
·C
) (6 .39)

The above result holds because A
×A
=
0 and implies

α(A
·B
) = −β(A
·C
) or
α

A·C

= 
−β

A·B

=λ

where λ is a scalar. This shows that the equation (6.37) can be expressed in the

form

A
×(B
×C
) = λ(A
·C
)B
−λ(A
·B
)C
 (6 .40)

which shows that the vectors A
×(B
×C
)and (A
·C
)B
−(A
·B
)C
are colinear. The

equation (6.40) must hold for all vectors A,
B,
C
and so it must be true in the special

case A
=ˆe 2 , B
=ˆe 1 , C
=ˆe 2 where equation (6.40) reduces to

ˆe 2 ×(ˆe 1 ׈e 2 ) = ˆe 1 =λˆe 1 which implies λ= 1